What is the intuitive meaning of having a linear relationship between the logs of two variables?What type of test to use to determine correlation/relationship between two non-continuous varaiblesWhat is the qualitative difference between a Michaelis-Menten model and a log-linear model?correlation with logarithmic transformationCan I use the correlation between two variables when observations on each variable are autocorrelated?What is the relationship between orthogonal, correlation and independence?Classifying according to relationship between variablesScatterplot dovetailing?Interpreting how much my linear model has improved after Box-Cox transformationGuess Relationship/Association between two quantitatives variablesCorrelation coefficients and range of residuals

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What is the intuitive meaning of having a linear relationship between the logs of two variables?


What type of test to use to determine correlation/relationship between two non-continuous varaiblesWhat is the qualitative difference between a Michaelis-Menten model and a log-linear model?correlation with logarithmic transformationCan I use the correlation between two variables when observations on each variable are autocorrelated?What is the relationship between orthogonal, correlation and independence?Classifying according to relationship between variablesScatterplot dovetailing?Interpreting how much my linear model has improved after Box-Cox transformationGuess Relationship/Association between two quantitatives variablesCorrelation coefficients and range of residuals













14












$begingroup$


I have two variables which don't show much correlation when plotted against each other as is, but a very clear linear relationship when I plot the logs of each variable agains the other.



So I would end up with a model of the type:



$$log(Y) = a log(X) + b$$ , which is great mathematically but doesn't seem to have the explanatory value of a regular linear model.



How can I interpret such a model?










share|cite|improve this question











$endgroup$







  • 4




    $begingroup$
    I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
    $endgroup$
    – Upper_Case
    yesterday










  • $begingroup$
    The interpretation of a log-log model, where the dependent variable is log(y) and the independent variable is log(x), is: $%Δ=β_1%Δx$.
    $endgroup$
    – C Thom
    10 hours ago







  • 1




    $begingroup$
    The complementary log-log link is an ideal GLM specification when the outcome is binary (risk model) and the exposure is cumulative, such as number of sexual partners vs. HIV infection. jstor.org/stable/2532454
    $endgroup$
    – AdamO
    9 hours ago










  • $begingroup$
    Thank you @AdamO ! Do you know if there are any circumstances where logit, probit, or complementary log-log links give substantively different results, either in point estimates, or in standard errors? As the authors you cited wrote "Since the logit of P closely approximates the complementary log- log of P over a wide range, such an analysis is likely to lead to qualitatively similar results to the methods described here," I wonder if there's a strong reason to prefer any in given circumstances? I know that the econometricians like probit links as natural for things like elasticities.
    $endgroup$
    – Alexis
    5 hours ago






  • 1




    $begingroup$
    @Alexis you can see the sticky points if you overlay the curves. Try curve(exp(-exp(x)), from=-5, to=5) vs curve(plogis(x), from=-5, to=5). The concavity accelerates. If the risk of event from a single encounter was $p$, then the risk after the second event should be $1-(1-p)^2$ and so on, that's a probabilistic shape logit won't capture. High high exposures would skew logistic regression results more dramatically (falsely according to the prior probability rule). Some simulation would show you this.
    $endgroup$
    – AdamO
    5 hours ago















14












$begingroup$


I have two variables which don't show much correlation when plotted against each other as is, but a very clear linear relationship when I plot the logs of each variable agains the other.



So I would end up with a model of the type:



$$log(Y) = a log(X) + b$$ , which is great mathematically but doesn't seem to have the explanatory value of a regular linear model.



How can I interpret such a model?










share|cite|improve this question











$endgroup$







  • 4




    $begingroup$
    I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
    $endgroup$
    – Upper_Case
    yesterday










  • $begingroup$
    The interpretation of a log-log model, where the dependent variable is log(y) and the independent variable is log(x), is: $%Δ=β_1%Δx$.
    $endgroup$
    – C Thom
    10 hours ago







  • 1




    $begingroup$
    The complementary log-log link is an ideal GLM specification when the outcome is binary (risk model) and the exposure is cumulative, such as number of sexual partners vs. HIV infection. jstor.org/stable/2532454
    $endgroup$
    – AdamO
    9 hours ago










  • $begingroup$
    Thank you @AdamO ! Do you know if there are any circumstances where logit, probit, or complementary log-log links give substantively different results, either in point estimates, or in standard errors? As the authors you cited wrote "Since the logit of P closely approximates the complementary log- log of P over a wide range, such an analysis is likely to lead to qualitatively similar results to the methods described here," I wonder if there's a strong reason to prefer any in given circumstances? I know that the econometricians like probit links as natural for things like elasticities.
    $endgroup$
    – Alexis
    5 hours ago






  • 1




    $begingroup$
    @Alexis you can see the sticky points if you overlay the curves. Try curve(exp(-exp(x)), from=-5, to=5) vs curve(plogis(x), from=-5, to=5). The concavity accelerates. If the risk of event from a single encounter was $p$, then the risk after the second event should be $1-(1-p)^2$ and so on, that's a probabilistic shape logit won't capture. High high exposures would skew logistic regression results more dramatically (falsely according to the prior probability rule). Some simulation would show you this.
    $endgroup$
    – AdamO
    5 hours ago













14












14








14


2



$begingroup$


I have two variables which don't show much correlation when plotted against each other as is, but a very clear linear relationship when I plot the logs of each variable agains the other.



So I would end up with a model of the type:



$$log(Y) = a log(X) + b$$ , which is great mathematically but doesn't seem to have the explanatory value of a regular linear model.



How can I interpret such a model?










share|cite|improve this question











$endgroup$




I have two variables which don't show much correlation when plotted against each other as is, but a very clear linear relationship when I plot the logs of each variable agains the other.



So I would end up with a model of the type:



$$log(Y) = a log(X) + b$$ , which is great mathematically but doesn't seem to have the explanatory value of a regular linear model.



How can I interpret such a model?







regression correlation log






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









StubbornAtom

2,8371532




2,8371532










asked yesterday









Akaike's ChildrenAkaike's Children

1086




1086







  • 4




    $begingroup$
    I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
    $endgroup$
    – Upper_Case
    yesterday










  • $begingroup$
    The interpretation of a log-log model, where the dependent variable is log(y) and the independent variable is log(x), is: $%Δ=β_1%Δx$.
    $endgroup$
    – C Thom
    10 hours ago







  • 1




    $begingroup$
    The complementary log-log link is an ideal GLM specification when the outcome is binary (risk model) and the exposure is cumulative, such as number of sexual partners vs. HIV infection. jstor.org/stable/2532454
    $endgroup$
    – AdamO
    9 hours ago










  • $begingroup$
    Thank you @AdamO ! Do you know if there are any circumstances where logit, probit, or complementary log-log links give substantively different results, either in point estimates, or in standard errors? As the authors you cited wrote "Since the logit of P closely approximates the complementary log- log of P over a wide range, such an analysis is likely to lead to qualitatively similar results to the methods described here," I wonder if there's a strong reason to prefer any in given circumstances? I know that the econometricians like probit links as natural for things like elasticities.
    $endgroup$
    – Alexis
    5 hours ago






  • 1




    $begingroup$
    @Alexis you can see the sticky points if you overlay the curves. Try curve(exp(-exp(x)), from=-5, to=5) vs curve(plogis(x), from=-5, to=5). The concavity accelerates. If the risk of event from a single encounter was $p$, then the risk after the second event should be $1-(1-p)^2$ and so on, that's a probabilistic shape logit won't capture. High high exposures would skew logistic regression results more dramatically (falsely according to the prior probability rule). Some simulation would show you this.
    $endgroup$
    – AdamO
    5 hours ago












  • 4




    $begingroup$
    I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
    $endgroup$
    – Upper_Case
    yesterday










  • $begingroup$
    The interpretation of a log-log model, where the dependent variable is log(y) and the independent variable is log(x), is: $%Δ=β_1%Δx$.
    $endgroup$
    – C Thom
    10 hours ago







  • 1




    $begingroup$
    The complementary log-log link is an ideal GLM specification when the outcome is binary (risk model) and the exposure is cumulative, such as number of sexual partners vs. HIV infection. jstor.org/stable/2532454
    $endgroup$
    – AdamO
    9 hours ago










  • $begingroup$
    Thank you @AdamO ! Do you know if there are any circumstances where logit, probit, or complementary log-log links give substantively different results, either in point estimates, or in standard errors? As the authors you cited wrote "Since the logit of P closely approximates the complementary log- log of P over a wide range, such an analysis is likely to lead to qualitatively similar results to the methods described here," I wonder if there's a strong reason to prefer any in given circumstances? I know that the econometricians like probit links as natural for things like elasticities.
    $endgroup$
    – Alexis
    5 hours ago






  • 1




    $begingroup$
    @Alexis you can see the sticky points if you overlay the curves. Try curve(exp(-exp(x)), from=-5, to=5) vs curve(plogis(x), from=-5, to=5). The concavity accelerates. If the risk of event from a single encounter was $p$, then the risk after the second event should be $1-(1-p)^2$ and so on, that's a probabilistic shape logit won't capture. High high exposures would skew logistic regression results more dramatically (falsely according to the prior probability rule). Some simulation would show you this.
    $endgroup$
    – AdamO
    5 hours ago







4




4




$begingroup$
I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
$endgroup$
– Upper_Case
yesterday




$begingroup$
I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
$endgroup$
– Upper_Case
yesterday












$begingroup$
The interpretation of a log-log model, where the dependent variable is log(y) and the independent variable is log(x), is: $%Δ=β_1%Δx$.
$endgroup$
– C Thom
10 hours ago





$begingroup$
The interpretation of a log-log model, where the dependent variable is log(y) and the independent variable is log(x), is: $%Δ=β_1%Δx$.
$endgroup$
– C Thom
10 hours ago





1




1




$begingroup$
The complementary log-log link is an ideal GLM specification when the outcome is binary (risk model) and the exposure is cumulative, such as number of sexual partners vs. HIV infection. jstor.org/stable/2532454
$endgroup$
– AdamO
9 hours ago




$begingroup$
The complementary log-log link is an ideal GLM specification when the outcome is binary (risk model) and the exposure is cumulative, such as number of sexual partners vs. HIV infection. jstor.org/stable/2532454
$endgroup$
– AdamO
9 hours ago












$begingroup$
Thank you @AdamO ! Do you know if there are any circumstances where logit, probit, or complementary log-log links give substantively different results, either in point estimates, or in standard errors? As the authors you cited wrote "Since the logit of P closely approximates the complementary log- log of P over a wide range, such an analysis is likely to lead to qualitatively similar results to the methods described here," I wonder if there's a strong reason to prefer any in given circumstances? I know that the econometricians like probit links as natural for things like elasticities.
$endgroup$
– Alexis
5 hours ago




$begingroup$
Thank you @AdamO ! Do you know if there are any circumstances where logit, probit, or complementary log-log links give substantively different results, either in point estimates, or in standard errors? As the authors you cited wrote "Since the logit of P closely approximates the complementary log- log of P over a wide range, such an analysis is likely to lead to qualitatively similar results to the methods described here," I wonder if there's a strong reason to prefer any in given circumstances? I know that the econometricians like probit links as natural for things like elasticities.
$endgroup$
– Alexis
5 hours ago




1




1




$begingroup$
@Alexis you can see the sticky points if you overlay the curves. Try curve(exp(-exp(x)), from=-5, to=5) vs curve(plogis(x), from=-5, to=5). The concavity accelerates. If the risk of event from a single encounter was $p$, then the risk after the second event should be $1-(1-p)^2$ and so on, that's a probabilistic shape logit won't capture. High high exposures would skew logistic regression results more dramatically (falsely according to the prior probability rule). Some simulation would show you this.
$endgroup$
– AdamO
5 hours ago




$begingroup$
@Alexis you can see the sticky points if you overlay the curves. Try curve(exp(-exp(x)), from=-5, to=5) vs curve(plogis(x), from=-5, to=5). The concavity accelerates. If the risk of event from a single encounter was $p$, then the risk after the second event should be $1-(1-p)^2$ and so on, that's a probabilistic shape logit won't capture. High high exposures would skew logistic regression results more dramatically (falsely according to the prior probability rule). Some simulation would show you this.
$endgroup$
– AdamO
5 hours ago










4 Answers
4






active

oldest

votes


















20












$begingroup$

You just need to take exponential of both sides of the equation and you will get a potential relation, that may make sense for some data.



$$log(Y) = alog(X) + b$$



$$exp(log(Y)) = exp(a log(X) + b)$$



$$Y = e^bcdot X^a$$



And since $e^b$ is just a parameter that can take any positive value, this model is equivalent to:



$$Y=c cdot X^a$$



It should be noted that model expression should include the error term, and these change of variables has interesting effects on it:



$$log(Y) = a log(X) + b + epsilon$$



$$Y = e^bcdot X^acdot exp(epsilon)$$



That is, your model with a additive errors abiding to the conditions for OLS (normally distributed errors with constant variance) is equivalent to a potential model with multiplicative errors whose logaritm follows a normal distribution with constant variance.






share|cite|improve this answer











$endgroup$








  • 2




    $begingroup$
    OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
    $endgroup$
    – gardenhead
    yesterday






  • 2




    $begingroup$
    What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
    $endgroup$
    – Stats
    yesterday



















11












$begingroup$

You can take your model $log(Y)=alog(X)+b$ and calculate the total differential, you will end up with something like :
$$frac1YdY=afrac1XdX$$
which yields to
$$fracdYdXfracXY=a$$



Hence one simple interpretation of the coefficient $a$ will be the percent change in $Y$ for a percent change in $X$.
This implies furthermore that the variable $Y$ growths at a constant fraction ($a$) of the growth rate of $X$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    So if the log-log plot is linear, that would imply a constant growth rate?
    $endgroup$
    – Dimitriy V. Masterov
    yesterday










  • $begingroup$
    Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
    $endgroup$
    – RScrlli
    yesterday










  • $begingroup$
    Not over time, the growth rate with respect to the growth in x.
    $endgroup$
    – Dimitriy V. Masterov
    yesterday










  • $begingroup$
    reordering doesn't help, i'd remove it
    $endgroup$
    – Aksakal
    yesterday






  • 1




    $begingroup$
    @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
    $endgroup$
    – RScrlli
    yesterday


















6












$begingroup$

Intuitively $log$ gives us the order of magnitude of a variable, so we can view the relationship as the orders of magnitudes of the two variables are linearly related. For example, increasing the predictor by one order of magnitude may be associated with an increase of three orders of magnitude of the response.



When plotting using a log-log plot we hope to see a linear relationship.
Using an example from this question, we can check the linear model assumptions:



log-log






share|cite|improve this answer











$endgroup$








  • 2




    $begingroup$
    +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
    $endgroup$
    – Frans Rodenburg
    18 hours ago







  • 1




    $begingroup$
    The answer is right, but authorship attribution is wrong. The image shouldn't be attributed to Google Images but, at least, to the web page where it is to be found, that can be find out just by clicking in Google images.
    $endgroup$
    – Pere
    17 hours ago










  • $begingroup$
    @Pere I cannot find the original source of the image unfortunately (at least using reverse image search)
    $endgroup$
    – qwr
    16 hours ago


















3












$begingroup$

Reconciling the answer by @Rscrill with actual discrete data, consider



$$log(Y_t) = alog(X_t) + b,;;; log(Y_t-1) = alog(X_t-1) + b$$



$$implies log(Y_t) - log(Y_t-1) = aleft[log(X_t)-log(X_t-1)right]$$



But



$$log(Y_t) - log(Y_t-1) = logleft(fracY_tY_t-1right) equiv logleft(fracY_t-1+Delta Y_tY_t-1right) = logleft(1+fracDelta Y_tY_t-1right)$$



$fracDelta Y_tY_t-1$ is the percentage change of $Y$ between periods $t-1$ and $t$, or the growth rate of $Y_t$, say $g_Y_t$. When it is smaller than $0.1$, we have that an acceptable approximation is



$$logleft(1+fracDelta Y_tY_t-1right) approx fracDelta Y_tY_t-1=g_Y_t$$



Therefore we get



$$g_Y_tapprox ag_X_t$$



which validates in empirical studies the theoretical treatment of @Rscrill.






share|cite|improve this answer









$endgroup$












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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    20












    $begingroup$

    You just need to take exponential of both sides of the equation and you will get a potential relation, that may make sense for some data.



    $$log(Y) = alog(X) + b$$



    $$exp(log(Y)) = exp(a log(X) + b)$$



    $$Y = e^bcdot X^a$$



    And since $e^b$ is just a parameter that can take any positive value, this model is equivalent to:



    $$Y=c cdot X^a$$



    It should be noted that model expression should include the error term, and these change of variables has interesting effects on it:



    $$log(Y) = a log(X) + b + epsilon$$



    $$Y = e^bcdot X^acdot exp(epsilon)$$



    That is, your model with a additive errors abiding to the conditions for OLS (normally distributed errors with constant variance) is equivalent to a potential model with multiplicative errors whose logaritm follows a normal distribution with constant variance.






    share|cite|improve this answer











    $endgroup$








    • 2




      $begingroup$
      OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
      $endgroup$
      – gardenhead
      yesterday






    • 2




      $begingroup$
      What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
      $endgroup$
      – Stats
      yesterday
















    20












    $begingroup$

    You just need to take exponential of both sides of the equation and you will get a potential relation, that may make sense for some data.



    $$log(Y) = alog(X) + b$$



    $$exp(log(Y)) = exp(a log(X) + b)$$



    $$Y = e^bcdot X^a$$



    And since $e^b$ is just a parameter that can take any positive value, this model is equivalent to:



    $$Y=c cdot X^a$$



    It should be noted that model expression should include the error term, and these change of variables has interesting effects on it:



    $$log(Y) = a log(X) + b + epsilon$$



    $$Y = e^bcdot X^acdot exp(epsilon)$$



    That is, your model with a additive errors abiding to the conditions for OLS (normally distributed errors with constant variance) is equivalent to a potential model with multiplicative errors whose logaritm follows a normal distribution with constant variance.






    share|cite|improve this answer











    $endgroup$








    • 2




      $begingroup$
      OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
      $endgroup$
      – gardenhead
      yesterday






    • 2




      $begingroup$
      What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
      $endgroup$
      – Stats
      yesterday














    20












    20








    20





    $begingroup$

    You just need to take exponential of both sides of the equation and you will get a potential relation, that may make sense for some data.



    $$log(Y) = alog(X) + b$$



    $$exp(log(Y)) = exp(a log(X) + b)$$



    $$Y = e^bcdot X^a$$



    And since $e^b$ is just a parameter that can take any positive value, this model is equivalent to:



    $$Y=c cdot X^a$$



    It should be noted that model expression should include the error term, and these change of variables has interesting effects on it:



    $$log(Y) = a log(X) + b + epsilon$$



    $$Y = e^bcdot X^acdot exp(epsilon)$$



    That is, your model with a additive errors abiding to the conditions for OLS (normally distributed errors with constant variance) is equivalent to a potential model with multiplicative errors whose logaritm follows a normal distribution with constant variance.






    share|cite|improve this answer











    $endgroup$



    You just need to take exponential of both sides of the equation and you will get a potential relation, that may make sense for some data.



    $$log(Y) = alog(X) + b$$



    $$exp(log(Y)) = exp(a log(X) + b)$$



    $$Y = e^bcdot X^a$$



    And since $e^b$ is just a parameter that can take any positive value, this model is equivalent to:



    $$Y=c cdot X^a$$



    It should be noted that model expression should include the error term, and these change of variables has interesting effects on it:



    $$log(Y) = a log(X) + b + epsilon$$



    $$Y = e^bcdot X^acdot exp(epsilon)$$



    That is, your model with a additive errors abiding to the conditions for OLS (normally distributed errors with constant variance) is equivalent to a potential model with multiplicative errors whose logaritm follows a normal distribution with constant variance.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    answered yesterday









    PerePere

    4,6071720




    4,6071720







    • 2




      $begingroup$
      OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
      $endgroup$
      – gardenhead
      yesterday






    • 2




      $begingroup$
      What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
      $endgroup$
      – Stats
      yesterday













    • 2




      $begingroup$
      OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
      $endgroup$
      – gardenhead
      yesterday






    • 2




      $begingroup$
      What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
      $endgroup$
      – Stats
      yesterday








    2




    2




    $begingroup$
    OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
    $endgroup$
    – gardenhead
    yesterday




    $begingroup$
    OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
    $endgroup$
    – gardenhead
    yesterday




    2




    2




    $begingroup$
    What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
    $endgroup$
    – Stats
    yesterday





    $begingroup$
    What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
    $endgroup$
    – Stats
    yesterday














    11












    $begingroup$

    You can take your model $log(Y)=alog(X)+b$ and calculate the total differential, you will end up with something like :
    $$frac1YdY=afrac1XdX$$
    which yields to
    $$fracdYdXfracXY=a$$



    Hence one simple interpretation of the coefficient $a$ will be the percent change in $Y$ for a percent change in $X$.
    This implies furthermore that the variable $Y$ growths at a constant fraction ($a$) of the growth rate of $X$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      So if the log-log plot is linear, that would imply a constant growth rate?
      $endgroup$
      – Dimitriy V. Masterov
      yesterday










    • $begingroup$
      Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
      $endgroup$
      – RScrlli
      yesterday










    • $begingroup$
      Not over time, the growth rate with respect to the growth in x.
      $endgroup$
      – Dimitriy V. Masterov
      yesterday










    • $begingroup$
      reordering doesn't help, i'd remove it
      $endgroup$
      – Aksakal
      yesterday






    • 1




      $begingroup$
      @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
      $endgroup$
      – RScrlli
      yesterday















    11












    $begingroup$

    You can take your model $log(Y)=alog(X)+b$ and calculate the total differential, you will end up with something like :
    $$frac1YdY=afrac1XdX$$
    which yields to
    $$fracdYdXfracXY=a$$



    Hence one simple interpretation of the coefficient $a$ will be the percent change in $Y$ for a percent change in $X$.
    This implies furthermore that the variable $Y$ growths at a constant fraction ($a$) of the growth rate of $X$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      So if the log-log plot is linear, that would imply a constant growth rate?
      $endgroup$
      – Dimitriy V. Masterov
      yesterday










    • $begingroup$
      Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
      $endgroup$
      – RScrlli
      yesterday










    • $begingroup$
      Not over time, the growth rate with respect to the growth in x.
      $endgroup$
      – Dimitriy V. Masterov
      yesterday










    • $begingroup$
      reordering doesn't help, i'd remove it
      $endgroup$
      – Aksakal
      yesterday






    • 1




      $begingroup$
      @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
      $endgroup$
      – RScrlli
      yesterday













    11












    11








    11





    $begingroup$

    You can take your model $log(Y)=alog(X)+b$ and calculate the total differential, you will end up with something like :
    $$frac1YdY=afrac1XdX$$
    which yields to
    $$fracdYdXfracXY=a$$



    Hence one simple interpretation of the coefficient $a$ will be the percent change in $Y$ for a percent change in $X$.
    This implies furthermore that the variable $Y$ growths at a constant fraction ($a$) of the growth rate of $X$.






    share|cite|improve this answer











    $endgroup$



    You can take your model $log(Y)=alog(X)+b$ and calculate the total differential, you will end up with something like :
    $$frac1YdY=afrac1XdX$$
    which yields to
    $$fracdYdXfracXY=a$$



    Hence one simple interpretation of the coefficient $a$ will be the percent change in $Y$ for a percent change in $X$.
    This implies furthermore that the variable $Y$ growths at a constant fraction ($a$) of the growth rate of $X$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    answered yesterday









    RScrlliRScrlli

    192112




    192112











    • $begingroup$
      So if the log-log plot is linear, that would imply a constant growth rate?
      $endgroup$
      – Dimitriy V. Masterov
      yesterday










    • $begingroup$
      Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
      $endgroup$
      – RScrlli
      yesterday










    • $begingroup$
      Not over time, the growth rate with respect to the growth in x.
      $endgroup$
      – Dimitriy V. Masterov
      yesterday










    • $begingroup$
      reordering doesn't help, i'd remove it
      $endgroup$
      – Aksakal
      yesterday






    • 1




      $begingroup$
      @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
      $endgroup$
      – RScrlli
      yesterday
















    • $begingroup$
      So if the log-log plot is linear, that would imply a constant growth rate?
      $endgroup$
      – Dimitriy V. Masterov
      yesterday










    • $begingroup$
      Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
      $endgroup$
      – RScrlli
      yesterday










    • $begingroup$
      Not over time, the growth rate with respect to the growth in x.
      $endgroup$
      – Dimitriy V. Masterov
      yesterday










    • $begingroup$
      reordering doesn't help, i'd remove it
      $endgroup$
      – Aksakal
      yesterday






    • 1




      $begingroup$
      @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
      $endgroup$
      – RScrlli
      yesterday















    $begingroup$
    So if the log-log plot is linear, that would imply a constant growth rate?
    $endgroup$
    – Dimitriy V. Masterov
    yesterday




    $begingroup$
    So if the log-log plot is linear, that would imply a constant growth rate?
    $endgroup$
    – Dimitriy V. Masterov
    yesterday












    $begingroup$
    Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
    $endgroup$
    – RScrlli
    yesterday




    $begingroup$
    Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
    $endgroup$
    – RScrlli
    yesterday












    $begingroup$
    Not over time, the growth rate with respect to the growth in x.
    $endgroup$
    – Dimitriy V. Masterov
    yesterday




    $begingroup$
    Not over time, the growth rate with respect to the growth in x.
    $endgroup$
    – Dimitriy V. Masterov
    yesterday












    $begingroup$
    reordering doesn't help, i'd remove it
    $endgroup$
    – Aksakal
    yesterday




    $begingroup$
    reordering doesn't help, i'd remove it
    $endgroup$
    – Aksakal
    yesterday




    1




    1




    $begingroup$
    @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
    $endgroup$
    – RScrlli
    yesterday




    $begingroup$
    @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
    $endgroup$
    – RScrlli
    yesterday











    6












    $begingroup$

    Intuitively $log$ gives us the order of magnitude of a variable, so we can view the relationship as the orders of magnitudes of the two variables are linearly related. For example, increasing the predictor by one order of magnitude may be associated with an increase of three orders of magnitude of the response.



    When plotting using a log-log plot we hope to see a linear relationship.
    Using an example from this question, we can check the linear model assumptions:



    log-log






    share|cite|improve this answer











    $endgroup$








    • 2




      $begingroup$
      +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
      $endgroup$
      – Frans Rodenburg
      18 hours ago







    • 1




      $begingroup$
      The answer is right, but authorship attribution is wrong. The image shouldn't be attributed to Google Images but, at least, to the web page where it is to be found, that can be find out just by clicking in Google images.
      $endgroup$
      – Pere
      17 hours ago










    • $begingroup$
      @Pere I cannot find the original source of the image unfortunately (at least using reverse image search)
      $endgroup$
      – qwr
      16 hours ago















    6












    $begingroup$

    Intuitively $log$ gives us the order of magnitude of a variable, so we can view the relationship as the orders of magnitudes of the two variables are linearly related. For example, increasing the predictor by one order of magnitude may be associated with an increase of three orders of magnitude of the response.



    When plotting using a log-log plot we hope to see a linear relationship.
    Using an example from this question, we can check the linear model assumptions:



    log-log






    share|cite|improve this answer











    $endgroup$








    • 2




      $begingroup$
      +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
      $endgroup$
      – Frans Rodenburg
      18 hours ago







    • 1




      $begingroup$
      The answer is right, but authorship attribution is wrong. The image shouldn't be attributed to Google Images but, at least, to the web page where it is to be found, that can be find out just by clicking in Google images.
      $endgroup$
      – Pere
      17 hours ago










    • $begingroup$
      @Pere I cannot find the original source of the image unfortunately (at least using reverse image search)
      $endgroup$
      – qwr
      16 hours ago













    6












    6








    6





    $begingroup$

    Intuitively $log$ gives us the order of magnitude of a variable, so we can view the relationship as the orders of magnitudes of the two variables are linearly related. For example, increasing the predictor by one order of magnitude may be associated with an increase of three orders of magnitude of the response.



    When plotting using a log-log plot we hope to see a linear relationship.
    Using an example from this question, we can check the linear model assumptions:



    log-log






    share|cite|improve this answer











    $endgroup$



    Intuitively $log$ gives us the order of magnitude of a variable, so we can view the relationship as the orders of magnitudes of the two variables are linearly related. For example, increasing the predictor by one order of magnitude may be associated with an increase of three orders of magnitude of the response.



    When plotting using a log-log plot we hope to see a linear relationship.
    Using an example from this question, we can check the linear model assumptions:



    log-log







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 10 hours ago









    Anon1759

    32




    32










    answered yesterday









    qwrqwr

    240112




    240112







    • 2




      $begingroup$
      +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
      $endgroup$
      – Frans Rodenburg
      18 hours ago







    • 1




      $begingroup$
      The answer is right, but authorship attribution is wrong. The image shouldn't be attributed to Google Images but, at least, to the web page where it is to be found, that can be find out just by clicking in Google images.
      $endgroup$
      – Pere
      17 hours ago










    • $begingroup$
      @Pere I cannot find the original source of the image unfortunately (at least using reverse image search)
      $endgroup$
      – qwr
      16 hours ago












    • 2




      $begingroup$
      +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
      $endgroup$
      – Frans Rodenburg
      18 hours ago







    • 1




      $begingroup$
      The answer is right, but authorship attribution is wrong. The image shouldn't be attributed to Google Images but, at least, to the web page where it is to be found, that can be find out just by clicking in Google images.
      $endgroup$
      – Pere
      17 hours ago










    • $begingroup$
      @Pere I cannot find the original source of the image unfortunately (at least using reverse image search)
      $endgroup$
      – qwr
      16 hours ago







    2




    2




    $begingroup$
    +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
    $endgroup$
    – Frans Rodenburg
    18 hours ago





    $begingroup$
    +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
    $endgroup$
    – Frans Rodenburg
    18 hours ago





    1




    1




    $begingroup$
    The answer is right, but authorship attribution is wrong. The image shouldn't be attributed to Google Images but, at least, to the web page where it is to be found, that can be find out just by clicking in Google images.
    $endgroup$
    – Pere
    17 hours ago




    $begingroup$
    The answer is right, but authorship attribution is wrong. The image shouldn't be attributed to Google Images but, at least, to the web page where it is to be found, that can be find out just by clicking in Google images.
    $endgroup$
    – Pere
    17 hours ago












    $begingroup$
    @Pere I cannot find the original source of the image unfortunately (at least using reverse image search)
    $endgroup$
    – qwr
    16 hours ago




    $begingroup$
    @Pere I cannot find the original source of the image unfortunately (at least using reverse image search)
    $endgroup$
    – qwr
    16 hours ago











    3












    $begingroup$

    Reconciling the answer by @Rscrill with actual discrete data, consider



    $$log(Y_t) = alog(X_t) + b,;;; log(Y_t-1) = alog(X_t-1) + b$$



    $$implies log(Y_t) - log(Y_t-1) = aleft[log(X_t)-log(X_t-1)right]$$



    But



    $$log(Y_t) - log(Y_t-1) = logleft(fracY_tY_t-1right) equiv logleft(fracY_t-1+Delta Y_tY_t-1right) = logleft(1+fracDelta Y_tY_t-1right)$$



    $fracDelta Y_tY_t-1$ is the percentage change of $Y$ between periods $t-1$ and $t$, or the growth rate of $Y_t$, say $g_Y_t$. When it is smaller than $0.1$, we have that an acceptable approximation is



    $$logleft(1+fracDelta Y_tY_t-1right) approx fracDelta Y_tY_t-1=g_Y_t$$



    Therefore we get



    $$g_Y_tapprox ag_X_t$$



    which validates in empirical studies the theoretical treatment of @Rscrill.






    share|cite|improve this answer









    $endgroup$

















      3












      $begingroup$

      Reconciling the answer by @Rscrill with actual discrete data, consider



      $$log(Y_t) = alog(X_t) + b,;;; log(Y_t-1) = alog(X_t-1) + b$$



      $$implies log(Y_t) - log(Y_t-1) = aleft[log(X_t)-log(X_t-1)right]$$



      But



      $$log(Y_t) - log(Y_t-1) = logleft(fracY_tY_t-1right) equiv logleft(fracY_t-1+Delta Y_tY_t-1right) = logleft(1+fracDelta Y_tY_t-1right)$$



      $fracDelta Y_tY_t-1$ is the percentage change of $Y$ between periods $t-1$ and $t$, or the growth rate of $Y_t$, say $g_Y_t$. When it is smaller than $0.1$, we have that an acceptable approximation is



      $$logleft(1+fracDelta Y_tY_t-1right) approx fracDelta Y_tY_t-1=g_Y_t$$



      Therefore we get



      $$g_Y_tapprox ag_X_t$$



      which validates in empirical studies the theoretical treatment of @Rscrill.






      share|cite|improve this answer









      $endgroup$















        3












        3








        3





        $begingroup$

        Reconciling the answer by @Rscrill with actual discrete data, consider



        $$log(Y_t) = alog(X_t) + b,;;; log(Y_t-1) = alog(X_t-1) + b$$



        $$implies log(Y_t) - log(Y_t-1) = aleft[log(X_t)-log(X_t-1)right]$$



        But



        $$log(Y_t) - log(Y_t-1) = logleft(fracY_tY_t-1right) equiv logleft(fracY_t-1+Delta Y_tY_t-1right) = logleft(1+fracDelta Y_tY_t-1right)$$



        $fracDelta Y_tY_t-1$ is the percentage change of $Y$ between periods $t-1$ and $t$, or the growth rate of $Y_t$, say $g_Y_t$. When it is smaller than $0.1$, we have that an acceptable approximation is



        $$logleft(1+fracDelta Y_tY_t-1right) approx fracDelta Y_tY_t-1=g_Y_t$$



        Therefore we get



        $$g_Y_tapprox ag_X_t$$



        which validates in empirical studies the theoretical treatment of @Rscrill.






        share|cite|improve this answer









        $endgroup$



        Reconciling the answer by @Rscrill with actual discrete data, consider



        $$log(Y_t) = alog(X_t) + b,;;; log(Y_t-1) = alog(X_t-1) + b$$



        $$implies log(Y_t) - log(Y_t-1) = aleft[log(X_t)-log(X_t-1)right]$$



        But



        $$log(Y_t) - log(Y_t-1) = logleft(fracY_tY_t-1right) equiv logleft(fracY_t-1+Delta Y_tY_t-1right) = logleft(1+fracDelta Y_tY_t-1right)$$



        $fracDelta Y_tY_t-1$ is the percentage change of $Y$ between periods $t-1$ and $t$, or the growth rate of $Y_t$, say $g_Y_t$. When it is smaller than $0.1$, we have that an acceptable approximation is



        $$logleft(1+fracDelta Y_tY_t-1right) approx fracDelta Y_tY_t-1=g_Y_t$$



        Therefore we get



        $$g_Y_tapprox ag_X_t$$



        which validates in empirical studies the theoretical treatment of @Rscrill.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Alecos PapadopoulosAlecos Papadopoulos

        42.7k296195




        42.7k296195



























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            Serbia Índice Etimología Historia Geografía Entorno natural División administrativa Política Demografía Economía Cultura Deportes Véase también Notas Referencias Bibliografía Enlaces externos Menú de navegación44°49′00″N 20°28′00″E / 44.816666666667, 20.46666666666744°49′00″N 20°28′00″E / 44.816666666667, 20.466666666667U.S. Department of Commerce (2015)«Informe sobre Desarrollo Humano 2018»Kosovo-Metohija.Neutralna Srbija u NATO okruzenju.The SerbsTheories on the Origin of the Serbs.Serbia.Earls: Webster's Quotations, Facts and Phrases.Egeo y Balcanes.Kalemegdan.Southern Pannonia during the age of the Great Migrations.Culture in Serbia.History.The Serbian Origin of the Montenegrins.Nemanjics' period (1186-1353).Stefan Uros (1355-1371).Serbian medieval history.Habsburg–Ottoman Wars (1525–1718).The Ottoman Empire, 1700-1922.The First Serbian Uprising.Miloš, prince of Serbia.3. 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Josip Broz.El nuevo orden y la resistencia.La conquista del poder.Algunos aspectos de la economía yugoslava a mediados de 1962.Albania-Kosovo crisis.De Kosovo a Kosova: una visión demográfica.La crisis de la economía yugoslava y la política de "estabilización".Milosevic: el poder de un absolutista."Serbia under Milošević: politics in the 1990s"Milosevic cavó en Kosovo la tumba de la antigua Yugoslavia.La ONU exculpa a Serbia de genocidio en la guerra de Bosnia.Slobodan Milosevic, el burócrata que supo usar el odio.Es la fuerza contra el sufrimiento de muchos inocentes.Matanza de civiles al bombardear la OTAN un puente mientras pasaba un tren.Las consecuencias negativas de los bombardeos de Yugoslavia se sentirán aún durante largo tiempo.Kostunica advierte que la misión de Europa en Kosovo es ilegal.Las 24 horas más largas en la vida de Slobodan Milosevic.Serbia declara la guerra a la mafia por matar a Djindjic.Tadic presentará "quizás en diciembre" la solicitud de entrada en la UE.Montenegro declara su independencia de Serbia.Serbia se declara estado soberano tras separación de Montenegro.«Accordance with International Law of the Unilateral Declaration of Independence by the Provisional Institutions of Self-Government of Kosovo (Request for Advisory Opinion)»Mladic pasa por el médico antes de la audiencia para extraditarloDatos de Serbia y Kosovo.The Carpathian Mountains.Position, Relief, Climate.Transport.Finding birds in Serbia.U Srbiji do 2010. godine 10% teritorije nacionalni parkovi.Geography.Serbia: Climate.Variability of Climate In Serbia In The Second Half of The 20thc Entury.BASIC CLIMATE CHARACTERISTICS FOR THE TERRITORY OF SERBIA.Fauna y flora: Serbia.Serbia and Montenegro.Información general sobre Serbia.Republic of Serbia Environmental Protection Agency (SEPA).Serbia recycling 15% of waste.Reform process of the Serbian energy sector.20-MW Wind Project Being Developed in Serbia.Las Naciones Unidas. Paz para Kosovo.Aniversario sin fiesta.Population by national or ethnic groups by Census 2002.Article 7. Coat of arms, flag and national anthem.Serbia, flag of.Historia.«Serbia and Montenegro in Pictures»Serbia.Serbia aprueba su nueva Constitución con un apoyo de más del 50%.Serbia. Population.«El nacionalista Nikolic gana las elecciones presidenciales en Serbia»El europeísta Borís Tadic gana la segunda vuelta de las presidenciales serbias.Aleksandar Vucic, de ultranacionalista serbio a fervoroso europeístaKostunica condena la declaración del "falso estado" de Kosovo.Comienza el debate sobre la independencia de Kosovo en el TIJ.La Corte Internacional de Justicia dice que Kosovo no violó el derecho internacional al declarar su independenciaKosovo: Enviado de la ONU advierte tensiones y fragilidad.«Bruselas recomienda negociar la adhesión de Serbia tras el acuerdo sobre Kosovo»Monografía de Serbia.Bez smanjivanja Vojske Srbije.Military statistics Serbia and Montenegro.Šutanovac: Vojni budžet za 2009. godinu 70 milijardi dinara.Serbia-Montenegro shortens obligatory military service to six months.No hay justicia para las víctimas de los bombardeos de la OTAN.Zapatero reitera la negativa de España a reconocer la independencia de Kosovo.Anniversary of the signing of the Stabilisation and Association Agreement.Detenido en Serbia Radovan Karadzic, el criminal de guerra más buscado de Europa."Serbia presentará su candidatura de acceso a la UE antes de fin de año".Serbia solicita la adhesión a la UE.Detenido el exgeneral serbobosnio Ratko Mladic, principal acusado del genocidio en los Balcanes«Lista de todos los Estados Miembros de las Naciones Unidas que son parte o signatarios en los diversos instrumentos de derechos humanos de las Naciones Unidas»versión pdfProtocolo Facultativo de la Convención sobre la Eliminación de todas las Formas de Discriminación contra la MujerConvención contra la tortura y otros tratos o penas crueles, inhumanos o degradantesversión pdfProtocolo Facultativo de la Convención sobre los Derechos de las Personas con DiscapacidadEl ACNUR recibe con beneplácito el envío de tropas de la OTAN a Kosovo y se prepara ante una posible llegada de refugiados a Serbia.Kosovo.- El jefe de la Minuk denuncia que los serbios boicotearon las legislativas por 'presiones'.Bosnia and Herzegovina. Population.Datos básicos de Montenegro, historia y evolución política.Serbia y Montenegro. Indicador: Tasa global de fecundidad (por 1000 habitantes).Serbia y Montenegro. Indicador: Tasa bruta de mortalidad (por 1000 habitantes).Population.Falleció el patriarca de la Iglesia Ortodoxa serbia.Atacan en Kosovo autobuses con peregrinos tras la investidura del patriarca serbio IrinejSerbian in Hungary.Tasas de cambio."Kosovo es de todos sus ciudadanos".Report for Serbia.Country groups by income.GROSS DOMESTIC PRODUCT (GDP) OF THE REPUBLIC OF SERBIA 1997–2007.Economic Trends in the Republic of Serbia 2006.National Accounts Statitics.Саопштења за јавност.GDP per inhabitant varied by one to six across the EU27 Member States.Un pacto de estabilidad para Serbia.Unemployment rate rises in Serbia.Serbia, Belarus agree free trade to woo investors.Serbia, Turkey call investors to Serbia.Success Stories.U.S. Private Investment in Serbia and Montenegro.Positive trend.Banks in Serbia.La Cámara de Comercio acompaña a empresas madrileñas a Serbia y Croacia.Serbia Industries.Energy and mining.Agriculture.Late crops, fruit and grapes output, 2008.Rebranding Serbia: A Hobby Shortly to Become a Full-Time Job.Final data on livestock statistics, 2008.Serbian cell-phone users.U Srbiji sve više računara.Телекомуникације.U Srbiji 27 odsto gradjana koristi Internet.Serbia and Montenegro.Тренд гледаности програма РТС-а у 2008. и 2009.години.Serbian railways.General Terms.El mercado del transporte aéreo en Serbia.Statistics.Vehículos de motor registrados.Planes ambiciosos para el transporte fluvial.Turismo.Turistički promet u Republici Srbiji u periodu januar-novembar 2007. godine.Your Guide to Culture.Novi Sad - city of culture.Nis - european crossroads.Serbia. Properties inscribed on the World Heritage List .Stari Ras and Sopoćani.Studenica Monastery.Medieval Monuments in Kosovo.Gamzigrad-Romuliana, Palace of Galerius.Skiing and snowboarding in Kopaonik.Tara.New7Wonders of Nature Finalists.Pilgrimage of Saint Sava.Exit Festival: Best european festival.Banje u Srbiji.«The Encyclopedia of world history»Culture.Centenario del arte serbio.«Djordje Andrejevic Kun: el único pintor de los brigadistas yugoslavos de la guerra civil española»About the museum.The collections.Miroslav Gospel – Manuscript from 1180.Historicity in the Serbo-Croatian Heroic Epic.Culture and Sport.Conversación con el rector del Seminario San Sava.'Reina Margot' funde drama, historia y gesto con música de Goran Bregovic.Serbia gana Eurovisión y España decepciona de nuevo con un vigésimo puesto.Home.Story.Emir Kusturica.Tercer oro para Paskaljevic.Nikola Tesla Year.Home.Tesla, un genio tomado por loco.Aniversario de la muerte de Nikola Tesla.El Museo Nikola Tesla en Belgrado.El inventor del mundo actual.República de Serbia.University of Belgrade official statistics.University of Novi Sad.University of Kragujevac.University of Nis.Comida. Cocina serbia.Cooking.Montenegro se convertirá en el miembro 204 del movimiento olímpico.España, campeona de Europa de baloncesto.El Partizan de Belgrado se corona campeón por octava vez consecutiva.Serbia se clasifica para el Mundial de 2010 de Sudáfrica.Serbia Name Squad For Northern Ireland And South Korea Tests.Fútbol.- El Partizán de Belgrado se proclama campeón de la Liga serbia.Clasificacion final Mundial de balonmano Croacia 2009.Serbia vence a España y se consagra campeón mundial de waterpolo.Novak Djokovic no convence pero gana en Australia.Gana Ana Ivanovic el Roland Garros.Serena Williams gana el US Open por tercera vez.Biography.Bradt Travel Guide SerbiaThe Encyclopedia of World War IGobierno de SerbiaPortal del Gobierno de SerbiaPresidencia de SerbiaAsamblea Nacional SerbiaMinisterio de Asuntos exteriores de SerbiaBanco Nacional de SerbiaAgencia Serbia para la Promoción de la Inversión y la ExportaciónOficina de Estadísticas de SerbiaCIA. Factbook 2008Organización nacional de turismo de SerbiaDiscover SerbiaConoce SerbiaNoticias de SerbiaSerbiaWorldCat1512028760000 0000 9526 67094054598-2n8519591900570825ge1309191004530741010url17413117006669D055771Serbia