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

How can a function with a hole (removable discontinuity) equal a function with no hole?

CREATE opcode: what does it really do?

How easy is it to start Magic from scratch?

Do sorcerers' Subtle Spells require a skill check to be unseen?

Why escape if the_content isnt?

Why, precisely, is argon used in neutrino experiments?

Detecting if an element is found inside a container

Customer Requests (Sometimes) Drive Me Bonkers!

Term for the "extreme-extension" version of a straw man fallacy?

How long to clear the 'suck zone' of a turbofan after start is initiated?

What is paid subscription needed for in Mortal Kombat 11?

Is the destination of a commercial flight important for the pilot?

How did Doctor Strange see the winning outcome in Avengers: Infinity War?

Proof of work - lottery approach

Large drywall patch supports

Failed to fetch jessie backports repository

How do we know the LHC results are robust?

Increase performance creating Mandelbrot set in python

How to Reset Passwords on Multiple Websites Easily?

Sequence of Tenses: Translating the subjunctive

Short story about space worker geeks who zone out by 'listening' to radiation from stars

How does it work when somebody invests in my business?

Method to test if a number is a perfect power?

What happens if you roll doubles 3 times then land on "Go to jail?"



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









$endgroup$












    Your Answer





    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "65"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f399527%2fwhat-is-the-intuitive-meaning-of-having-a-linear-relationship-between-the-logs-o%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    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



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Cross Validated!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f399527%2fwhat-is-the-intuitive-meaning-of-having-a-linear-relationship-between-the-logs-o%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Triangular numbers and gcdProving sum of a set is $0 pmod n$ if $n$ is odd, or $fracn2 pmod n$ if $n$ is even?Is greatest common divisor of two numbers really their smallest linear combination?GCD, LCM RelationshipProve a set of nonnegative integers with greatest common divisor 1 and closed under addition has all but finite many nonnegative integers.all pairs of a and b in an equation containing gcdTriangular Numbers Modulo $k$ - Hit All Values?Understanding the Existence and Uniqueness of the GCDGCD and LCM with logical symbolsThe greatest common divisor of two positive integers less than 100 is equal to 3. Their least common multiple is twelve times one of the integers.Suppose that for all integers $x$, $x|a$ and $x|b$ if and only if $x|c$. Then $c = gcd(a,b)$Which is the gcd of 2 numbers which are multiplied and the result is 600000?

            Ingelân Ynhâld Etymology | Geografy | Skiednis | Polityk en bestjoer | Ekonomy | Demografy | Kultuer | Klimaat | Sjoch ek | Keppelings om utens | Boarnen, noaten en referinsjes Navigaasjemenuwww.gov.ukOffisjele webside fan it regear fan it Feriene KeninkrykOffisjele webside fan it Britske FerkearsburoNederlânsktalige ynformaasje fan it Britske FerkearsburoOffisjele webside fan English Heritage, de organisaasje dy't him ynset foar it behâld fan it Ingelske kultuergoedYnwennertallen fan alle Britske stêden út 'e folkstelling fan 2011Notes en References, op dizze sideEngland

            Հադիս Բովանդակություն Անվանում և նշանակություն | Դասակարգում | Աղբյուրներ | Նավարկման ցանկ