Joint CDF's of both continuous and discrete random variables The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)joint probability distribution of one discrete, one continuous random variableCalculating joint probability distribution of two random variables.conditional probability combining discrete and continous random variablesFinding the Expected Value for a joint discrete/continuous systemDefinition of Random VariablesSeemingly Identical Random Variables with Different VariancesJoint PMF of 2 discrete random variablesHow is conditional probability defined for the joint distribution of two random continuous variables?Calculating Simple Functions of Discrete Random VariablesProbability of random variables
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Joint CDF's of both continuous and discrete random variables
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)joint probability distribution of one discrete, one continuous random variableCalculating joint probability distribution of two random variables.conditional probability combining discrete and continous random variablesFinding the Expected Value for a joint discrete/continuous systemDefinition of Random VariablesSeemingly Identical Random Variables with Different VariancesJoint PMF of 2 discrete random variablesHow is conditional probability defined for the joint distribution of two random continuous variables?Calculating Simple Functions of Discrete Random VariablesProbability of random variables
$begingroup$
I am working on some homework and have arrived at a problem that has me stumped. I am trying to find the conditional probability of Y given a discrete variable X or:
$$F_Y(y)$$ which I know is equal to:
$$
F_Y(y) = frac mathbbP(Y leq y, X = x)mathbbP(X = x)=fracF_Y,X(y,x)mathbbP(X=x)
$$
In the problem we are given that X is a random variable representing a dice roll and that Y is a continuous random variable that is distributed exponential with parameter X.
My first thought was to sum , to the given x value, the exponential distribution of Y with the corresponding value of X:
$$sum_i=1^x 1-e^(-iy)$$
However this gave me a super nasty looking partial sum:
$$frac(e^(-(1+x) y) (e^y+e^((2+x) y) x-e^((1+x) y) (1+x)))(-1+e^y)$$
which does not look right.
Next I thought about trying to find the conditional probability with a given X. This lead to me to try:
$$mathbbP(X=x)*F_Y(y, X=x)= frac16*(1-e^-xy)$$
This formula, however limited the values I could get to ~.1666 for any value of y.
I am not looking for a complete answer but if anyone could help me correct my thinking that would be great, thanks!
probability probability-distributions
asked Jan 22 '16 at 4:48
SamSam
1314
$endgroup$
add a comment |
$begingroup$
I am working on some homework and have arrived at a problem that has me stumped. I am trying to find the conditional probability of Y given a discrete variable X or:
$$F_Y(y)$$ which I know is equal to:
$$
F_Y(y) = frac mathbbP(Y leq y, X = x)mathbbP(X = x)=fracF_Y,X(y,x)mathbbP(X=x)
$$
In the problem we are given that X is a random variable representing a dice roll and that Y is a continuous random variable that is distributed exponential with parameter X.
My first thought was to sum , to the given x value, the exponential distribution of Y with the corresponding value of X:
$$sum_i=1^x 1-e^(-iy)$$
However this gave me a super nasty looking partial sum:
$$frac(e^(-(1+x) y) (e^y+e^((2+x) y) x-e^((1+x) y) (1+x)))(-1+e^y)$$
which does not look right.
Next I thought about trying to find the conditional probability with a given X. This lead to me to try:
$$mathbbP(X=x)*F_Y(y, X=x)= frac16*(1-e^-xy)$$
This formula, however limited the values I could get to ~.1666 for any value of y.
I am not looking for a complete answer but if anyone could help me correct my thinking that would be great, thanks!
probability probability-distributions
asked Jan 22 '16 at 4:48
SamSam
1314
$endgroup$
$begingroup$
It looks like the problem is a lot easier than you are trying to make it be. If random variate $y$ is chosen by rolling a discrete uniform random $x$ on $[1,6]$ and then finding $y$ by an exponential distribution with mean $x$, then the conditional distribution of $y$ given $x$ is just that exponential distribution (using that value of $x$ as the mean).
$endgroup$
– Mark Fischler
Jan 22 '16 at 4:55
$begingroup$
So my second answer would be more correct, the conditional probability would just be without the 1/6. Would that make what I have written in my second answer actually be the joint CDF of the two variables then?
$endgroup$
– Sam
Jan 22 '16 at 5:11
add a comment |
$begingroup$
I am working on some homework and have arrived at a problem that has me stumped. I am trying to find the conditional probability of Y given a discrete variable X or:
$$F_Y(y)$$ which I know is equal to:
$$
F_Y(y) = frac mathbbP(Y leq y, X = x)mathbbP(X = x)=fracF_Y,X(y,x)mathbbP(X=x)
$$
In the problem we are given that X is a random variable representing a dice roll and that Y is a continuous random variable that is distributed exponential with parameter X.
My first thought was to sum , to the given x value, the exponential distribution of Y with the corresponding value of X:
$$sum_i=1^x 1-e^(-iy)$$
However this gave me a super nasty looking partial sum:
$$frac(e^(-(1+x) y) (e^y+e^((2+x) y) x-e^((1+x) y) (1+x)))(-1+e^y)$$
which does not look right.
Next I thought about trying to find the conditional probability with a given X. This lead to me to try:
$$mathbbP(X=x)*F_Y(y, X=x)= frac16*(1-e^-xy)$$
This formula, however limited the values I could get to ~.1666 for any value of y.
I am not looking for a complete answer but if anyone could help me correct my thinking that would be great, thanks!
probability probability-distributions
asked Jan 22 '16 at 4:48
SamSam
1314
$endgroup$
I am working on some homework and have arrived at a problem that has me stumped. I am trying to find the conditional probability of Y given a discrete variable X or:
$$F_Y(y)$$ which I know is equal to:
$$
F_Y(y) = frac mathbbP(Y leq y, X = x)mathbbP(X = x)=fracF_Y,X(y,x)mathbbP(X=x)
$$
In the problem we are given that X is a random variable representing a dice roll and that Y is a continuous random variable that is distributed exponential with parameter X.
My first thought was to sum , to the given x value, the exponential distribution of Y with the corresponding value of X:
$$sum_i=1^x 1-e^(-iy)$$
However this gave me a super nasty looking partial sum:
$$frac(e^(-(1+x) y) (e^y+e^((2+x) y) x-e^((1+x) y) (1+x)))(-1+e^y)$$
which does not look right.
Next I thought about trying to find the conditional probability with a given X. This lead to me to try:
$$mathbbP(X=x)*F_Y(y, X=x)= frac16*(1-e^-xy)$$
This formula, however limited the values I could get to ~.1666 for any value of y.
I am not looking for a complete answer but if anyone could help me correct my thinking that would be great, thanks!
probability probability-distributions
probability probability-distributions
asked Jan 22 '16 at 4:48
SamSam
1314
asked Jan 22 '16 at 4:48
SamSam
1314
asked Jan 22 '16 at 4:48
SamSam
1314
asked Jan 22 '16 at 4:48
SamSam
1314
asked Jan 22 '16 at 4:48
SamSam
1314
1314
$begingroup$
It looks like the problem is a lot easier than you are trying to make it be. If random variate $y$ is chosen by rolling a discrete uniform random $x$ on $[1,6]$ and then finding $y$ by an exponential distribution with mean $x$, then the conditional distribution of $y$ given $x$ is just that exponential distribution (using that value of $x$ as the mean).
$endgroup$
– Mark Fischler
Jan 22 '16 at 4:55
$begingroup$
So my second answer would be more correct, the conditional probability would just be without the 1/6. Would that make what I have written in my second answer actually be the joint CDF of the two variables then?
$endgroup$
– Sam
Jan 22 '16 at 5:11
add a comment |
$begingroup$
It looks like the problem is a lot easier than you are trying to make it be. If random variate $y$ is chosen by rolling a discrete uniform random $x$ on $[1,6]$ and then finding $y$ by an exponential distribution with mean $x$, then the conditional distribution of $y$ given $x$ is just that exponential distribution (using that value of $x$ as the mean).
$endgroup$
– Mark Fischler
Jan 22 '16 at 4:55
$begingroup$
So my second answer would be more correct, the conditional probability would just be without the 1/6. Would that make what I have written in my second answer actually be the joint CDF of the two variables then?
$endgroup$
– Sam
Jan 22 '16 at 5:11
$begingroup$
It looks like the problem is a lot easier than you are trying to make it be. If random variate $y$ is chosen by rolling a discrete uniform random $x$ on $[1,6]$ and then finding $y$ by an exponential distribution with mean $x$, then the conditional distribution of $y$ given $x$ is just that exponential distribution (using that value of $x$ as the mean).
$endgroup$
– Mark Fischler
Jan 22 '16 at 4:55
$begingroup$
It looks like the problem is a lot easier than you are trying to make it be. If random variate $y$ is chosen by rolling a discrete uniform random $x$ on $[1,6]$ and then finding $y$ by an exponential distribution with mean $x$, then the conditional distribution of $y$ given $x$ is just that exponential distribution (using that value of $x$ as the mean).
$endgroup$
– Mark Fischler
Jan 22 '16 at 4:55
$begingroup$
So my second answer would be more correct, the conditional probability would just be without the 1/6. Would that make what I have written in my second answer actually be the joint CDF of the two variables then?
$endgroup$
– Sam
Jan 22 '16 at 5:11
$begingroup$
So my second answer would be more correct, the conditional probability would just be without the 1/6. Would that make what I have written in my second answer actually be the joint CDF of the two variables then?
$endgroup$
– Sam
Jan 22 '16 at 5:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think you first answer reverses what you want.
You could say $$displaystyle F_Y(y) = sum_x mathbbP(X=x) F_Y(y)$$ and possibly even turn this into a density $displaystyle f_Y(y) = sum_x mathbbP(X=x) f_Y(y)$
So your particular example would give $displaystyle F_Y(y) = 1 - tfrac16(e^-y+e^-2y+e^-3y+e^-4y+e^-5y+e^-6y)=1-frace^-y(1-e^-6y )6(1-e^-y)$
answered Jan 22 '16 at 8:55
HenryHenry
101k482170
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add a comment |
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1 Answer
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$begingroup$
I think you first answer reverses what you want.
You could say $$displaystyle F_Y(y) = sum_x mathbbP(X=x) F_Y(y)$$ and possibly even turn this into a density $displaystyle f_Y(y) = sum_x mathbbP(X=x) f_Y(y)$
So your particular example would give $displaystyle F_Y(y) = 1 - tfrac16(e^-y+e^-2y+e^-3y+e^-4y+e^-5y+e^-6y)=1-frace^-y(1-e^-6y )6(1-e^-y)$
answered Jan 22 '16 at 8:55
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
I think you first answer reverses what you want.
You could say $$displaystyle F_Y(y) = sum_x mathbbP(X=x) F_Y(y)$$ and possibly even turn this into a density $displaystyle f_Y(y) = sum_x mathbbP(X=x) f_Y(y)$
So your particular example would give $displaystyle F_Y(y) = 1 - tfrac16(e^-y+e^-2y+e^-3y+e^-4y+e^-5y+e^-6y)=1-frace^-y(1-e^-6y )6(1-e^-y)$
answered Jan 22 '16 at 8:55
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
I think you first answer reverses what you want.
You could say $$displaystyle F_Y(y) = sum_x mathbbP(X=x) F_Y(y)$$ and possibly even turn this into a density $displaystyle f_Y(y) = sum_x mathbbP(X=x) f_Y(y)$
So your particular example would give $displaystyle F_Y(y) = 1 - tfrac16(e^-y+e^-2y+e^-3y+e^-4y+e^-5y+e^-6y)=1-frace^-y(1-e^-6y )6(1-e^-y)$
answered Jan 22 '16 at 8:55
HenryHenry
101k482170
$endgroup$
I think you first answer reverses what you want.
You could say $$displaystyle F_Y(y) = sum_x mathbbP(X=x) F_Y(y)$$ and possibly even turn this into a density $displaystyle f_Y(y) = sum_x mathbbP(X=x) f_Y(y)$
So your particular example would give $displaystyle F_Y(y) = 1 - tfrac16(e^-y+e^-2y+e^-3y+e^-4y+e^-5y+e^-6y)=1-frace^-y(1-e^-6y )6(1-e^-y)$
answered Jan 22 '16 at 8:55
HenryHenry
101k482170
answered Jan 22 '16 at 8:55
HenryHenry
101k482170
answered Jan 22 '16 at 8:55
HenryHenry
101k482170
answered Jan 22 '16 at 8:55
HenryHenry
101k482170
101k482170
add a comment |
add a comment |
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$begingroup$
It looks like the problem is a lot easier than you are trying to make it be. If random variate $y$ is chosen by rolling a discrete uniform random $x$ on $[1,6]$ and then finding $y$ by an exponential distribution with mean $x$, then the conditional distribution of $y$ given $x$ is just that exponential distribution (using that value of $x$ as the mean).
$endgroup$
– Mark Fischler
Jan 22 '16 at 4:55
$begingroup$
So my second answer would be more correct, the conditional probability would just be without the 1/6. Would that make what I have written in my second answer actually be the joint CDF of the two variables then?
$endgroup$
– Sam
Jan 22 '16 at 5:11