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As the title says, I am wondering how to find the joint distribution of two variables when only given the conditional distribution.

An example problem is, Suppose, $Y$ given $X = x$ follows Exponential($1/x$): $f_{Y \mid X}(y\mid X = x) ~=~ (1/x)e^{-(y/x)}$ iff $y > 0$ and $X$ follows Exponential(1) distribution. Find joint distribution of $Y$ and $X$.

I know the joint distribution of two variables is equal to the conditional distribution multiplied by the marginal distribution of the 'given' variable, but I am not sure how to find the marginals from the information given.

  • Is the marginal of $X$ equal to the probability density function of an Exponential(1) distribution?

  • Also, how can I find the variance of $Y$?

  • Would I just use moment generating functions to solve for the variance of Y?

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  • $\begingroup$ The marginal of $X$ is already given to you. To find the marginal of $Y$, find the joint distribution (you have described the procedure in your question), and then integrate the joint density $f_{X,Y}(x,y)$ over $x$. $\endgroup$ – angryavian Nov 29 '17 at 23:23
  • $\begingroup$ @angryavian where is the marginal of X given? Is it determined from being told that X follows Exponential(1) distribution? And if so, that would mean the marginal of X is just e^-x correct? $\endgroup$ – a.nas Nov 29 '17 at 23:54
  • $\begingroup$ Yes, the marginal distribution of $X$ is Exponential(1), which has pdf $f(x)=e^{-x}$. $\endgroup$ – angryavian Nov 29 '17 at 23:59
  • $\begingroup$ @a.nas It´s a good practice to give a reply and to mark a good answer as accepted. Please check your other questions as well. $\endgroup$ – callculus Dec 4 '17 at 0:30
  • $\begingroup$ @callculus i apologize, didn't realize there was an option to click the check mark next to given answers. Also, I meant to thank the person who answered but when replying/commenting it says to refrain from just saying "thank you", which is why I did not respond. Thanks for letting me know how to go about this. $\endgroup$ – a.nas Dec 4 '17 at 5:02
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The joint distribution is simply: $f_{Y,X}(y, x) ~{=~ f_{Y\mid X}(y\mid x)\cdot f_X(x)\\=~ \tfrac 1x \mathsf e^{-y/x}\mathbf 1_{0\leqslant y} \cdot \mathsf e^{-x} \mathbf 1_{0\leqslant x} \\ = \tfrac 1x\mathsf e^{-x-y/x}~\mathbf 1_{0\leqslant y, 0\leqslant x} }$

That is all.   You know $f_X(x)=\mathsf e^{-x}\mathbf 1_{0\leqslant x}$ because you were told that $X\sim\mathcal{Exp}(1)$ .   Everything else is just the definition of conditional probability density function (for jointly continuous random variables).


Don't try to find the marginal pdf for $Y$; it looks unlikely to be expressable in terms of elementary functions.


The variance for $Y$ is easiest found via the Law of Total Variance: $$\mathsf{Var}(Y)~=~\mathsf E(\mathsf {Var}(Y\mid X))+\mathsf{Var}(\mathsf E(Y\mid X))$$

Since $Y\mid X\sim\mathcal{Exp}(1/X)$ we know $\mathsf{E}(Y\mid X)=X, \mathsf {Var}(Y\mid X)=X^2$, and since $X\sim\mathcal{Exp}(1)$ we know $\mathsf E(X)=1$ and $\mathsf {E}(X^2) = \mathsf{Var}(X)+\mathsf E(X)^2 = 2$. Put it together.


Otherwise use $\mathsf{Var}(Y) =\int_0^\infty \int_0^\infty y^2~f_{Y,X}(y,x)\mathsf d y\mathsf d x- \left(\int_0^\infty \int_0^\infty y~f_{Y,X}(y,x)\mathsf d y\mathsf d x\right)^2$


Also, similarly the expectation for $Y$ is $\mathsf E(Y)~{=\mathsf E(\mathsf E(Y\mid X)) \\= \mathsf E(X) \\= 1}$

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