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?

  • $\begingroup$ 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. $\endgroup$ 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

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 $\mathsf E(Y)~{=\mathsf E(\mathsf E(Y\mid X)) \\= \mathsf E(X) \\= 1}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.