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Let $X$ , $Y$ be two independent, random variables each follows the exponential law with parameter $\lambda$

a. Find the joint density function $f(u,v)$ where $V=X+Y$ and $U=X$?

b. Find the marginal density function $f(v)$?

Solution:

a. I got

$f(u,v)= {\lambda}^2 e^{-v}$ is this the right answer ? if yes should the double integral be 1 in this case ... I don't think it is 1.

b. the joint density function doesn't depend on $u$ explicitly, what should we do in this case?

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Hint:

The issue has to do with the support of the joint density function of $U,V$.

Where is it $\lambda^2 \mathsf e^{-\lambda v}$ and where is it $0$?

$$f_{U,V}(u,v) =\begin{cases}\lambda^2 \mathsf e^{-\lambda v} & : v\in \color{gray}{\boxed{\qquad ?}}, u\in \color{gray}{\boxed{\qquad ?}}\\[0.75ex] 0 & : \textsf{otherwise}\end{cases}$$

This will tell you the proper bounds for integration, to obtain the marginal $f_V(v)$ and to verify that the total integral is $1$.


Hint 2:

The joint support for $X,Y$ is: $x\in[0;\infty), y\in[0;\infty)$, so by substituting $x=u, y=v-u$ , then...

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  • $\begingroup$ you mean the limits of the double integral for $u$ are 0 to $+\infty$ for $v$ from $u$ to $+\infty$ $\endgroup$ – Klaus82 Jul 31 '16 at 3:16
  • $\begingroup$ @Klaus82 Yes, though it is more helpful for this problem to set it up so $v$ is the outer integral. $0\leq u \leq v <\infty$ $\endgroup$ – Graham Kemp Jul 31 '16 at 3:22

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