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Suppose $U$ and $V$ are jointly distributed continuous r.v's with $U \sim Uni(1,3)$ and $V$ given $U = u$ follows an exponential distribution with mean $\frac{1}{u}$. Calculate $Var(V \vert U)$.

Attempt:

Since $U \sim Uni(1,3)$, $f_U(u) = \frac{1}{3-1} = \frac{1}{2} ,\forall u \in (1,3)$. Also $f_{V\vert U}(v \vert u) = ue^{-uv}$ (not sure with this). Then

$$E(V \vert U = u) = \int_{}^{}v f_{V\vert U}(v\vert u)dv$$ $$Var(V\vert U) = \int_{-\infty}^{\infty} (v - E(V \vert U = u))^2f_{V \vert U}(v \vert u) dv$$

Witht that, how do we determine the bounds of integration wrt $v$?

860009898987
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  • That looks too complicated. I would start with $\operatorname{Var}(V \mid U=u)$ which, since this is an exponential distribution $\operatorname{Exp}(u)$, is $\frac1{u^2}$. That makes $\operatorname{Var}(V \mid U) = \frac1{U^2}$ – Henry Dec 06 '21 at 14:36
  • I edited the question since i copied a wrong phrase – 860009898987 Dec 06 '21 at 14:42
  • Changing from rate parameter $u$ to mean parameter $\frac1u$ for the exponential distribution does not affect my comment or tommik's answer – Henry Dec 06 '21 at 14:49

1 Answers1

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sorry but your question is immediate...and you do not need to know how $U$ is distributed. Are you sure about the question?

You are given that

$$(V|U=u)\sim \text{Exp}(u)$$

thus

$$\mathbb{V}[V|U=u]=\frac{1}{u^2}$$

Sure that you are not asked to find $\mathbb{V}[V]$?

tommik
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