0
$\begingroup$

Let $U$ be uniformly distributed on the interval [$\frac{1}{3},1$]. Let $X$ be a random variable such that the conditional distribution of $X$ given $U=p$ is Geometric with parameter $p$.

(a) Find expressions for $E(X|U)$ and $Var(X|U)$

(b) Using your answer for (a) explain why $E(X)=1.65$ and $Var(X)=1.64$

I'm getting confused about the two different distributions in this problem. How do I approach this question?

$\endgroup$
0
$\begingroup$

a) It is given that $X_{|U=p} \sim Geo(p)$, so $E[X|U=p] = \frac{1}{p}$ and $Var[X|U=p] = \frac{1-p}{p^2}$.

b) Using the total expectation and total variance:

$$ E[X] = E[E[X|U]] = E\left[\frac{1}{u} \right] = \int_{1/3}^1\frac{1}{u}\frac{3}{2}du = \frac{3}{2}(0 + \ln(3)) \approx 1.65 $$ For the variance of $X$ use the total variance formula, i.e., \begin{align} Var(X) &= E(Var(X|U)) + Var(E(X|U))\\ &= E\left[\frac{1-u}{u^2}\right] + Var\left[\frac{1}{u}\right]\\ & =\int_{1/3}^1\frac{1-u}{u^2}\frac{3}{2}du + \int_{1/3}^1\frac{1}{u^2}\frac{3}{2}du - 1.65^2 \end{align}

$\endgroup$
  • $\begingroup$ Is $E_U$ the same as $E[U]$? $\endgroup$ – Lindsey G Feb 3 '16 at 17:52
  • 1
    $\begingroup$ It was to emphasize that the Expectation or the Variance were taken w.r.t. the distribution of $U$. Anyway, I deleted this notation. It is not necessarily and clear from the context. $\endgroup$ – V. Vancak Feb 3 '16 at 18:10

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.