0
$\begingroup$

Suppose that $X$ is a discrete random variable and $Y$ is a continuous random variable. The conditional pdf of $X$ given $Y$ is $g_1(x|y)=\frac{(2y)^x}{x!} exp(-2y)$, $x=0,1,2,...$ (poisson dist. with $λ=2y$)

The conditional PDF of $Y$ given $X$ is $g_2(y|x)=\frac{5^{x+2}}{(x+1)!} y^{x+1}exp(-5y)$, y>0 (gamma dist. with $k=x+2$, $θ=\frac{1}{5}$

Q. Find the expected value of $x$ ($E(x)$)

I tried to use of "double expectation" concept to solve, but I realized that I still needed PDF of $x$.

How can I find $E(x)$?

$\endgroup$
  • 1
    $\begingroup$ Welcome to Math.SE. Please use Mathjax for better clarity $\endgroup$ – Shailesh Jul 29 '16 at 7:38
1
$\begingroup$

You know the mean of a Poisson distribution so

  • $E[X \mid Y=y]=\lambda=2y$

    • $E[X \mid Y]=2Y$
    • $E[X]=E[E[X\mid Y]]=2E[Y]$

Similarly you know the mean of a Gamma distribution so

  • $E[Y \mid X=x]=k\theta =\dfrac{x+2}{5}$
    • $E[Y \mid X]=\dfrac{X+2}{5}$
    • $E[Y]=E[E[Y\mid X]]=\dfrac{E[X]+2}{5} $

Solving these simultaneous equations gives

  • $E[X]=\dfrac43$
  • $E[Y]=\dfrac23$
$\endgroup$
  • $\begingroup$ oh i was almost on the end. $\endgroup$ – StatisticBang Jul 29 '16 at 8:20
  • $\begingroup$ Its really helpful! Thank you very much! $\endgroup$ – StatisticBang Jul 29 '16 at 8:20

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.