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 $E(x)$?

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

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

  • $E[X]=\dfrac43$
  • $E[Y]=\dfrac23$
  • $\begingroup$ 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

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