# expectation of X only given conditional pdf

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 $$E(x)$$?

• Welcome to Math.SE. Please use Mathjax for better clarity – 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

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