Iterative expectation of continuous and discrete distributions

[cielo]

Homework Statement

Suppose X ~ uniform (0,1) and the conditional distribution of Y given X = x is binomial (n, p=x), i.e. P(Y=y|X=x) = nCy x^{y} (1-x)^{n-y} for y = 0, 1,..., n.

Homework Equations

FInd E(y) and the distribution of Y.

The Attempt at a Solution

f(x) = \frac{1}{b-a} = \frac{1}{1-0} =1E[Y] = E [E[Y|X=x]
= \int E[Y|X=x] f(x) dx where the integral is from o to 1
= \int [\Sigma y f(y|x)] f(x) dx
= \int [\Sigma y nCy x^{y} (1-x)^{n-y}] f(x) dx

...but I do not know anymore what to do next...please help.

 

[statdad]

Homework Statement

Suppose X ~ uniform (0,1) and the conditional distribution of Y given X = x is binomial (n, p=x), i.e. P(Y=y|X=x) = nCy x^{y} (1-x)^{n-y} for y = 0, 1,..., n.

Homework Equations

FInd E(y) and the distribution of Y.

The Attempt at a Solution

f(x) = \frac{1}{b-a} = \frac{1}{1-0} =1


E[Y] = E [E[Y|X=x]
= \int E[Y|X=x] f(x) dx where the integral is from o to 1
= \int [\Sigma y f(y|x)] f(x) dx
= \int [\Sigma y nCy x^{y} (1-x)^{n-y}] f(x) dx

...but I do not know anymore what to do next...please help.

You know the conditional distribution of Y given X. Use that to find E[Y | X]. The answer is a function of X - find its expectation with respect to X to get E[E[Y |X]] = E[Y]

 

[cielo]

Thank you so much for your very good idea. Because of that, I already got the E[Y].

Can you still help me in finding the distribution of Y?

I am confused about this one I made:

P[Y] = \int^{0}_{1} \left[nCy x^{y} (1-x)^{n-y} dx\right]

I understand that is a a beta function if we ignore the constant. But can you help me find the final distribution of Y?