1. ## Double Expectation Theorems

A man selects a ball from an urn containing n balls numbered from 1 to n. The number he selects is the number of true coins he tosses to determine his prize. The variance of the number of heads he obtains is 55/16. If his prize is $1000 times the number of heads he obtains, determine his expected prize. I'm having a hard time with this problem. I think you're supposed to use the double expectation theorems, i.e. Var(Y) = Var[E(Y|X)] + E[Var(Y|X)] and I think X~Unif[1,n] and Y~Bin(n=x,p=1/2) , but I'm not sure where to go from here. Thanks for any help! 2. Hello, Originally Posted by tbl9301 A man selects a ball from an urn containing n balls numbered from 1 to n. The number he selects is the number of true coins he tosses to determine his prize. The variance of the number of heads he obtains is 55/16. If his prize is$1000 times the number of heads he obtains, determine his expected prize.

I'm having a hard time with this problem. I think you're supposed to use the double expectation theorems, i.e. Var(Y) = Var[E(Y|X)] + E[Var(Y|X)] and I think X~Unif[1,n] and Y~Bin(n=x,p=1/2) , but I'm not sure where to go from here. Thanks for any help!
Note that we're looking for 1000E[Y]

The red part should actually be X : the conditional law of Y given that X=x is a binomial (x,1/2)
So Y|X follows a binomial(X,1/2)

Hence E[Y|X]=X/2 and Var[Y|X]=X/4.

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For the solution, we'll need E[X] and Var[X].

So see here : Uniform distribution (discrete) - Wikipedia, the free encyclopedia
And we get E[X]=(n+1)/2 and Var[X]=(n²-1)/12

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Using the double expectation theorem, we get E[Y]=E[E[Y|X]]=E[X/2]=(n+1)/4

But you have to find n, and that's where the formula of the double expectation that you wrote and the 55/16 parts intervene.

So $\tfrac{55}{16}=Var\left[\tfrac X2\right]+E\left[\tfrac X4\right]=\tfrac 14\cdot \left(Var[X]+E[X]\right)=\frac{n^2+6n+5}{48}$

Hence $n^2+6n+5=(n-1)(n-5)=55\times 3=15\times 11$
Which gives a unique positive integer solution : n=16.

And the result follows...