I have a problem that I am trying to solve. I think my answer is correct, but I am not 100% sure. Any help you could provide would be greatly appreciated.

My Question:
Let's start flipping a fair coin. When the coin lands on heads, I will pay you $10. When the coin lands on tales, I will pay you$20. However, you do not know how many times I will flip the coin. Write an expression for the expected value of the total pay out for this coin toss game.

My intuition is that the expected value will be the following:

However, I am not sure how to prove that this is correct. I've attached my first stab (N is the total number of tosses, Nh is the number of tosses that came up heads, and Nt is the number of tosses that came up tales). Am I on the right tracks?

It seems correct.

I guess the reason I am skeptical is that I can't really defend step 3. When I take the expectation of the expression in Step 2, I am left with the ratio of two expected values (e.g. E(Nh)/E(N)). For clarity, I've pasted the expression below. If N were a constant, then I could confidently show that this ratio reduced to 1/2 (because E(Nh)=1/2*N and E(N)=N). Since N is a random variable, I am not sure if this ratio actually reduces to 1/2 or not. Again, any help would be appreciated!

It is true that the ratio of two expected values doesn't have to be equal to the expected value of the ratio of the same two variables. Since and are dependent you can't in general say that . Instead you should evaluate .

hmmm you're right. So how do I show ?

PS* Thanks so much for helping me out Englund.

hi,
I think the 3rd step is indeed wrong, because you are assuming E(YX)=E(Y)E(X)

Maybe you can directly express the payoff as a function of n?
regards

rogojel, what's wrong with that assumption is this case?

I think you have a formula like V=f(n)*n and step 3 is E(V)=E(f(n)*n)=E(f(n))*E(n) which is wrong IMO

My math skills are quite rusty but my guess for the payoff would be:

V=Sum(Combinations(N,k)*(10*k+(N-k)*20) )/2^N where the summation is over k from 0 to N.

regards

Originally Posted by NCSUStudent
hmmm you're right. So how do I show ?
I'm not sure. It's probably easy to show that, but the only way I can think of right now is to treat N as a constant.

Honestly it would probably just be easiest to not muck around with multiplying by N/N and instead just take the expectation using the law of iterated expectations.

E[X] = E[E[X|Y]] or in this case E[Nh] = E[E[Nh|N]] = E[N/2] = E[N]/2 which you can use easily to get what you want.

hi,
I think I figured it out - the expected payoff if one plays N rounds is simply 15N If N is random, it is now easy to work out the expected value.

Originally Posted by rogojel
hi,
I think I figured it out - the expected payoff if one plays N rounds is simply 15N
Yes of course. But I think the problem has been to prove that

Originally Posted by Dason
E[Nh] = E[E[Nh|N]] = E[N/2] = E[N]/2
Nice and simple, just as I like it. But out of curiosity, if we'd take the expectation of the equation in step two, how do we prove that E[Nh/N] equals one half when N also is a random variable?

Well Nh and N aren't independent. So you could do a similar thing

E[Nh/N] = E[E[Nh/N | N]] = E[ ((1/2)*N)/N] = 1/2

But getting to the point where you take that expectation would be less than trivial since in general you can't just say that E[XY] = E[X]E[Y] without showing independence.

Usually for these kinds of question, you need to specify the model "correctly" so that you can facilitate the calculation.

Although the number of flips is random, you still have the following:

(these kind of set up should be known as "compound distribution"). And then using double expectation as instructed by Dason, you can see the answer quickly.