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Thread: An odd payout question...Please Help!

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    An odd payout question...Please Help!




    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 Answer:
    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?
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    Last edited by NCSUStudent; 11-03-2015 at 07:31 AM.

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    Re: An odd payout question...Please Help!

    It seems correct.

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    Re: An odd payout question...Please Help!

    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!


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    Re: An odd payout question...Please Help!

    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 N_H and N are dependent you can't in general say that E[N_H/N] = E[N_H]/E[N]. Instead you should evaluate E[N_H/N].

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    Re: An odd payout question...Please Help!

    hmmm you're right. So how do I show E[N_H/N]=1/2?

    PS* Thanks so much for helping me out Englund.

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    Re: An odd payout question...Please Help!

    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

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    Re: An odd payout question...Please Help!

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

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    Re: An odd payout question...Please Help!

    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

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    Re: An odd payout question...Please Help!

    Quote Originally Posted by NCSUStudent View Post
    hmmm you're right. So how do I show E[N_H/N]=1/2?
    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.

    E[N_H/N]=1/N*E[N_H]=1/N*E[\sum_{i=1}^N{x_{i_H}}]=N/N*E[X] =\sum_{j=0}^1{P(H)j} = 1/2

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    Re: An odd payout question...Please Help!

    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.
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    Re: An odd payout question...Please Help!

    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.

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    Re: An odd payout question...Please Help!

    Quote Originally Posted by rogojel View Post
    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

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    Re: An odd payout question...Please Help!

    Quote Originally Posted by Dason View Post
    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?

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    Re: An odd payout question...Please Help!

    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.
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    Re: An odd payout question...Please Help!


    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 N is random, you still have the following:

    N_H|N \sim \text{Binomial}\left(N, \frac {1} {2}\right)

    (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.

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