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Thread: Expected value and variance of profit

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    Expected value and variance of profit




    Hello all, I have this question, which I think I partially knows how to solve, but need some completion.

    "A man is playing versus a machine in the following way: The machine chooses 2 numbers randomly from the set of numbers 1,2,3,4,5, where a number can be chosen twice (with replacement). If the multiplication of the 2 chosen numbers is even, the man gets 5 dollars Calculate the expected value (mean) and variance of the profit after 100 games, if for every game he pays 2 dollars to play."

    What I did to start with, is to calculate the probability of having an even multiplication and I got p=16/25. Now I know I can calculate the profit for a single game, get a probability function with 2 values, and find E(X), and multiply it by 100. I am not sure if I can do the same with the variance, and I think I might be able to do it with Binomial distribution, but not sure how. Maybe a linear transformation is also possible?

    Thank you

    Addition:

    to be more specific, if X is the profit in 1 game, then it can get the values 3 and -2 only. The matching probabilities are 16/25 and 9/25 respectively. Therefore it's easy to find E(X) and V(X). However I need the expected value and variance after 100 games. I know that for E it's 100*E(X), what about V ?
    Last edited by Yankel; 01-06-2015 at 02:56 AM.

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    Re: Expected value and variance of profit


    So you know the distribution of the profit in a single game.

    Let X_i be the profit of the i-th game, i = 1, 2, \ldots, 100. Then the total profit of n games will just be the sum

    \sum_{i=1}^n X_i

    Therefore you get

    E\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n E[X_i] = n E[X_1]

    The last equality holds as you are assuming they have identical distribution. You should also know a similar identity for the variance, assuming they are independent.

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