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    Problem regarding expectation, cdf




    There are two random variables X1, X2 that are iid and continuous. Their cdf is F(x)= P(Xi<x).

    Show P(X1<X2) = E(F(X1))
    Show that E(F(X1)) = 1/2. Can you find another way?

    That E(F(X1)) is giving me a hard time. Please help!

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    Re: Problem regarding expectation, cdf

    E[F(X_1)]

    = \int_{-\infty}^{+\infty} F(x_2)f(x_2)dx_2

    = \int_{-\infty}^{+\infty} \int_{-\infty}^{x_2} f(x_1)dx_1 f(x_2)dx_2

    = \int_{-\infty}^{+\infty} \int_{-\infty}^{x_2} f(x_1) f(x_2)dx_1dx_2

    = \int_{-\infty}^{+\infty} \int_{-\infty}^{x_2} f_{X_1, X_2}(x_1, x_2)dx_1dx_2

    = \Pr\{X_1 < X_2\}

    For the next part, I guess you can either use a symmetric argument, or consider the standard uniform random variable.

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    Re: Problem regarding expectation, cdf

    Quote Originally Posted by BGM View Post
    E[F(X_1)]

    = \int_{-\infty}^{+\infty} F(x_2)f(x_2)dx_2

    = \int_{-\infty}^{+\infty} \int_{-\infty}^{x_2} f(x_1)dx_1 f(x_2)dx_2

    = \int_{-\infty}^{+\infty} \int_{-\infty}^{x_2} f(x_1) f(x_2)dx_1dx_2

    = \int_{-\infty}^{+\infty} \int_{-\infty}^{x_2} f_{X_1, X_2}(x_1, x_2)dx_1dx_2

    = \Pr\{X_1 < X_2\}

    For the next part, I guess you can either use a symmetric argument, or consider the standard uniform random variable.
    Thank you so much! But if you don't mind me asking about the process that you took, how did you go from E[F(X_1)] to \int_{-\infty}^{+\infty} F(x_2)f(x_2)dx_2?
    I know the formula for the expectation of a continuous variable, but what I don't understand is how you were able to put x2 instead of x1.

    Are x1 and x2 interchangeable when X1 and X2 are iid random variables?

    Similar question about how you got \int_{-\infty}^{x_2} f(x_1)dx_1 from \int_{-\infty}^{+\infty} F(x_2)dx_2.

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    Re: Problem regarding expectation, cdf


    The first step is hold for any measurable function F, so it will hold for CDF as well

    The choice of the dummy variable is arbitrary; you can choose u, v, x_3, z as you like. The choice I made is simply for better illustration purpose in the latter step.

    Please note that upper case letter here is reserved for random variable, while the lower case is not.

    The last question follows directly from the relationship between CDF and pdf.

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