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Thread: Difference of expectations of RV's given arbtrary pdfs

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    Difference of expectations of RV's given arbtrary pdfs




    I've been beating my head on this for a while...

    Let X and Y have pdf's f & g respectively such that \begin{cases}f(x) >= g(x) & \mbox{if } x \leq a \\f(x) <= g(x) & \mbox{if } x > a\\\end{cases}. Show that E[X] \leq E[Y].

    As the expectation is a probability weighted average, it stands to reason that if there is more "weight" given to larger values (i.e. larger values are more likely to occur), then the expectation is larger. I just don't know how to show mathematically. So far, I've tried looking at the difference in the pdfs:
    E[Y]-E[X] = \int_{-\infty}^{\infty} t g(t)dt - \int_{-\infty}^{\infty} t f(t)dt = \int_{-\infty}^{\infty} t \cdot [g(t)-f(t)]dt = \int_{-\infty}^{\infty} t \cdot h(t)dt. I can show that h(t) integrates to zero (as the difference of the integrals of two pdf's) so \int_{-\infty}^a h(t)dt=-\int_a^{\infty} h(t)dt, but then I don't know where to go from there.

    Any thoughts are appreciated!

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    Re: Difference of expectations of RV's given arbtrary pdfs


    There maybe a simpler and more direct method, but you may need to make use of the following formula:

    E[X] = -\int_{-\infty}^0 F(x)dx + \int_0^{+\infty} [1 - F(x)]dx

    Then you may obtain the following:

    E[Y] - E[X] = \int_{-\infty}^{+\infty} [F(t) - G(t)]dt

    where F, G are the CDF of X, Y respectively.

    Next you may show that the integrand tends to zero when it goes to infinity (property of CDF) and has a unique global and local maximum at a by usual calculus check with the given property. Thus showing that the integrand, and therefore the integral is non-negative which completes the proof.

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