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Math Help - Need help with another proof.

  1. #1
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    Need help with another proof.

    I've got a problem that I don't know how to solve. The question: Suppose that Y is a continuous random variable with density f(y) that is positive only if y> or equal to zero. If F(y) is the distribution function, show that

    E(Y) = \int_{0}^{infty}{} yf(y) dy = \int_{0}^{infty}{} [1-F(y)] dy.

    The book also gives a hint: If y > 0, y=\int_{0}^{y}{} dt and  E(Y) = \int_{0}^{+\infty} yf(y) dy = \int_{0}^{+\infty} [\int_{0}^{y} dt] f(y) dy

    Exchange the order of integration to obtain the desired result.

    I've tried to change the integration order like this:

    E(Y) = \int_{0}^{+\infty} yf(y) dy = - \int_{+\infty}^{0} [\int_{0}^{y} dt] f(y) dy

    but how do I go on from there?

    Thanks in advance.
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  2. #2
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    In order to change the order of integration, you must remove the y from the bound. Here is how:
    E(Y) = \int_{0}^{+\infty} yf(y) dy = \int_{0}^{+\infty} \left(\int_{0}^{y} dt\right) f(y) dy
    .... = \int_0^\infty \int_0^\infty {\bf 1}_{(t\leq y)}dt f(y) dy = \int_0^\infty\left(\int_0^\infty {\bf 1}_{(t\leq y)}f(y) dy\right)dt
    .... = \int_0^\infty \left(\int_t^{+\infty} f(y) dy\right)dt =\int_0^\infty (1-F(t))dt.
    (I denote by {\bf 1}_{(t\leq y)} an indicator function: it is equal to 1 if t\leq y and to 0 otherwise)
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  3. #3
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    Thank you, Laurent. I'm not sure that I understand the use of an indicator function. Can you please try to explain how I use it to go from:

    \int_0^\infty\left(\int_0^\infty {\bf 1}_{(t\leq y)}f(y) dy\right)dt


    to:

    \int_0^\infty \left(\int_t^{+\infty} f(y) dy\right)dt

    ?
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  4. #4
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    Quote Originally Posted by approx View Post
    Thank you, Laurent. I'm not sure that I understand the use of an indicator function. Can you please try to explain how I use it to go from:

    \int_0^\infty\left(\int_0^\infty {\bf 1}_{(t\leq y)}f(y) dy\right)dt


    to:

    \int_0^\infty \left(\int_t^{+\infty} f(y) dy\right)dt

    ?
    Consider (for some fixed t>0) \int_0^\infty {\bf 1}_{(t\leq y)}f(y) dy. The function that is being integrated is equal to f(y) when y\geq t (because the indicator equals 1), and it is equal to 0 when y<t (because the indicator equals 0). So it simplifies into: \int_t^{+\infty} f(y) dy (we integrate only on the values of y where the integrand is not zero, that is for y greater than t). Is it clear? I use this operation twice: once to remove the y from the integration sign, and once to put back a t.
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  5. #5
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    There is also a "visual" way to understand this exchange in the order of integration. Draw the set of all (t,y) such that 0\leq t\leq y (with the y-axis as the horizontal axis): this is the part of the plane delimited by the positive y-axis and the half-line t=y, t\geq 0. When you compute \int_0^\infty\int_0^y f(y) dt\,dy, you integrate the function f(y) along the vertical slices corresponding to fixed y's, and then you "sum" the slices together. These slices are bounded: from 0 to y. And, by Fubini Theorem, the value of the integral does not change if you sum the horizontal slices instead: for each t, the slice goes from t to +\infty. So that the integral becomes: \int_0^\infty\int_t^\infty f(y) dy\,dt. That's clearer when drawn on a paper actually, but I hope you got it anyway.
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  6. #6
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    I think I understand now. Thank you for that explanation.
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