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EX proof

  1. Sep 30, 2005 #1
    I have a problem understanding the following proof:
    EX = integral from 0 to infinity of 1 - F(x) dx
    Say this integral can be:
    EX = integral from 0 to infinity of P(X > x), then
    EX = integral from 0 to infinity of integral from x to infinity of f(y)dydx
    = integral from 0 to infinity of integral from 0 to y of dxf(y)dy = EX

    Can somebody explain me how and why the last integral is included?
    Thanks.
     
  2. jcsd
  3. Oct 1, 2005 #2

    EnumaElish

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    [tex]\int_0^\infty \left(\int_x^\infty f(y)dy\right) dx \overset{?}{=} \int_0^\infty \left(\int_0^y dx\right) f(y)dy[/tex]

    Think of this as integrating the area of the unit square. You could first integrate along the y axis then integrate that integral along the x axis. Alternatively you could first integrate over x then integrate that over y.

    Or suppose you are to integrate function f over the area that lies below the y = x line on the unit square. You can either integrate f from 0 to x on the y axis then integrate that from 0 to 1 on the x axis. Alternatively you could integrate f from 0 to y on the x axis then integrate that from 0 to 1 on the y axis.

    This isn't an exact or perfect explanation but I hope that it will be useful to some degree.
     
  4. Aug 15, 2008 #3

    statdad

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    The integral

    [tex]
    \int_0^\infty \left(1-F(x)\right) \, dx
    [/tex]

    can be rewritten with this step.

    [tex]
    \int_0^\infty \left(1-F(x)\right) \, dx = \int_0^\infty \left(\int_x^\infty f(y)\dy\right) \,dx
    [/tex]

    The region over which we are integrating is the portion of the first quadrant that is on and to the right of line [tex] y = x [/tex]

    This can also be calculated by reversing the order of integration. In this case the inner integral goes from [tex] x=0[/tex] to [tex] x = y [/tex]. Here is the work all in a single location.

    [tex]
    \begin{align*}
    \int_0^\infty \left(1-F(x)\right) \, dx & = \int_0^\infty \left(\int_x^\infty f(y) \,dy\right) \,dx\\
    & = \int_0^\infty \left(\int_0^y \, dy \right) f(x) \,dx = \int_0^\infty y f(y) \,dy
    \end{align*}
    [/tex]

    If you remember that the variable names are simply placeholders in this work you realize that the final integral is simply the expected value of [tex] X [/tex].
     
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