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P(X>x)=e^-ax, x less/equal 0

  1. Nov 5, 2009 #1
    Let X be a continuous random variable with

    P(X>x)=e^-ax, x less/equal 0

    Where a is a positive constant. Find EX and Var(x)
     
  2. jcsd
  3. Nov 5, 2009 #2

    HallsofIvy

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    If P(X> x)= e^{-ax}, x less than or equal to 0, then the probability density function is p(x)= -a e^{-ax}.

    Now just use the definitions:
    [tex]E(x)= -a \int_x^0 xe^{-ax}dx[/itex]
    and
    [tex]Var(x)= -a \int_x^0 (x- E(x))^2e^{-ax}dx= -aE(x)\int_x^0 x^2e^{-ax}- E^2(x)[/tex]
     
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