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nuk64

  • one year ago

pdf and expected value question

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  1. nuk64
    • one year ago
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    if we have a pdf: f(x) = 0 when x<0 1 - 0.5x when 0=<x =<2 0 for x > 2 We are asked to find c given E(X+c) = 4E(X-c)

  2. nuk64
    • one year ago
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    cdf is: F(x) = 0 when x<0 x - 0.25x^2 when x is between 0 to 2 1 when x > 2

  3. kirbykirby
    • one year ago
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    E(X+c)=E(X)+c, 4E(X-c) = 4E(X)-4c E(X)+c=4E(X)-4c => 3E(X) - 5c E(X) = 5c/3 Since E(X) = 5c/3, we know that E(X) = integral (x*pdf x) from 0 to 2 and you know that the integral of (pdf) from 0 to 2 =1 by the property of a pdf

  4. nuk64
    • one year ago
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    ahhhhh thank you so much! so since c is just a constant, it can be taken out? it doesn't become: E(X+c)?

  5. kirbykirby
    • one year ago
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    Yeah the exectation of a constant is just the constant itself. E(c)=c

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