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anonymous
 3 years ago
pdf and expected value question
anonymous
 3 years ago
pdf and expected value question

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if 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(Xc)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0cdf is: F(x) = 0 when x<0 x  0.25x^2 when x is between 0 to 2 1 when x > 2

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1E(X+c)=E(X)+c, 4E(Xc) = 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ahhhhh thank you so much! so since c is just a constant, it can be taken out? it doesn't become: E(X+c)?

kirbykirby
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah the exectation of a constant is just the constant itself. E(c)=c
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