anonymous
  • anonymous
pdf and expected value question
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
if we have a pdf: f(x) = 0 when x<0 1 - 0.5x when 0= 2 We are asked to find c given E(X+c) = 4E(X-c)
anonymous
  • anonymous
cdf is: F(x) = 0 when x<0 x - 0.25x^2 when x is between 0 to 2 1 when x > 2
kirbykirby
  • kirbykirby
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

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anonymous
  • anonymous
ahhhhh thank you so much! so since c is just a constant, it can be taken out? it doesn't become: E(X+c)?
kirbykirby
  • kirbykirby
Yeah the exectation of a constant is just the constant itself. E(c)=c

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