Question

pdf and expected value question

Answer 1

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)

Answer 2

cdf is:
F(x) = 0 when x<0
x - 0.25x^2 when x is between 0 to 2
1 when x > 2

Answer 3

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

Answer 4

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

Answer 5

Yeah the exectation of a constant is just the constant itself. E(c)=c