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Rup
December 20th 2005, 09:34 PM
Can anybody try to solve and explain this problem?

The CDF of X is

F(x) = 0 x<0
x/3 0<=x<2
1 x>=2

Calculate E(X) and E(X^2).

Ans. 4/3, and 20/9

Thanks
Rup

SirVLCIV
December 20th 2005, 10:29 PM
There's a jump discontinuity at F(2), so the P(2) is 1/3.

It's a mixed distribution, so just add the moments of the partial distributions.


f(x)=F'(x)=1/3
E(x)=int(x*1/3), 0<x<2, x^2/6 0<x<2=2/3

2/3+(1/3)*2=4/3

E(x^2)=int(x^2*1/3)=x^3/9 0<x<2=8/9

8/9+(1/3)*(2^2=4)=8/9+(4/3=12/9)=20/9.

Rup
December 21st 2005, 10:20 PM
Thanks for your help SIRVLC!