Uniform distribution

evinda

Junior Member
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Apr 13, 2013
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57
Let X1,X2,X3 be independent random values with uniform distribution at [0,1]. Calculate E[(X1-2*X2+X3)^2].
I show you what I've done, could you tell me if it's right???
-> E[(X1-2*X2+X3)^2]=E((X1)^2-4*X1*X2+2*X1*X3+4*(X2)^2-4*X2*X3+(X3)^2)=E((X1)^2)+4*E((X2)^2)+E((X3)^2)-4*E(X1)*E(X2)+2*E(X1)*E(X3)-4*E(X2)*E(X3), where E(Xi)=(0+1)/2 and E((Xi)^2)=integral((xi)^2)dxi

Thanks in advance!!!
;)
 
Let X1,X2,X3 be independent random values with uniform distribution at [0,1].
Calculate E[(X1 - 2*X2 + X3)^2].
I show you what I've done, could you tell me if it's right???
-> E[(X1 - 2*X2 + X3)^2] = E{(X1)^2 - 4*X1*X2 + 2*X1*X3 + 4*(X2)^2 - 4*X2*X3 + (X3)^2}
...= E((X1)^2)+4*E((X2)^2)+E((X3)^2) - 4*E(X1)*E(X2)+2*E(X1)*E(X3)-4*E(X2)*E(X3),
where E[Xi] = (0+1)/2
and E[(Xi)^2)] = integral((xi)^2)dxi
= (1/3)*[(1^3 - 0^3]

Thanks in advance!!!
;)
E[X1] = E[X2] = E[X3] = 1/2
E[X1^2] = E[X2^2] = E[X3^2] = 1/3
E[X1X2] = E[X1X3] = E[X2X3] = 1/4

I think you have all the correct terms - go ahead and put in the numbers.
 
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