Question

X1 and X2 are independently distributed random variables with
P(X1=Ɵ+1) = P(X1=Ɵ-1) = 1/2
P(X2=Ɵ-2) = P(X2=Ɵ+2) = 1/2

i)Find the values of a and b which minimize the variance of Y=aX1 + bX2 subject to the condition that E[Y]=Ɵ.
ii)What is the minimum value of this variance?



The answers at the back of the book says that a and b is 4/5 and 1/5 respectively and the variance is 4/5.
I don't know how they got that answer and I'm not even sure where to start...

Answer 1

check this I worked out that E[X1]=E[X2]=θ
and then i got var(X1)=1 and var(X2) = 4 using var(X) = E[X^2] - (E[X])^2

by equating E[Y]=θ=E[aX1+bX2] i get a+b=1

Answer 2

don't know what to do next

Answer 3

Yes, i to the exact same and im stuck on that same point to...

Answer 4

Var(aX+bY)= a^2 Var(X) + b^2 Var(Y) + 2ab Cov(XY)
independently distributed random variables means Cov(XY)=0
so for this problem
Var(Y)= 1*a^2 + 4*b^2
also, as noted above, a+b=1
so a= 1-b. substitute into the equation for Var(Y):
Var(Y)= (1-b)^2 + 4b^2
simplify, take the derivative wrt b, set to zero and solve for b
to find the value of b that minimizes Var(Y)

Answer 5

thanks so much for the help!^-^