Weighted verage of two variables with minimal variance

[sara_87]

Homework Statement

X1 and X2 are independent random variables. They both have the same mean (mue). Their variances are s1^2 and s2^2 respectively, where s1^2 and s2^2 are known constants. It is proposed to estimate mue by an estimator T of the form T=c1X1 + c2X2.
Show that T will be unbiased if c1 + c2=1
and find an expression for var(T) in terms of c1, s1^2 and s2^2.
(assuming c1+c2=1)

Homework Equations

The Attempt at a Solution

I showed that T will be unbiased if c1+c2=1
For the next part this is what i did:

var(T) = var(c1X1+c2X2)
var(c1X1+c2X2) = E[(c1X1+c2X2)^2] + {E[c1X1+c2X2]}^2

and then after expanding and simplifying, i got:
var(T) = 2(mue)^2(c1^2 + 2c1c2 + c2^2)

I can easily change c2 in terms of c1 but how do put in terms of s1^2 and s2^2 as this is what they are asking for??

Thank you

 

[statdad]

If X, Y are independent random variables, and a, b are real numbers, then

Var(aX + bY) = a^2 Var(X) + b^2 Var(Y)

Apply this to the setting of your problem.

Note that, relating to your work,

Var(W)

does not equal

E(W^2) + (E(W))^2

so your formula would not get you to the desired result.

 

[sara_87]

Thank you v much.
I should have known that Var(aX + bY) = a^2 Var(X) + b^2 Var(Y) !

But how come for this question
Var(W)

does not equal

E(W^2) + (E(W))^2

?

 

[statdad]

Var(W) = E((W - mu_w)^2) = E(W^2 - 2Wmu_w + (mu_w)^2) = E(W^2) - 2(mu_w)^2 + (mu_w)^2 = E(W^2) - (mu_w)^2

for any random variable W. I believe you just missed a sign.

Sometimes, after staring at a problem for some time, our minds see what we want them too rather than what we've actually written - it happens to me a lot.

 

[sara_87]

Oh ofcourse...it's minus...silly me.

What u said is SO TRUE.
Thanks v much.