# Variance

by sara_87
Tags: variance
 P: 774 1. The problem statement, all variables and given/known data 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) 2. Relevant equations 3. 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
 HW Helper P: 1,334 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.
 P: 774 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 ?
HW Helper
P: 1,334

## Variance

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.
 P: 774 Oh ofcourse...it's minus...silly me. What u said is SO TRUE. Thanks v much.

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