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Weighted verage of two variables with minimal varianceby sara_87
Tags: variance 
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#1
Oct2108, 01:51 PM

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 


#2
Oct2108, 03:11 PM

HW Helper
P: 1,361

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. 


#3
Oct2108, 03:18 PM

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 ? 


#4
Oct2108, 03:25 PM

HW Helper
P: 1,361

Weighted verage of two variables with minimal 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. 


#5
Oct2108, 03:31 PM

P: 774

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


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