I'm given this problem. σ²(t) = V((1-t)X + tY) = at² + 2bt+c where

a=? b=? c=?

E(X) = 2

E(Y) = 3

E(X^2) = 53

E(Y^2) = 45

E(XY) = -21

Can someone please explain what σ²(t) = V((1-t)X + tY) means and how to solve it?

Thanks.

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- Feb 21st 2008, 07:47 AMtaypezStatistics Question
I'm given this problem. σ²(t) = V((1-t)X + tY) = at² + 2bt+c where

a=? b=? c=?

E(X) = 2

E(Y) = 3

E(X^2) = 53

E(Y^2) = 45

E(XY) = -21

Can someone please explain what σ²(t) = V((1-t)X + tY) means and how to solve it?

Thanks. - Feb 21st 2008, 08:06 AMheathrowjohnny
I think $\displaystyle V(X) $ is the variance and it is equaled to $\displaystyle E[X^2] - (E[X])^{2} $. This is for one random variable $\displaystyle X $.

But $\displaystyle V(X+Y) = V(X) + 2 \text{Cov}(X,Y) + V(Y) $ for two random variables.

But then there are the $\displaystyle t $ terms. So plug in $\displaystyle t = 0 $ and $\displaystyle t = 1 $ to solve for $\displaystyle a,b,c $. So for $\displaystyle t = 0 $ , $\displaystyle V(X) = c $. For $\displaystyle t = 1 $, $\displaystyle V(Y) = a+2b+V(X) $. - Feb 21st 2008, 08:19 AMtaypez
Alright, does this make any sense...

I calculated:

cov(X,Y) = -27

V(X) = 49

V(Y) = 54

so, for: at^2 + 2bt + c

a = V(x)= 49

b = cov(X,Y) = -27

c = V(Y) = 54

I'm not sure that what he's looking for.

Then how would I find the value of t* that minimizes this? t* = -σy/σx