# Statistics Question

• Feb 21st 2008, 07:47 AM
taypez
Statistics 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 AM
heathrowjohnny
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 AM
taypez
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