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.
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.
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) $.
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