Let $W = aX + bY$, where $(X,Y)$ are jointly normal random variables (bivariate). Also let $E[X] = E[Y] = 0$; so we know $E[W] = 0$, as well. Furthermore, we assume $Var(X) = 4$ and $Var(Y) = 25$, and $Cor(X,Y) = 0.3$ (which implies that $Cov(X,Y) = 3$).
Find values for $a$ and $b$ so that $W$ and $X$ are independent and $Var(W) > 0$.
The part I get stuck on is that $Cor(W,X) = 0$ does not necessarily mean that the variables $W$ and $X$ are independent, just that they do not correlate.
The first equation for my system of equations is $4a + 3b = 0$.
My next step was to define $W$ and $X$ as independent and Let $Y = (1/b)W - (a/b)X = cW + dX$.
This gives an equation for $Var(Y) = 25 = c^2Var(W) + 4d^2$, which resolves to:
$Var(W) = 25b^2 - 4a^2$.
Choosing $a = -3$ and $b = 4$ works just fine for both of these equations, but my equation for $Var(W)$ will always be satisfied (greater than 0) so long as $a$ and $b$ satisfy $4a + 3b = 0$. So it isn't helping to limit my choices of $a$ and $b$ at all!
Please help! I have no idea how to ensure that W and X are independent if $W = aX + bY$ besides making $a = b = 0$, and this gives a $Var(W) = 0$...
you are almost done. You just need the following result, it is also stated here:
But be careful: it is important that the joint distribution is normal. If $X$ has a normal distribution and $Y$ has a normal distribution but the joint distribution is not normal, then the above result does not hold. There is also an example which illustrates this if you follow the link above. So $a=-3$ and $b=4$ is a perfect solution. You just have to argue why the joint normal distribution of $X$ and $Y$ implies the joint normal distribution of $X$ and $W$.