I have a question for class that asks: Let $X$ and $Y$ be i.i.d. Unif$(0,1)$. Are $X+Y$ and $X-Y$ independent?
In an earlier part of the question I found that the covariance between (X+Y) and (X-Y) was 0, but I know this does not necessarily mean independence. My professor advisd that I use their MGFs, because I know that if they are independent then:
$$ M_{(X+Y)+(X-Y)}(t) = M_{X+Y}(t) \cdot M_{X-Y}(t) $$
I think that if I let W = X+Y: $$ M_{W}(t) = M_{w_1}(t) + M_{W_2}(t) = M_{X}(t) + M_Y(t) $$
and if I let U = X-Y: $$ M_{U}(t) = M_{U_1}(t) + M_{U_2}(t) = M_{X}(t) + M_{-Y}(t) $$ (but I'm not really sure how to handle the negative).
and finally, if I let Z = (X+Y)+(X-Y) = X + Y + X - Y = X + X: $$ M_Z(t) = M_{Z_1}(t) + M_{Z_2}(t) = M_{X}(t) + M_{X}(t) $$
I also know that because X and Y are i.i.d. their MGFs should be the same. So it looks like:
$$ M_{(X+Y)+(X-Y)}(t) = 2 \cdot M_X(t) $$
and
$$ M_{X+Y}(t) \cdot M_{X-Y}(t) = 3 \cdot M_{X}(t) + M_{-Y}(t) $$
BUT, if that negative signed in $M_{U}$ worked out differently, so that it was $M_X(t) - M_Y(t)$ when I know the MGFs are the same, then the whole thing would come out differently! Both MGFs would ultimately be $2 \cdot M_X(t)$ and they would be independent. (Based on a passage in my text, I think they're supposed to come out independent). So is this whole approach bad, or is it possible I mishandled the negative?
The MGF of $X+Y$ is $$M_{X+Y}(t)=M_X(t)M_Y(t)=\left(\frac{e^t-1}{t}\right)^2$$ Similarly, the MGF of $X-Y$ is $$M_{X-Y}(t)=M_X(t)M_Y(-t)=\left(\frac{e^t-1}{t}\right)\left(\frac{e^{-t}-1}{-t}\right)=\frac{e^t+e^{-t}-2}{t^2}$$
Now, the $MGF$ of $(X+Y)+(X-Y)$ is equal to the MGF of $2X$, i.e., $$MGF_{(X+Y)+(X-Y)}=M_X(2t)=\frac{e^{2t}-1}{2t}$$ Just putting this together shows that $$M_{(X+Y)+(X-Y)}(t)\neq M_{X+Y}(t)\cdot M_{X-Y}(t)$$ hence $X+Y$ and $X-Y$ are not independent. However, as said in the comments, this is not the fastest way to do it. It suffices to use your intuition (that if you know $X+Y$ then you already know something about $X-Y$ as well) and find an event that violates the independence condition.