# Sample Mean & Variance

Let $X, Y$ be IID $\sim N(\mu, \sigma^2)$.

$$M = \frac12(X + Y),\qquad V = (X - M)^2 + (Y - M)^2$$

Consider the joint moment generating function of $(M, X - M, Y - M)$, show that $M$ and $V$ are independent.

We haven't learn about stuff like Cochran's Theorm or multivariable normal distribution.

I'm confused in that how do you find the MGF of something whos pmf is not give? Or should I be able to work out the pmf? Any help appreciated, thanks!

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Must be getting late, sorry yes there shouldn't be a square - M is the sample mean & V is the sample varience –  Rcwt Nov 10 '11 at 22:45
Here is a sketch of the proof: (1) show that X+Y and X-Y are independent, (2) show that M is a function of X+Y and V a function of X-Y, (3) conclude. –  Did Nov 10 '11 at 22:49
(2) & (3) are trivial. For (1), I managed to do it using f(x,y) = f(x)f(y) since X, Y independent. Then did a transformation to g(x+y,x-y) to show that g(x+y,x-y)=g(x+y)g(x-y) thus showing (X+Y) and (X-Y) independent. But is there a way to use MGF since the question says so? –  Rcwt Nov 10 '11 at 23:16
I have no idea of what you mean by using f(x,y) = f(x)f(y) or did a transformation to g(x+y,x-y) to show that g(x+y,x-y)=g(x+y)g(x-y). Which f, which g, which transformation? You seem to rely on some (unstated) characterization of independence which eludes me. –  Did Nov 11 '11 at 7:10

$X-M$ is the same thing as $(X-Y)/2$ and $Y-M$ is the same as $(Y-X)/2$.
The requested mgf is: \begin{align} & {}\qquad M_{(M,X-M,Y-M)}(s,t,u) = \mathbb{E}\left( e^{sM+t(X-M)+u(Y-M)} \right) = \mathbb{E}\left( e^{s(X+Y)+t(X-Y)/2 + u(Y-X)/2} \right) \\ \\ \\ & = \mathbb{E}\left(e^{(s+t/2-u/2)X+(s-t/2+u/2)Y} \right) =M_{(X,Y)}\left(s+\frac {t-u}{2},s-\frac {t-u}{2}\right) \\ \\ \\ & = M_X\left(s+\frac {t-u}{2}\right) M_Y\left(s-\frac {t-u}{2}\right) = M_X\left(s+\frac {t-u}{2}\right) M_X\left(s-\frac {t-u}{2}\right) \\ \\ \\ & = \exp\left( \left(s+\frac{t-u}{2}\right)\mu + \left(s+\frac{t-u}{2}\right)^2\frac{\sigma^2}{2} \right)\cdot \exp\left( \left(s-\frac{t-u}{2}\right)\mu + \left(s+\frac{t-u}{2}\right)^2\frac{\sigma^2}{2} \right) \\ \\ & = \exp\left( s(2\mu) + s^2\frac{2\sigma^2}{2} \right)\cdot\exp \left( (t-u)^2\frac{\sigma^2/2}{2} \right) = M_{X+Y}(s) M_{N(0,\sigma^2/2)} (t-u) \end{align} Conclude that the thing that originally got multiplied by $s$ is independent of the thing that originally got multiplied by $t-u$. The former is $X+Y$. The latter is $(X-Y)/2$. You can also conclude that the former is distributed as $N(2\mu,2\sigma^2)$ and the latter as $N(0,\sigma^2/2)$.