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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 varience2011-11-10
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    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.2011-11-10
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    (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?2011-11-10
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    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.2011-11-11

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$X-M$ 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 $N(0,\sigma^2/2)$.

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    Amazing simple. Thanks alot, can't believe I missed that!2011-11-11