Proving that the squares of normal rv's is Chi-square distributed

Question

I start with three independent random variables, $X_1, X_2, X_3$. They are each normally distributed with:

$$X_i \sim N(\mu_i, \sigma^2), i = 1, 2, 3.$$

I then have three transformations,

$$\eqalign{ Y_1 &= -X_1/\sqrt{2} + X_2/\sqrt{2} \cr Y_2 &= -X_1/\sqrt{3} - X_2/\sqrt{3} + X_3/\sqrt{3} \cr Y_3 &= X_1/\sqrt{6} + X_2/\sqrt{6} + 2X_3 / \sqrt{6} \cr }$$

I am supposed to show that when $\mu_i = 0,$ $i = 1, 2, 3,$ $(Y_1^2 + Y_2^2 + Y_3^2)/\sigma^2 \sim \chi^2(3)$. I have also shown the transformations to preserve the independence, as the transformation matrix is orthogonal.

I have already shown that the expectations of $Y_1, Y_2, Y_3$ is 0 and their variances are all the same. Using the normal pdf, I have shown that:

$$Y_i^2 \sim \frac{1}{2\pi\sigma^2} \exp(-2x^2 / 2\sigma^2).$$

I thought about applying a substitution of $z = 2x^2 / \sigma^2$ to get the exponent into a similar form as the chi-square's $\exp(-x/2)$ form, but I'm stuck on what to do with the constants outside to get them to look similar. Could someone offer a hand?

Answer 1

We have $X_1\sim N(\mu_1,\sigma^2)$ and $X_2\sim N(\mu_2,\sigma^2)$, hence

$$EY_1=E(-X_1/\sqrt{2}+X_2/\sqrt{2})=-1/\sqrt{2}EX_1+1/\sqrt{2}EX_2=0$$

\begin{align*} EY_1^2&=E(-X_1/\sqrt{2}+X_2/\sqrt{2})^2\\\\ &=E(X_1/\sqrt{2})^2-2E(X_1X_2/2)+E(X_2/\sqrt{2})^2\\\\ &=1/2\sigma^2+1/2\sigma^2=\sigma^2 \end{align*}

Hence $Y_1\sim N(0,\sigma^2)$ since it is the linear combination of normal variables.

Similarly we get $Y_2\sim N(0,\sigma^2)$ and $Y_3\sim N(0,\sigma^2)$

Now

$$EY_1Y_2=1/\sqrt{6}E(X_1)^2-1/\sqrt{6}EX_2^2=0$$

and similarly $EY_2Y_3=EY_1Y_3=0$, hence $Y_1$, $Y_2$ and $Y_3$ are independent, since for normal variables independece coincided with zero correlation.

Having established that we have

$$(Y_1^2+Y_2^2+Y_3^2)/\sigma^2=\left(\frac{Y_1}{\sigma}\right)^2+\left(\frac{Y_2}{\sigma}\right)^2+\left(\frac{Y_3}{\sigma}\right)^2=Z_1^2+Z_2^2+Z_3^2$$,

where $Z_i=Y_i/\sigma$. Since $Y_i\sim N(0,\sigma^2)$, we have $Z_i\sim N(0,1)$.

We have showed that our quantity of interest is a sum of squares of 3 independent standard normal variables, which by definition is $\chi^2$ with 3 degrees of freedom.

As I've said in the comments you do not need to calculate the densities. If you on the other hand want to do that, your formula is wrong. Here is why. Denote by $G(x)$ distribution of $Y_1^2$ and $F(x)$ the distribution of $Y_1$. Then we have

$$G(x)=P(Y_1^2<x)=P(-\sqrt{x}<Y_1<\sqrt{x})=F(\sqrt{x})-F(-\sqrt{x})$$

Now the density of $Y_1^2$ is $G'(x)$, so

$$G'(x)=\frac{1}{2\sqrt{x}}(F'(\sqrt{x})+F'(-\sqrt{x})$$

We have that

$$F'(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^2}{\sigma^2}},$$

so

$$G'(x)=\frac{1}{\sigma\sqrt{2\pi x}}e^{-\frac{x}{2}}$$

If $\sigma^2=1$ we have a pdf of $\chi^2$ with one degree of freedom. (Note that for $Z_1$ instead of $Y_1$ the calculation is similar and $\sigma^2=1$ ) As @whuber pointed out, this is gamma distribution, and sums of independent gamma distributions is again gamma, the exact formula is provided in the wikipedia page.

Answer 2

I want to offer a solution in the view of matrix algebra.

Let $$ U' = \left[\begin{matrix} -1/\sqrt{2} && 1/\sqrt{2} && 0 \\ -1/\sqrt{3} && -1/\sqrt{3} && 1/\sqrt{3} \\ 1/\sqrt{6} && 1/\sqrt{6} && 2/\sqrt{6} \end{matrix}\right]$$

As $Y = U'X$,

$$E(Y) = E(U'X) = U' \cdot 0 = 0$$ $$Var(Y) = U' Var(X) U = \sigma^2 U' U = \sigma^2 I $$

You can see $Y_1, Y_2, Y_3$ in $Y$ are all follow $N(0, \sigma^2)$, and are independent to each other. Then you can prove $Y_i^2/\sigma^2 \sim \chi^2(1)$, and finally, $(Y_1^2 + Y_2^2 + Y_3^2)/\sigma^2 \sim \chi^2(3)$

Answer 3

Have you tried simply multiplying out the squared Y^2's in terms of the X[1:3] terms. I suspect that when you are all done that you will see that you simply have: (1/2 +1/3 +1/6)* X1^2 + (1/2 +1/3 +1/6)*X2^2 + (1/2 +1/3 +1/6)*X3^2 . This, of course, assumes that X1X3=X3X1, i.e. that your random variable algebra is commutative, but unless you are working on complex variables in particle physics, that assumption should hold. So far I have gotten about halfway there, and my approach seems to be holding up. It would seem to be useful that you go through the exercise, rather than for me to display it.