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?

  • $\begingroup$ If you have shown independence and zero means, you only need to check whether the standard errors are ones. In your case if we put $\sigma^2$ inside the parenthesis, your statement will follow if you prove that $EY_i^2/\sigma^2=1$ $\endgroup$
    – mpiktas
    Commented Jan 9, 2011 at 20:53
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    $\begingroup$ @Christopher (I TeXified your question to make it readable.) You should double-check the distributional form for $Y_i^2$, because it is incorrect: it is not Gaussian, but exponential. (That is, the argument of exp should be linear in $x$, not $x^2$.) There also needs to be a factor of $x^{-1/2}$. Once you fix that, all you need to show is that a sum of Gamma distributions is also Gamma. $\endgroup$
    – whuber
    Commented Jan 10, 2011 at 2:05
  • $\begingroup$ @Whuber: Thanks for TeXing me out! $Y_i$ is distributed normal, being a linear transformation of two random normal variables. I didn't specify the distribution of $Y_i$, but I'd figure it to be $\chi^2(1)$. The sqrt(x) also confused me. I was not sure where that should come from. No one in the normal pdf could I see a sqrt(x) coming from. I was hoping someone would have pointers. $\endgroup$ Commented Jan 10, 2011 at 4:13
  • $\begingroup$ @Christopher. If Yi are the sum of normal variates with mean zero then they cannot be χ2(1). $\endgroup$
    – DWin
    Commented Jan 10, 2011 at 5:33
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    $\begingroup$ @Christopher I suspect that in computing the pdf of $Y_i^2$ you made a substitution but did not back-substitute again. Letting $y=x^2$, you can rewrite $C\exp(-2x^2/(2\sigma^2))dx$ = $C\exp(-y/\sigma^2)dy/(2\sqrt{y}),$ which is a (scaled) gamma(1/2) distribution. $\endgroup$
    – whuber
    Commented Jan 10, 2011 at 13:46

3 Answers 3


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


\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 $Y_3\sim N(0,\sigma^2)$



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


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


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


We have that



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

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.

  • $\begingroup$ Thanks, mpiktas. I figured there must've been a mistake somewhere in my pdf. $\endgroup$ Commented Jan 10, 2011 at 8:31

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, $(Y_1^2 + Y_2^2 + Y_3^2)/\sigma^2 \sim \chi^2(3)$

  • $\begingroup$ Can you give some references to how to approach problems like these using matrices? $\endgroup$
    – Dom Jo
    Commented Aug 31, 2020 at 6:38

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.

  • $\begingroup$ This method is equivalent, and would probably be the way to go, had I not already calculated the means and variances of the Yi's. Either way, it gets stuck at the same point, I believe. $\endgroup$ Commented Jan 9, 2011 at 21:22
  • $\begingroup$ Not really. You have (or at least I now have) proved that ((Y1^2 + Y2^2 + Y3^2) == (X1^2 + X2^2 + X3^2) and surely that is sufficient to establish the chi-sq-ness when you have Xi ~ N(0, sigma^2) $\endgroup$
    – DWin
    Commented Jan 9, 2011 at 21:43
  • $\begingroup$ That's where the problem lies. I'm looking for how to do that. I'm not sure how to go from having a sum of three squared normal pdfs to a chi-square pdf with df=3. $\endgroup$ Commented Jan 10, 2011 at 0:32
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    $\begingroup$ I'm confused. Isn't that precisely the definition of a chi-sq variate with 3 df? If you are worried about the sigma being different than unity, it's just a constant and so easily factors out. en.wikipedia.org/wiki/Chi-square_distribution $\endgroup$
    – DWin
    Commented Jan 10, 2011 at 1:09

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