# Probability that one sum of squared standard normals is greater than a constant times another such sum

Let $Y_1,Y_2,..., Y_n$ $i.i.d$ $\mathcal{N}(0,\sigma^2)$ random variables and $n>4$. Find the probability $\mathbb{P}\{Y_3^2 + Y_4^2+ ... + Y_n^2 \geq \alpha ( Y_1^2 + Y_2^2)\}$ for $\alpha >0$.

Probably I have to use the fact that the sum of squared standard gaussians is chi-distributed random variable but the $\alpha$ messes things up. Something else that I noticed is the symmetry i.e. this probability should be the same for any $2$ random variables that I have on the right side of the inequality.

• The fact that the standard normals were independent was important, since this is what allowed you to group them together as chi-squared. Note also that your chi-squared variables were independent as they were the sum of different (independent) $Y_i^2$ and it's this which allows you to make an F variable. May 2, 2015 at 22:35
• @Silverfish I think that response is probably good enough to stand as an answer. You can't do much better without actually answering the whole question. May 3, 2015 at 0:41
• @Glen_b Thanks, I've added a wee bit more but with more focus on the solution strategy and mechanics of the question. May 22, 2015 at 12:14

You can use the relation of the F-distribution to the chi-squared

$$F_{m,n}=\frac{\chi_m^2/m}{\chi_n^2/n}$$

$P\{Y_3^2 + Y_4^2+ ... + Y_n^2 \geq \alpha ( Y_1^2 + Y_2^2)\}=P(\chi_{n-2}^2\ge\alpha\chi_2^2)=P(\frac{\chi_{n-2}^2}{\chi_2^2}\ge\alpha)$

Now you can adjust it to be in the form of an $F$.

$$P\Bigg(F_{n-2,2}\ge\frac{2}{n-2}\alpha\Bigg)$$

• Can you say more about why the F distribution might be the right one for the OP? May 2, 2015 at 22:01
• Hi @gung I do.. May 2, 2015 at 22:11
• I think the answer is correct May 2, 2015 at 22:16
• Yes! it is correct. May 2, 2015 at 22:16
• You're saying it's okay to just do someone's homework ... if someone demands it? I expect gung was looking for a very small change in your answer, just something that would explain the connection. May 3, 2015 at 16:04

In keeping with the self-study policy I will leave some hints rather than post a complete answer, but also try to explore a little about why this type of question "works".

Probably I have to use the fact that the sum of squared standard gaussians is chi-distributed random variable

Yes, but you need to make that "the sum of independent squared standard gaussians is a chi-distributed random variable" — the independence is an important part of this question. Fortunately you were told in the beginning that $Y_i$ were "i.i.d.", which means independent and identically distributed.

but the $\alpha$ messes things up

Actually it's not really the $\alpha$ which is causing you the problem. Consider the special case that $\alpha = 1$. Does that improve things? We still get a chi-squared variable on the left-hand side and another chi-squared variable on the right-hand side. But as before, we are unable to combine them into a single chi-squared variable, because they are on different sides of the equation. If you had moved the $Y_1^2$ and $Y_2^2$ across from the right to the left by subtraction, you would still have been unable to combine into a single chi-squared variable, because the coefficients on $Y_1^2$ and $Y_2^2$ would have been $-1$ while the coefficients on the the other $Y_i^2$ would have been $+1$.

So the problem is no harder with or without the $\alpha$. You are bound to get two different chi-squared variables, and so your solution strategy should be to exploit this fact, rather than try to reduce it to a single chi-squared.

Something else that I noticed is the symmetry i.e. this probability should be the same for any 2 random variables that I have on the right side of the inequality.

Moreover, note that they are different $Y_i$s on the right-hand and left-hand sides, and therefore are independent of each other.

As a result, the chi-squared variables you get on the left-hand and right-hand sides are also independent of each other. Let's write $X_1 \sim \chi^2_{\nu_1}$ and $X_2 \sim \chi^2_{\nu_2}$ with $X_1$ and $X_2$ independent; we are seeking $\Pr(X_1 \geq \alpha X_2)$.

Can you think of a distributional fact that relates two independent chi-squared variables to each other?

Perhaps look at the list of relationships in the chi-squared article in Wikipedia. A couple of candidates present themselves.

$\frac{X_1}{X_1 + X_2} \sim \text{Beta}(\frac{\nu_1}{2}, \frac{\nu_2}{2})$ looks a little tricky to apply since we don't have the sum of two chi-squared variables at the moment. But we could e.g. add $\alpha X_1$ to both sides to get: $$(\alpha + 1) X_1 \geq \alpha (X_1 + X_2)$$

$\frac{X_1/\nu_1}{X_2/\nu_2} \sim F(\nu_1, \nu_2)$ could be applied more easily; it requires the ratio of two independent chi-squared variables, and since we have a chi-squared as a factor on each side, the desired fraction is only a short manipulation away.

One neat thing about the Beta distribution method, though, is that (particularly with the nice numbers given in the question) you will obtain a probability that you can easily integrate out to give you a formula for the probability as a rational function of $\alpha$. The Beta distribution's density function will turn out to be "nice" because the two degrees of freedom on the right-hand chi-squared become one degree of freedom in the Beta distribution, and in the PDF we only raise the corresponding factor to a power one less than the degrees of freedom; this ease of integration will also apply to the normalizing factor, which is a Beta function. Such a solution feels, at least to me, more satisfying than a solution in terms of the $F$ distribution's CDF. It's not even that much harder to keep $n$ general, rather than substituting a specific value, to yield a formula in terms of $\alpha$ and $n$.

Try it!