# Show that $Y = \sum_{i=1}^n Y_i$ is distributed as $\chi _{2n}^2$.

The Statement of the Problem:

Suppose that $X_1,\ldots, X_n$ is a random sample from the $U(0,1)$ distribution and

$$Y_i = -2\log X_i.$$

Show that $Y = \sum_{i=1}^n Y_i$ is distributed as $\chi _{2n}^2$.

Where I Am:

Ok. So, I've got some dots that I can't seem to connect. Upon consideration of the cdf of $Y_i$...

$$P(Y_i < x) = P(-2\log X_i < x) = P(\log X_i > -\frac{1}{2}x) = P(X_i > e^{-x/2}) = 1- P(X_i < e^{-x/2})$$

[EDITED MATERIAL BEGINS HERE]

So, we see that

$$P(Y_i < x ) = 1 - P(X_i < e^{-x/2}) = 1 - F_{X_i}(e^{-x/2})$$

and because $X_i \sim U(0,1)$

$$1 - F_{X_i}(e^{-x/2}) = 1 - \frac{e^{-x/2}-0}{1-0} = 1 - e^{-x/2}$$

which is the cdf of a chi-squared r.v. with $2$ degrees of freedom. Therefore

$$Y_i \sim \chi_{2}^2$$

[EDITED MATERIAL ENDS HERE]

Here, I feel like I'm getting close to the special case of the cdf of a chi-squared r.v. in which $k=2$, i.e.

$$F(x;2) = 1-e^{-x/2} \qquad \text{(thanks Wikipedia)}.$$

However, I assume it's not the case that

$$P(X_i > e^{-x/2}) = P(X_i < 1 - e^{-x/2}). \quad (*)$$

Or is that actually true? And, if it is, does that suffice to show that, indeed,

$$X_{i} \sim \chi _{2}^2?$$

I'm aware that

$$X \sim U(0,1) \implies -2\log(X) \sim \chi _{2}^2 \qquad \text{(thanks again Wikipedia)}$$

But, I don't know if this has sufficiently shown that.

Finally, I know that the sum of independent chi-squared r.v.'s is also has a chi-squared distribution. From this property, it's clear that I have a sum of $n$ independent chi-squared r.v.'s, each with $2$ degrees of freedom, meaning, of course, that

$$Y \sim \chi _{2n}^2$$

completing the proof.

So, yeah, it's just mostly putting this together into a coherent proof without relying too much on properties given by Wikipedia. Any help, of course, would be appreciated. Thanks!

I wrote what appears below under the impression that what you had was not $\vphantom{\frac \int\int}P(X_i > e^{-x/2}) = P(X_i < 1 - e^{-x/2})$ but $P(X_i > e^{-x/2}) = 1 - P(X_i < e^{-x/2})$. But maybe that last is what you actually need.

$$\Pr(X>x) = 1-\Pr(X\not>x)=1-\Pr(X\le x) \overset{\huge\text{?}} = 1 - \Pr(X<x).$$ So is it true that $\Pr(X\le x) = \Pr(X<x)$? That seems to be what your question reduces to.

$$\Pr(X\le x) = \Pr(X<x\text{ or }X=x) = \Pr(X<x)+\Pr(X=x).$$ So now the question reduces to whether $\Pr(X=x)=0$.

And if the cumulative distribution function is continuous, then that is true. One circumstance in which the cumulative distribution function is continuous is when the distribution has a density function. But in your case, you've identified the cumulative distribution function, so you don't need to think about densities.

Theorem:

If $F(x) = \Pr(X\le x)$ is a continuous function of $x$, then for every $x$, $\Pr(X=x)=0$.

Proof:

First observe that for $u\le x$, \begin{align} F(u) + \Pr (u<X\le x) & =\Pr(X\le u) + \Pr (u<X\le x) \\[10pt] & = \Pr(X\le u\text{ or }u<X\le x) = \Pr(X\le x) = F(x), \end{align} and therefore $$\Pr(u < X \le x) = F(x)-F(u).$$ Next: $$\Pr(X=x) \le \Pr(u<X\le x) = F(x) - F(u).$$ and since $F$ is continuous, that last item can be made $<$ any $\varepsilon>0$ by making $u$ close enough to $x$ but still $<x$. Then $P(X=x)$ is smaller than every positive number; hence equal to $0$. ${}\qquad\blacksquare$

• Right. I was actually just being sloppy with my inequalities because I'm only dealing with continuous distributions and don't really consider a difference between strict and non-strict inequalities (because, obviously, $P(X=x)=0$ for cts distributions). In light of that, my question is, rather, if the above equation (I'll mark it $*$) is actually true (being aware of the fact that $P(X=x)=0$). And, also, if it suffices to show that if two rv's have the same cdf, then those rv's have the same distribution. Commented Jul 27, 2015 at 0:25
• Oh, wow. I just realized the (rather dumb) mistake I'd made. I'll put it in my post. Thanks for the help! Commented Jul 27, 2015 at 1:21