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!
 A: 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 $0$. ${}\qquad\blacksquare$
