Distribution of $-2\ln Y$ if Y is uniform in $[0,1]$

Question

Distribution of $-2\ln Y$ if Y is uniform in $[0,1]$.
There are lots of similar questions about this. But I got a proposition seems different to all these answers to the question.
That is

If Y is uniform in $[0,1]$, then $-2\ln Y \sim \chi^2(2) $.

My intuition is to find the CDF. But when I can't related it to the standard normal distribution since we already knew that chi-square distribution with $k$ degrees is the sum of the squares of $k$ standard normal random variables.

Answer 1

Lets consider the situation where we use the PDF. We know:

$$f_X(x)=1,\space y=-2\ln x \implies x=e^{-\frac y2}$$

Lets take the derivative with respect to $y$:

$$\frac {dx}{dy}=-\frac 12e^{-\frac y2}$$

Under the inverse transform method, we can use the fact that:

$$f_Y(y)=f_X(y) \left| \frac {dx}{dy} \right|$$ $$=f_X(-2 \ln x)\left|-\frac 12e^{-\frac y2} \right|$$ $$=1\cdot \frac 12e^{-\frac y2}$$ $$=\frac 12e^{-\frac y2}$$

The only part left is to find our bounds for the random variable $Y$.

From the equation $y=-2\ln x$:

$$x=0 \implies y=-2 \ln 0=\infty$$ $$x=1 \implies y=-2 \ln 1=0$$

We can check the density of $Y$ integrates to $1$:

$$\int_0^\infty f_Y(y)dy=\int_0^\infty \frac 12e^{-\frac y2}dy$$

$$=\left[-e^{- \frac y2}\right]_0^\infty 0-(-1)=1$$

As you have postulated, this is the form of a Chi-Squared distribution with $2$ degrees of freedom, with probability density function:

$$\frac 1{2^\frac k2 \Gamma(\frac k2)}y^{\frac k2-1}e^{-\frac y2}$$

Setting $k=2$ we end up with:

$$=\frac 12e^{-\frac y2}, \ y \in [0,\infty)$$

It is NOT true that the only way to get a Chi-Squared distribution is through the sum of Normal Squares (as seen in the other answer).

Answer 2

It is a general fact that if $U$ is uniformly distributed on $(0,1)$ and $Y=-\frac{1}{\lambda}\log U$ for some $\lambda>0$, then $Y$ is exponentially distributed with parameter $\lambda$. Indeed, we have $$ \mathbb{P}(Y>y)=\mathbb{P}(-\log U>\lambda y)=\mathbb{P}(U<e^{-\lambda y})=e^{-\lambda y}$$ for $y>0$.

In your problem $\lambda=\frac{1}{2}$, so $Y\sim \exp(\frac{1}{2})$, which happens to be the same distribution as $\chi^2(2)$. This is because both the exponential and chi-squared distributions are special cases of the gamma distribution, and the parameters in this case coincide.