Proving Transformation of Random Variable is Chi-Square with 2 Degrees of Freedom

Question

Let the cumulative distribution function for the random variable $T$ be given by $$P(T \leq t) = \begin{cases} 1 - e^{-\frac{t}{30}} & t \geq 0 \\ 0 & t < 0 \end{cases}$$ Prove that the transformation $\frac{2}{\beta}T$ where $\beta > 0$ is a parameter, is chi-square distributed with 2 degrees of freedom.

My attempt: I know that a chi square distribution with two degrees of freedom corresponds to an exponential distribution with parameter $\frac{1}{2}$, so I wish to find the pdf of $T$ and show that it is an exponential distribution with parameter $\frac{1}{2}$.

Let $Y = \frac{2}{\beta}T$, then solve $$P(Y \leq y) = P(\frac{2}{\beta}T \leq y) = P(T \leq \frac{\beta y}{2}) = 1 - e^{-\frac{\beta y}{60}}$$ Differentiating this then gives $$f_T(t) = \frac{\beta}{60} e^{-\frac{\beta y}{60}}, y \geq 0$$ which is not chi square distributed with two degrees of freedom for all $\beta > 0$. Is anyone able to see where I go wrong?

Answer 1

your text is wrong. If your CDF was

$$F_T(t)=1-e^{-t/\beta}$$

that is

$$f_T(t)=\frac{1}{\beta}e^{-t/\beta}\sim \text{exp}(1;1/\beta)$$

where $1/\beta$ is the rate parameter,

Then setting $Y=\frac{2}{\beta}T$ you immediately get

$$f_Y(y)=\frac{1}{2}e^{-y/2}$$

that is a $\chi_{(2)}^2$

this is also a negative exponential with mean 2