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

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?

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

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
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