How to find the CDF of $Y=\frac{1}{2}X$?

Question

Let $X$ have an exponential distribution with rate parameter $\lambda=\frac{1}{2}$

I believe that the probability density of $X$ is

$$f_X(x) = e^{-x/2}$$

So the CDF of $X$ is then

$$F_X(x)=-2e^{-x/2}$$

Plugging in $Y=\frac{1}{2}X$, I get

$$F_Y(Y) = -2e^{x}$$

Is my thought process correct? I haven't dealt with a problem like this, and we didn't do any examples that were similar, other than being given a PDF $f_X(x)$ and then working from there.

Answer 1

Your $f_X$ is not a density function (since it does not integrate to $1$) and $F_X$ is not a distribution function (because it takes negative values).

$f_X(x)=\frac 12 e^{-x/2}, 0 <x<\infty$ and $F_X(x)=1-e^{-x/2}$ for $x>0$, $F_X(x)=0$ for $x \leq 0$.

$P(Y \leq y)=P(X\leq 2y)=F_X(2y)=1-e^{-y}$ for $y >0$.