I was studying probability theory and came across something I didn't quite understand.
Let's say that we have a random variable $X$ that is distributed according to the Exponential distribution with $\lambda = 1$ (i.e. $X \sim$ Expo($1$)).
I know that the PDF of $X$ is
$$f(x) = \lambda e^{-\lambda x} $$
but what would the distribution look like for $-X$?
Any feedback is appreciated. Thank you.
If $X$ has density function $f(x)$, then $-X$ has density function $f(-x)$. For an exponential variable of mean $1$ it is incorrect to say its density is $f(x)=e^{-x}$. Rather $$f(x)= \begin{cases} e^{-x}& \text{if $x\ge0$,}\\ 0& \text{if $x<0$.} \end{cases}$$ Then the density function of $-X$ will be $$f(-x)= \begin{cases} 0& \text{if $x>0$,}\\ e^{x}& \text{if $x\le0$.} \end{cases}$$