0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your $f(x)$ hasn't got an $x$ in it! $\endgroup$ – Lord Shark the Unknown Nov 1 '18 at 9:24
  • $\begingroup$ Thanks @LordSharktheUnknown made the changes. $\endgroup$ – Seankala Nov 1 '18 at 9:25
  • $\begingroup$ Is $\lambda$ really equal to $1$? $\endgroup$ – Lord Shark the Unknown Nov 1 '18 at 9:25
  • $\begingroup$ In this particular case, yes. The specific problem that I'm referring to sets $\lambda = 1$. $\endgroup$ – Seankala Nov 1 '18 at 9:32
1
$\begingroup$

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 $$f(-x)= \begin{cases} 0& \text{if $x>0$,}\\ e^{x}& \text{if $x\le0$.} \end{cases}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.