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.

  • $\begingroup$ 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

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


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