# What would be the negative counterpart of a random variable's distribution?

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.

• Your $f(x)$ hasn't got an $x$ in it! – Lord Shark the Unknown Nov 1 '18 at 9:24
• Thanks @LordSharktheUnknown made the changes. – Seankala Nov 1 '18 at 9:25
• Is $\lambda$ really equal to $1$? – Lord Shark the Unknown Nov 1 '18 at 9:25
• In this particular case, yes. The specific problem that I'm referring to sets $\lambda = 1$. – 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 $$f(-x)= \begin{cases} 0& \text{if x>0,}\\ e^{x}& \text{if x\le0.} \end{cases}$$