2
$\begingroup$

Given $X$, a random variable with CDF (cumulative distribution function) $F$, how do you find the CDF $G$ of $Y$ which is a function of $X$?

I've read my course materials on CDFs but I'm not finding many resources on this kind of problem. For example if $Y = e^X$ then my intuition tells me that in the cases where $u \leq 0$, $G = 0$ since $Y$'s lower limit is $0$. And for cases where $u > 0$ then $G = e^F$? I'm not sure about the latter...

Another type of problem is $Y = mX+b$ (where $m$ is non-zero). Again my intuition is that there are 2 cases: 1 for which m>0 and 1 for m<0. One of the given hints for this type of problem is that the $m<0$ case might need to be expressed in terms of both the CDF and PMF of X, not just the CDF.

Any help in understanding how to break these types of problems down would be appreciated.

$\endgroup$
3
$\begingroup$

If $Y=h(X)$ where $h$ is strictly increasing, then: $G(y)=P(h(X)<y)=P(X<h^{-1}(y))$ where $h^{-1}(y)=-\infty$ if $y<h(x)$ for all $x$ and $h^{-1}(y)=+\infty$ if $F(x)>y$ or all $x$.

So if $Y=e^X$, then if $y>0$ then $G(y)=P(Y<y)=P(X<\log y)=F(\log y)$.

If $h$ is strictly decreasing, then $P(Y<y)=P(X>h^{-1}(y))$. So:

$$P(Y<y)=1-P(X<h^{-1}(y)) - P(X=h^{-1}(y)) = 1-F(h^{-1}(y)) - P(X=h^{-1}(y))$$

If $X$ is a continuous random variable, then $P(X=h^{-1}(y))=0$ and you just have:

$$P(Y<y)=1-F(h^{-1}(y)) $$

It becomes more complicated when $h$ is not $1-1$, for example, if $h(x)=x^2$ or $h(x)=\sin x$.

$\endgroup$
  • $\begingroup$ sorry for evoking this old thread. But what happens when h is not one to one? Thanks Thomas! $\endgroup$ – VJune Oct 12 '15 at 4:11
  • $\begingroup$ And where can i find more discussions regarding these topics? $\endgroup$ – VJune Oct 12 '15 at 4:12
  • $\begingroup$ Another counter-intuitive result is when G is F i.e. G is cumulative distribution function of X. $\endgroup$ – VJune Oct 12 '15 at 5:16

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.