0
$\begingroup$

I have a cumulative function/cdf defined as follows, where $X$ is a random variable:

$\ F_X(x) = \left\{\begin{aligned} &0 &&: x \le 0\\ &1-e^{-x} &&: x > 0 \end{aligned} \right.$

How would I go about solving $P(0 \le e^X \le 4) $ ?

I understand that if we are given a probability range and a CDF we can find the probability as follows:

$P(a \le X \le b) =F_X(b)-F_X(a)$

Would I just need to do some simplification like taking the natural log of both sides of the range of the probability?

$\endgroup$
1
$\begingroup$

We know that $e^X \geq 0$, the event $e^X \leq 4$ is equivalent to $X \leq \ln 4$, hence

$$P(0 \leq e^X \leq 4)= P(X \leq \ln 4)=F_X(\ln 4) = 1- \exp(-\ln 4)=\frac34$$

$\endgroup$
  • $\begingroup$ How would we take into consideration 0? or is that already 'implied' since we know that e^x is always >0 $\endgroup$ – Anas Yousef Feb 24 '17 at 23:50
  • $\begingroup$ Yup, $e^X >0 $ for sure. $\endgroup$ – Siong Thye Goh Feb 24 '17 at 23:51
1
$\begingroup$

$P(0 \le e^X \le 4) = P(e^X \le 4) = P(X \le \ln 4) = F(\ln 4)$

$\endgroup$

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.