0
$\begingroup$

Let's $X$ be a random variable with exponential distribution.
That means that:

  1. $F_X(t) = 1 - e^{1- \lambda t}$ which is the distribution function for $X$,
  2. $f_X(t) = \lambda e^{- \lambda t}$ which is the density function for $X$.

Now let's define another random variable $Y = \log X$. Our task is to find it's distribution and density function.


My attempt

Firstly I want to find the distribution function: $$F_{y}(t) := P(Y < t) = P(\log X < t) = P(X < e^t) = \int\limits_{0}^{e^t} \lambda e^{- \lambda x} \mbox{d}x$$ $$F_Y(t) = 1 - e^{- \lambda e^{t}}$$ Here's my question: how do I know what the integration limits should be? Is my reasoning correct?

$\endgroup$
  • 1
    $\begingroup$ Distribution of $X$ is $1-e^{\lambda t} $, for $t>0$. $F_Y(t)=1-e^{\lambda e^t}$ $\endgroup$ – Jakobian Jun 2 '18 at 14:53
  • 1
    $\begingroup$ I'd add that we can take $\log$ of $X$ because $P(X\leq 0)=0$. You know what integration limits are, because you know distribution function of $X$. Notice that $P(X<e^t)=F_X(e^t)$. Reasoning is overall ok $\endgroup$ – Jakobian Jun 2 '18 at 15:00
  • $\begingroup$ @Adam Thank you! So in general the limits will be given by the function $F_X(t)$ for any random variable $X$? $\endgroup$ – Hendrra Jun 2 '18 at 15:48
  • $\begingroup$ If $Y=\log X$ like here, then it will be analogous $\endgroup$ – Jakobian Jun 2 '18 at 15:52
  • $\begingroup$ @Adam Thank you! One more question: what can I say about the expected value of $\log X$? $\endgroup$ – Hendrra Jun 2 '18 at 15:53

Your Answer

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

Browse other questions tagged or ask your own question.