8
$\begingroup$

Let $X$ be a real valued random variable with exponential distribution. Let $a$ be a complex number. What is the distribution of $Y = e^{aX}$? Can Y be written in the form of another known distribution?

NOTE: based on the answer of Deep North (below) I note that solving the above problem is equivalent to solve this one:

$Y = e^{aX} = e^{(a_r + i a_i) X} = e^{a_rX} \cos(a_i X) + i e^{a_rX} \sin(a_i X)$. So the answer might also be a pair of distributions if it is not possible to write it as a single answer in the complex plane.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$
    – gung - Reinstate Monica
    Commented Sep 1, 2015 at 4:16
  • $\begingroup$ No it is not. It is a question that resulted from my research. Basically, I can generate X on a computer but I am trying to speed up my method by generating Y directly, but I need to know what is Y so I can do that. Thanks for your comment. $\endgroup$
    – Blue
    Commented Sep 1, 2015 at 4:20
  • $\begingroup$ How will knowing the distribution make it faster to generate? Is $X$ standard exponential, or exponential with rate parameter $\lambda$, say? $\endgroup$
    – Glen_b
    Commented Sep 1, 2015 at 4:32
  • $\begingroup$ Good question. I don't have the answer to your question yet. Because I don't know the distribution. But I can give two arguments: i) if the answer is simple enough I can even do it theoretically, and ii) if not, I can avoid the calculation of the exponential of a million random numbers of kind X if I know Y distribution. In that case I calculate a million random numbers of Y right from the start and avoid the exponential. In principle, it should be faster (but I may be wrong). $\endgroup$
    – Blue
    Commented Sep 1, 2015 at 4:38
  • $\begingroup$ Sorry, I did not answer your second question. It is exponential with rate parameter $\lambda$, as can be seen here: en.wikipedia.org/wiki/Exponential_distribution $\endgroup$
    – Blue
    Commented Sep 1, 2015 at 4:42

1 Answer 1

Reset to default
3
$\begingroup$

\begin{align} F(y) &= P(Y<y) \\ &= P(e^{aX}<y) \\ &= P(aX<\ln y) \\ &= P(X<\frac{\ln y}{a}) \\ &=\int_0^{\frac{\ln y}{a}}\lambda e^{-\lambda y}dy \\ &= \left.-e^{-\lambda y}\right\vert_{0}^{\frac{\ln y}{a}}\\ &=1-y^{-\lambda/a} \end{align}

We take derivatives of both side:

$$f(y)=\frac{\lambda}{a}y^{-\lambda/a -1}$$

A Beta distribution when $0<y<1$ with $\beta=1$?

Improve this answer
$\endgroup$
15
  • 1
    $\begingroup$ Thank you. I just have one simple question. Does this take into account that $a$ is a complex number? I ask this because the less or equal sign in the complex domain seems a bit triky. What you think? :) $\endgroup$
    – Blue
    Commented Sep 1, 2015 at 5:01
  • $\begingroup$ This a hard question for me I don't know if $a$ can be a complex number or not. I learned complex analysis but I have not able to apply that to statistics yet. $\endgroup$
    – Deep North
    Commented Sep 1, 2015 at 5:06
  • $\begingroup$ Oh ok. But thanks a lot because your answer is correct for the case when a is real. It is useful (i tried to give you +1 but i think i don't have enough reputation, this is my first quesiton). The problem is that in my case a is always complex. I like this answer, but I don't know how to extend it also :) $\endgroup$
    – Blue
    Commented Sep 1, 2015 at 5:11
  • 1
    $\begingroup$ @ Clarinetist - 1) the problem is that in the complex plane the sign < makes no sense. You can imagine a bidimensional plane (the complex plane is a bidimensional plane with some structure on it) and ask yourself if one point P1 is bigger or smaller than another P2. You see, makes no sense. But it makes sense to ask if |P1| is bigger or smaller than |P2|. 2) yes you can, the complex numbers form a division algebra. So each complex number (except $0 + i 0$) has an inverse and so $a/b$ is the same as $a * b^{-1}$ $\endgroup$
    – Blue
    Commented Sep 1, 2015 at 5:43
  • 1
    $\begingroup$ This answer is incorrect: you do not obtain a beta distribution because you don't have the very important constraint $Y \in [0,1]$ $\endgroup$
    – Guillaume Dehaene
    Commented Sep 1, 2015 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.