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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.

Blue
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    Is this a question from a course or textbook? If so, please add the `[self-study]` tag & read its [wiki](http://stats.stackexchange.com/tags/self-study/info). – gung - Reinstate Monica Sep 01 '15 at 04:16
  • 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. – Blue Sep 01 '15 at 04:20
  • How will knowing the distribution make it faster to generate? Is $X$ standard exponential, or exponential with rate parameter $\lambda$, say? – Glen_b Sep 01 '15 at 04:32
  • 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). – Blue Sep 01 '15 at 04:38
  • Sorry, I did not answer your second question. It is exponential with rate parameter $\lambda$, as can be seen here: https://en.wikipedia.org/wiki/Exponential_distribution – Blue Sep 01 '15 at 04:42
  • @blue what's your research? This is interesting – Neil G Sep 01 '15 at 05:54
  • @ Neil G - I am developing a monte carlo method to solve problems in electromagnetics. This particular problem arrised in the solution of a problem related to transmission lines. – Blue Sep 01 '15 at 12:54

1 Answers1

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\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 $\beta=1$?

Deep North
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    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? :) – Blue Sep 01 '15 at 05:01
  • 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. – Deep North Sep 01 '15 at 05:06
  • 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 :) – Blue Sep 01 '15 at 05:11
  • @Blue Some things to consider (FYI, I know absolutely **no** complex analysis): 1) Does $e^{aX} < y \implies aX < \ln(y)$ in $\mathbb{C}$? 2) Can you "divide" by a complex number $a \in \mathbb{C}$ which has a nonzero imaginary part? – Clarinetist Sep 01 '15 at 05:37
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    @ 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 $a * b^{-1}$ – Blue Sep 01 '15 at 05:43
  • @ Deep North - I just edited my post based on your answer. I was trying to extend you answer to real and imaginary part of my edit but I could not do it yet. I feel that you are very strong with calculus. would you give a look to it? :) – Blue Sep 01 '15 at 05:46
  • @Blue, thanks, this is beyond my current knowledge, hope someone else can help. – Deep North Sep 01 '15 at 05:54
  • This looks great (+1), but I don't understand why $Y$ must be less than 1? – Neil G Sep 01 '15 at 06:02
  • Why do you want complex $a$? how would you interpret a complex probability density function? – kjetil b halvorsen Sep 01 '15 at 08:27
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    This answer is incorrect: you do not obtain a beta distribution because you don't have the very important constraint $Y \in [0,1]$ – Guillaume Dehaene Sep 01 '15 at 11:41
  • @ Neil G - I don't think it makes sense to say if Y is or is not less than one. Y is a complex random variable, the signs < and > make no sense in the complex plane. – Blue Sep 01 '15 at 13:01
  • @ kjetil - it is the particular problem I am solving that imposes $a$ as a complex number (of course it can be real in particular case). Your second question was in fact the main reason I created this post. So here is my interpretation: $Y$ can be written as $Y(X) = f_r(X) + i f_i(X)$ (see my question above for more details). So basically, let x be a random number from X distribuion. Then calculate y_r = f_r(x) and y_i = f_i(x). The question is then what is the distribution of random variables $Y_r$ and $Y_i$ whose random numbers are for example $y_r$ and $y_i$. – Blue Sep 01 '15 at 13:15
  • @ Guillaume - you are correct. Thank you for clarifying this further. If it helps, as far as I can tell $Y$ can be any complex number as $X \in [0, \infty[$ – Blue Sep 01 '15 at 13:24
  • when $a$ is real I put $\int\limits_{y = 0}^{y = 1} {\frac{\lambda }{a}{y^{ - \lambda /a - 1}}} dy$ into mathematica but the program said that this integration does not converge on $(0,1)$ does that mean that $Y$ is not a random variable ? – Tuong Nguyen Minh Jun 08 '21 at 13:26
  • This answer should be prefixed by **only for the case $a>0$** – kjetil b halvorsen Nov 27 '21 at 15:03