I'm really struggling to figure out this problem from one of my practice exercises for a probability course. I know that the probability distribution function $f_x(x)$ is related to the cumulative distribution function $F_x(x)$ by integration, and from what I can tell based on my professor's notes and a few online resources I have looked into, the probability distribution function of an exponential function is of the form $f_x(x)=\lambda e^{-\lambda x}$.

From this I suppose $f_y(x)=\lambda e^{-\lambda x}$

Integrating would give me $F_y(x)=-e^{-ax}+C$

But how does this relate to the original $F_x(x)$ mentioned in the question?


Your question is unclear. If I were to answer the question you posed in your title, then the transformation $$Y = g(X) = e^X$$ is monotone, hence $$F_Y(y) = \Pr[Y \le y] = \Pr[e^X \le y] = \Pr[X \le \log y] = F_X(\log y).$$ We also have $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dg^{-1}}{dy} \right| = f_X(\log y)/y.$$

If $X$ is exponentially distributed--specifically, $f_X(x) = \lambda e^{-\lambda x}$, then the density of $Y = e^X$ is given by $$f_Y(y) = f_X(\log y)/y = \lambda e^{-\lambda \log y}/y = \lambda y^{-\lambda}/y = \lambda y^{-\lambda-1}, \quad y > 1.$$

  • $\begingroup$ I thought it was pretty unclear as well - the title is the question exactly as it is posted on my assignment $\endgroup$ – Yak Attack Jan 27 '15 at 3:47
  • $\begingroup$ I think your solution is a correct one for an interpretation of the question, but it seems to go into greater complexity than we have reached at this point in the course (i.e. I think the solution I am looking for is a much simpler one) - maybe by the end of the semester I will be able to understand it :) very well explained though, and thanks for the answer. $\endgroup$ – Yak Attack Jan 27 '15 at 3:58
  • $\begingroup$ I just noticed that your answer is exactly what I sought, but you also showed the inverse side of it - I'm starting to get this now - thanks again, sorry it took me a few re-reads to catch it. $\endgroup$ – Yak Attack Jan 27 '15 at 4:15

$F_Y(y)=\mathbb P(Y\le y)=\mathbb P(e^X\le y)=\mathbb P(X\le\ln y)=F_X(\ln y).$


Please consider the comment, if $Y=X^2$ then

$F_Y(y)=\mathbb P(Y \le y)=\mathbb P(X^2\le y)=\mathbb P(\pm X\leq \sqrt y),$

here you would have to be more careful, possibly.

  • $\begingroup$ I might include a little more explanation, but I think I follow, and this seems like what they would be looking for. $\endgroup$ – Yak Attack Jan 27 '15 at 3:56
  • $\begingroup$ @YakAttack when you do these sort of exercises where $Y=g(X)$, which means that $Y$ is a function of $X$, in this case $g(x)=e^x$, you have to make sure that the inequality $g(X)\leq x$ holds for every possible value of $X$!!! Also, when you apply the inverse of the function $g$ make sure that $X\le g^{-1}(x)$ holds for every value of $X$. This is very important!!!!!!!!! $\endgroup$ – Vladimir Vargas Jan 27 '15 at 4:03
  • $\begingroup$ I see, thank you for the explanation. I think you are getting at what @heropup showed in his answer with the inverse functions, correct? $\endgroup$ – Yak Attack Jan 27 '15 at 4:16
  • $\begingroup$ @YakAttack heropup showed the general approach. However I'm telling you to be careful when you apply the inverse functions. $\endgroup$ – Vladimir Vargas Jan 27 '15 at 4:19
  • $\begingroup$ I don't think the last equality is correct. The stated probability is impossible, it should be $\mathbb P(X \leq -\sqrt y\text{ or } X \geq \sqrt y)$, shouldn't it? $\endgroup$ – rubik Jul 19 '16 at 7:41

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