I'm trying to deduce the marginal cdf of $Y$ in Exponentiated Weibull-logarithmic Distribution from this paper: Exponentiated Weibull-logarithmic Distribution: Model, Properties and Applications In page 3 of the paper I find this:

Given $N$, let $X_1,...,X_N$ be $iid$ from Exponentiated Weibull Distribution. Let $N$ is distributed according to the logarithmic distribution with pdf

$P(N=n) = \frac{\theta^n}{-nlog(1-\theta)}, n = 1,2,..., \theta > 0$

Let $Y = max(X_1,...,X_N)$ then the conditional cdf of $Y|N = n$ is given by

$$F_{Y|N=n}(y) = [1-e^{-(\beta y)^{\gamma}}]^{n \alpha}$$

So what I'm doing is trying to find the joint cdf of $Y$ and $N$ and then obtain the marginal cdf of $Y$ doing this

$F_{Y|N=n}(y) = \frac{F_{Y,N}}{F_N}$

$[1-e^{-(\beta y)^{\gamma}}]^{n \alpha} = \frac{F_{Y,N}}{\frac{\theta^n}{-nlog(1-\theta)}}$

$F_{Y,N} = {\frac{\theta^n}{-nlog(1-\theta)}}\cdot [1-e^{-(\beta y)^{\gamma}}]^{n \alpha}$

How I can go from $F_{Y,N}$ to $F_Y$. Based on the paper, $F_Y$ should be equal to:

$F(y) = \frac{log[1-\theta(1-e^{-(\beta y)^{\gamma}})^\alpha]}{log(1-\theta)}$

Sorry If there's something obvious there but I don't know how to approach this problem.

  • 1
    $\begingroup$ Bayes formula applies to pdf's not cdf's. $\endgroup$
    – Xi'an
    Jul 11 at 18:05
  • $\begingroup$ Thanks! my mistake $\endgroup$
    – Seb
    Jul 11 at 19:26


$$ P(Y\leq y | N=n) = \left[F_X(y)\right]^n $$

$$ P(Y\leq y) = \sum_{n=1}^\infty P(Y\leq y | N=n) P(N=n) $$

$$ = \sum_{n=1}^\infty \left[F_X(y)\right]^n P(N=n) $$

$$ = (\ln(1-\theta))^{-1}\sum_{n=1}^\infty -\left[\theta F_X(y)\right]^n/n $$

  • $\begingroup$ thanks for the hint, sorry for my inexperience how can compute that sum over n? Is there a property for this that I don't know about? $\endgroup$
    – Seb
    Jul 11 at 19:08
  • $\begingroup$ Have a look at Taylor series for $\ln (1-x)$ en.wikipedia.org/wiki/Taylor_series#Natural_logarithm . $\endgroup$
    – ir7
    Jul 11 at 19:15
  • $\begingroup$ Thanks, I couldn't remember where I saw that expression. But by looking at the wikipedia link I managed to solve it. $\endgroup$
    – Seb
    Jul 11 at 19:30
  • $\begingroup$ Great. Glad it helped. $\endgroup$
    – ir7
    Jul 11 at 19:36

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.