Exponentiated Weibull-logarithmic Distribution

Question

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.

Answer 1

Hint:

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

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

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

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