# Exponentiated Weibull-logarithmic Distribution

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.

• Bayes formula applies to pdf's not cdf's. Jul 11 at 18:05
• Thanks! my mistake
– Seb
Jul 11 at 19:26

Hint:

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

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

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

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

• 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?
– Seb
Jul 11 at 19:08
• Have a look at Taylor series for $\ln (1-x)$ en.wikipedia.org/wiki/Taylor_series#Natural_logarithm .
– ir7
Jul 11 at 19:15
• Thanks, I couldn't remember where I saw that expression. But by looking at the wikipedia link I managed to solve it.
– Seb
Jul 11 at 19:30