1. ## Weibulls additive?

Are random variables for a Weibull distribution additive?

Let's say there are random variables $\displaystyle X$ and $\displaystyle Y$ with different Weibull distributions. The Weibulls are very similar, but not one and the same as both parameters vary slightly.

How would the distribution of $\displaystyle Z = X + Y$ look like? Would it be Weibull, and if so, what are the parameters?

2. Originally Posted by colby2152
Are random variables for a Weibull distribution additive?

Let's say there are random variables $\displaystyle X$ and $\displaystyle Y$ with different Weibull distributions. The Weibulls are very similar, but not one and the same as both parameters vary slightly.

How would the distribution of $\displaystyle Z = X + Y$ look like? Would it be Weibull, and if so, what are the parameters?
How to get the distribution is explained in http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf

3. Originally Posted by The Second Solution
How to get the distribution is explained in http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf
That is more for discrete distributions. How about I lay down my problem here for everyone.

I actually have four random variables based on slightly different three-parameter (beta shift) Weibull distributions.

$\displaystyle f(x_i)=\frac{\gamma_i}{\alpha_i}\left(\frac{x_i-\beta}{\alpha_i}\right)^{\gamma_i-1} e^{-\left(\frac{x_i-\beta}{\alpha_i}\right)^{\gamma_i}}, i: 1 \rightarrow 4$

Notice that $\displaystyle \beta$ is the same for all PDFs. I want to find the PDF, and more importantly, the CDF of the random variable $\displaystyle z=\sqrt{x_1*x_4}+\sqrt{x_2*x_3}$.

It seems complicated, but treating z with the same parameters as one of the x's would lose a lot of possible outcomes. In my calculations, forcing z to follow an x would lose 44% of possible outcomes.

Therefore, I either need a whole new distribution that is dependent upon the four Weibulls OR I need to form a new Weibull with new parameters dependent upon the parameters from each of the four Weibulls.

Thanks for anyone willing to look into this!

4. Hello,
Originally Posted by colby2152
Are random variables for a Weibull distribution additive?

Let's say there are random variables $\displaystyle X$ and $\displaystyle Y$ with different Weibull distributions. The Weibulls are very similar, but not one and the same as both parameters vary slightly.

How would the distribution of $\displaystyle Z = X + Y$ look like? Would it be Weibull, and if so, what are the parameters?
Are they independent ?

If so, I don't think it is additive, because I can barely see the product of the mgf to be the mgf of a Weibull

(right... I just saw there had been 2 replies after that -_- anyway, question stands )

5. Originally Posted by Moo
Hello,

Are they independent ?

If so, I don't think it is additive, because I can barely see the product of the mgf to be the mgf of a Weibull

(right... I just saw there had been 2 replies after that -_- anyway, question stands )
Generally, all of the random variables are dependent upon each other, but I am assuming that they are independent for simplicity.