Weibulls additive?

**colby2152** · April 21st 2009, 10:16 AM

Are random variables for a Weibull distribution additive?

Let's say there are random variables $X$ and $Y$ 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 $Z = X + Y$ look like? Would it be Weibull, and if so, what are the parameters?

**The Second Solution** · April 22nd 2009, 04:47 AM

Originally Posted by colby2152

Are random variables for a Weibull distribution additive?

Let's say there are random variables $X$ and $Y$ 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 $Z = X + Y$ 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

**colby2152** · April 22nd 2009, 06:18 AM

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.

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

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

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!

**Moo** · April 22nd 2009, 11:37 AM

Hello,

Originally Posted by colby2152

Are random variables for a Weibull distribution additive?

Let's say there are random variables $X$ and $Y$ 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 $Z = X + Y$ 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

)

**colby2152** · April 22nd 2009, 11:50 AM

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.

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