• April 21st 2009, 11:16 AM
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?
• April 22nd 2009, 05:47 AM
The Second Solution
Quote:

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
• April 22nd 2009, 07:18 AM
colby2152
Quote:

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!(Cool)
• April 22nd 2009, 12:37 PM
Moo
Hello,
Quote:

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 (Surprised)

(right... I just saw there had been 2 replies after that -_- anyway, question stands :p)
• April 22nd 2009, 12:50 PM
colby2152
Quote:

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 (Surprised)

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

Generally, all of the random variables are dependent upon each other, but I am assuming that they are independent for simplicity.