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Math Help - Weibulls additive?

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    Weibulls additive?

    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?
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    Quote Originally Posted by colby2152 View Post
    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
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    Quote Originally Posted by The Second Solution View Post
    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!
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    Hello,
    Quote Originally Posted by colby2152 View Post
    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 )
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    Quote Originally Posted by Moo View Post
    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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