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someone told me that the derived distribution of the following function of RVs is a power law distribution. The special thing is that such conclusion was said to me without having knowledge of the original pdfs of X and Y ...I believe such a conclusion is not possible to establish without knowing information about X and Y distributions. What do you reckon??...the function is: \begin{equation} Z = g(X,Y)=X+(Y-X)^b \end{equation} Where X and Y are positive independent RVs with general pdf $f_x(x)$ and $f_y(y)$ respectively and $0 < b < 1$ is a positive constant.....

I've tried to get the derived pdf..but you reach a point where you need the pdfs of X and Y to establish whether or not the derived distribution of g(X,Y) follows a power law distribution. \begin{equation} P(X+(Y-X)^b \le z) = P(Y \le (z-X)^{1/b}+X) \end{equation} After some algebra: \begin{equation} P(Y \le (z-X)^{1/b}+X) = \int_0^\infty\int_0^{(z-X)^{1/b}+X}f_x(x)f_y(y)dydx \end{equation} I think here the only way go further is knowing the pdfs....

Thanks...

Philip
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It's quite obvious that the distribution of $g(X,Y)$ will depend on the distributions of $X$ and $Y$. Think of the case where $X$ and $Y$ are constants. Then $X + (Y-X)^b$ is a constant, but you need to know the values of $X$ and $Y$ to tell which constant.

By the way, if $0 < b < 1$ you'll have a problem if $Y < X$ with nonzero probability.

Robert Israel
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