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\(Z=XY\) and \(W=Y\) and solve I have \(Y=W , X=Z/W\)

so \(|J|=1/W\) and,

\(f_X,_Y(x,y) = \frac{\beta^{\alpha}x^{\alpha-1}e^{- \beta x}}{\Gamma{(\alpha)}}* \frac{{\Gamma(\beta)}y^{\gamma-1}(1-y)^{\beta - \gamma-1}}{{\Gamma(\gamma)}{\Gamma(\beta - \gamma)}}*\frac{1}{y}\)

now

\(f_Z,_W(z,w)= \frac{\beta^{\alpha}\Gamma(\beta)}{\Gamma(\alpha)\Gamma(\gamma)\Gamma(\beta - \gamma)} *{(\frac{z}{w})}^{\alpha-1}*e^{- \beta (\frac{z}{w})}*w^{\gamma-1}*(1-w)^{\beta - \gamma-1}*\frac{1}{w}\)

so

\(f_Z(z)= \frac{\beta^{\alpha}\Gamma(\beta)}{\Gamma(\alpha)\Gamma(\gamma)\Gamma(\beta - \gamma)}* \int_{0}^{1}{(\frac{z}{w})}^{\alpha-1}*e^{- \beta (\frac{z}{w})}*w^{\gamma-1}*(1-w)^{\beta - \gamma-1}*\frac{1}{w}dw \)

The support of Beta distribution is [0,1] and W=Y~Beta.

Now I do not know how to continue.

Let \( X \sim \text{Gamma}(\alpha, \beta), Y \sim \text{Beta}(\gamma, \alpha - \gamma) \) and they are independent. Then \( Z = XY \sim \text{Gamma}(\gamma, \beta) \)

The key fact here is that the sum of the two Beta's parameters need to equal to the shape parameter of the Gamma distribution. Otherwise there will be no nice result.

Continue your integration with some careful simplifications and a change of variable, then it will be done. I have just verified it once.

Let \( X \sim \text{Gamma}(\alpha, \beta), Y \sim \text{Beta}(\gamma, \alpha - \gamma) \) and they are independent. Then \( Z = XY \sim \text{Gamma}(\gamma, \beta) \)

The key fact here is that the sum of the two Beta's parameters need to equal to the shape parameter of the Gamma distribution. Otherwise there will be no nice result.

Continue your integration with some careful simplifications and a change of variable, then it will be done. I have just verified it once.

Note that it will be also easy to see that

\( XY|Y = y \) has the same distribution as \( yX \sim \text{Gamma} \left (\alpha, \frac {\beta} {y}\right)\)

as \( X,Y \) are independent.