I'm trying to transform a gamma distributed variable with $\alpha=n$ and $\beta=15$ using the following formula:

$$U=\frac{2S}{\beta}$$

S is actuallya summation of n exponential variables ($Y_i$), with $\beta=15$

When trying to derive the probability distribution for U I get different values for $\beta$ in my calculations and in SAS. I tried to change the equation, but I'm not sure this is valid...

I'm using the method of transformations:

$$h^{-1}(U)=\frac{U\beta}{2}$$ $$\frac {dh^{-1}}{du}=\frac{\beta}{2}$$

The gamma density function is: $$\frac{y^{\alpha -1}e^{-\frac{y}{\beta}}}{\beta^\alpha\Gamma(\alpha)}$$

$$h^{-1}(u)=\frac{\beta}{2}{\sum U_i}$$

$$fu(u)=fs[h^{-1}(u)]|\frac{dh^{-1}}{du}|$$

substituting into the gamma function, using the method of transformation

$$\sum \frac {\frac {\beta}{2}u_i^{\alpha^{-1}}e^-{\frac {\sum u_i}{2}}}{\beta^\alpha \Gamma(n)}\frac{\beta}{2}$$

$$\frac {2}{\beta}\frac{\beta}{2}\sum \frac {u_i^{\alpha^{-1}}e^-{\frac {\sum u_i}{2}}}{\beta^\alpha \Gamma(n)}$$

$$\sum \frac {u_i^{\alpha^{-1}}e^-{\frac {\sum u_i}{2}}}{\beta^\alpha \Gamma(n)}$$

I draw the conclusion that $\beta=2$, which is the answer i get from SAS.

Is this the right approach? Have I done the transformation correctly?

I'm trying to transform a gamma distributed variable with $\alpha=n$ and $\beta=15$ using the following formula:

$$U=\frac{2S}{\beta}$$

S is actually a summation of n exponential variables ($Y_i$), with $\beta=15$

When trying to derive the probability distribution for U I get different values for $\beta$ in my calculations and in SAS. I tried to change the equation, but I'm not sure this is valid...

I'm using the method of transformations:

$$h^{-1}(U)=\frac{U\beta}{2}$$ $$\frac {dh^{-1}}{du}=\frac{\beta}{2}$$

The gamma density function is: $$\frac{y^{\alpha -1}e^{-\frac{y}{\beta}}}{\beta^\alpha\Gamma(\alpha)}$$

$$h^{-1}(u)=\frac{\beta}{2}{\sum U_i}$$

$$f_u(u)=f_s[h^{-1}(u)]|\frac{dh^{-1}}{du}|$$

substituting into the gamma function, using the method of transformation

$$\sum \frac {\frac {\beta}{2}u_i^{\alpha^{-1}}e^-{\frac {\sum u_i}{2}}}{\beta^\alpha \Gamma(n)}\frac{\beta}{2}$$

$$\frac {2}{\beta}\frac{\beta}{2}\sum \frac {u_i^{\alpha^{-1}}e^-{\frac {\sum u_i}{2}}}{\beta^\alpha \Gamma(n)}$$

$$\sum \frac {u_i^{\alpha^{-1}}e^-{\frac {\sum u_i}{2}}}{\beta^\alpha \Gamma(n)}$$

I draw the conclusion that $\beta=2$, which is the answer i get from SAS.

Is this the right approach? Have I done the transformation correctly?