Beta of transformed gamma variable

Question

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?

Answer 1

This is much simpler than you make it. First, as gamma (also exponential) distributions have different parametrizations in use, you need to specify which you mean. From your formulas, you seem to assume that $Y_i$ has exponential distributions with scale parameter $\beta$, and then the sum $S$ of $n$ independent copies is $\mathcal{Gamma}(n, \beta)$, using the shape/scale parametrization.

The transformation to $U$ is now simply a multiplication with $2/\beta$, and this simply multiplies the scale parameter with the same constant, giving that $$ U=\frac{2S}{\beta} \sim \mathcal{Gamma}(n, 2) .$$ There is no need for your use of Jacobians!