Question

Let X ~ Gamma(k, β) and Y ~ Gamma(k, 1) Gamma( α, 3) Cx Show that Y = 스 is a pivot

Answer 1

Given the Gamma PDF $f\left ( x|\alpha ,\beta \right )=\frac{1}{\Gamma \left ( \alpha \right )\beta ^{\alpha }}x^{\alpha -1}x^{-x/\beta };x>0$ . Now we will find the PDF of $Y=\frac{X}{\beta }$ .

Here $X=\beta Y$ . The PDF of $Y=\frac{X}{\beta }$ is

$f_Y\left ( y|\alpha ,\beta \right )=\frac{1}{\Gamma \left ( \alpha \right )\beta ^{\alpha }}\left ( \beta y \right )^{\alpha -1}x^{-y }\left | \frac{\mathrm{d} \left ( \beta y \right )}{\mathrm{d} y} \right |\\ f_Y\left ( y|\alpha ,\beta \right )=\frac{1}{\Gamma \left ( \alpha \right )\beta ^{\alpha }}\left ( \beta y \right )^{\alpha -1}x^{-y }\left | \beta \right |\\ f_Y\left ( y|\alpha ,\beta \right )=\frac{\beta^{\alpha }}{\Gamma \left ( \alpha \right )\beta ^{\alpha }} y ^{\alpha -1}x^{-y }\\ f_Y\left ( y|\alpha \right )=\frac{1}{\Gamma \left ( \alpha \right )} y ^{\alpha -1}x^{-y };y>0$

Thus the distribution of $Y\sim Gamma\left ( \alpha ,1 \right )$ . Since the distribution is independent of $\beta$ , the quantity $Y=\frac{X}{\beta }$ is a pivot.