# Gamma distribution proof

Consider the pdf of of $$Gamma(\alpha,\beta)$$

\begin{align} f(x;\alpha,\beta) & = \frac{ \beta^\alpha x^{\alpha-1} e^{-\beta x}}{\Gamma(\alpha)} \quad \text{ for } x > 0 \quad \alpha, \beta > 0, \\[6pt] \end{align}

My goal is to show that $$P(a \leq X_{\alpha,\beta} \leq b)=\frac{1}{\beta}\, (f_{X_{\alpha,\beta}}(a)-f_{X_{\alpha,\beta}}(b))+P(a < X_{\alpha-1,\beta} < b)$$

I am writing things out but not sure where to go from here:

$$P(a \leq X_{\alpha,\beta} \leq b)=F_{X_{\alpha,\beta}}(b)-F_{X_{\alpha,\beta}}(a)$$

Also know,

$$\Gamma(\alpha)= \int_{0}^{\infty} e^{-x} x^{\alpha-1} \, dx$$ $$\Gamma(\alpha)=(\alpha-1) \Gamma(\alpha-1)$$

Not sure where to go from here, any help is appreciated.

• If you mean $X_{\alpha, \beta} \sim \text{Gamma}(\alpha, \beta)$, then $$P(a\le X_{\alpha, \beta} \le b) = F_{X_{\alpha,\beta}}(b) - F_{X_{\alpha,\beta}}(a)$$ and $$F_{X_{\alpha, \beta}}(t)= \int_{-\infty}^t \dfrac{\beta^{\alpha} w^{\alpha-1}e^{-\beta w}}{\Gamma (\alpha)} dw \\ \implies F_{X_{\alpha,\beta}}(b) - F_{X_{\alpha,\beta}}(a)=\int_{-\infty}^b \dfrac{\beta^{\alpha} w^{\alpha-1}e^{-\beta w}}{\Gamma (\alpha)} dw - \int_{-\infty}^a \dfrac{\beta^{\alpha} w^{\alpha-1}e^{-\beta w}}{\Gamma (\alpha)} dw \\ = \int_a^b \dfrac{ \beta^{\alpha} w^{\alpha-1}e^{-\beta w} }{\Gamma (\alpha)} dw$$ Commented Oct 9, 2020 at 4:17
• Maybe, integration by parts would be a way to go from there. You could also look at the integral form of $$P(a \leq X_{\alpha,\beta} \leq b)-P(a < X_{\alpha-1,\beta} < b)$$ and try to show that it equals $$\beta \, (f_{X_{\alpha,\beta}}(a)-f_{X_{\alpha,\beta}}(b))$$ Commented Oct 9, 2020 at 4:20

It is very very easy...just to be solved in a couple of passages.

Observe that

$$P(a\leq X_{\alpha, \beta} \leq b)= \int_{a}^{b}\underbrace{\frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}}_{=f}\times \underbrace{e^{-x \beta} }_{=g'}dx=f\times g\Bigg]_a^b -\int_a^b f' \times g=$$

$$=\frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}\cdot \frac{e^{-x\beta}}{-\beta}\Bigg]_a^b+\int_a^b \frac{\beta^{\alpha}(\alpha-1)}{\Gamma(\alpha)}x^{\alpha-2}\cdot \frac{e^{-x\beta}}{\beta}dx$$

Simplify the expression and get immediately your proof

EDIT: it results to me

$$\frac{1}{\beta}[f_{X_{\alpha,\beta}}(a)-f_{X_{\alpha,\beta}}(b) ]+P(a

Can it be a typo in the posted solution which shows $$\beta$$ instead of $$\frac{1}{\beta}$$?

• Yes, It is supposed to be $\frac{1}{\beta}$. I have corrected it in the question. Thank you!
Following my second comment, \begin{aligned} &P(a \leq X_{\alpha,\beta} \leq b)-P(a < X_{\alpha-1,\beta} < b) \\ =& \int_a^b \dfrac{ \beta^{\alpha} w^{\alpha-1}e^{-\beta w} }{\Gamma (\alpha)} dw - \int_a^b \dfrac{ \beta^{\alpha-1} w^{\alpha-2}e^{-\beta w} }{\Gamma (\alpha-1)} dw \qquad \qquad (*) \\ =& \int_a^b \left(\dfrac{ \beta^{\alpha} w^{\alpha-1}e^{-\beta w} }{\Gamma (\alpha)}- \dfrac{ \beta^{\alpha-1} w^{\alpha-2}e^{-\beta w} }{\Gamma (\alpha-1)}\right)dw \\=& \dfrac{\beta^{\alpha-1}}{\alpha-1} \int_a^b \left(\dfrac{ \beta w^{\alpha-1}e^{-\beta w} }{\Gamma (\alpha-1)}- \dfrac{ (\alpha-1)w^{\alpha-2}e^{-\beta w} }{\Gamma (\alpha-1)}\right)dw \\ = & \dfrac{\beta^{\alpha-1}}{(\alpha-1)\Gamma(\alpha-1)} \int_a^b w^{\alpha-2} e^{-\beta w} \left( \beta w - (\alpha-1)\right)dw \\ = & \int_a^b \left(\dfrac{\beta^{\alpha-1} w^{\alpha-2} e^{-\beta w}}{\Gamma(\alpha)}\right) \left( \beta w - (\alpha-1)\right)dw \end{aligned} In the last line, take $$u(w)=\beta w - (\alpha-1),\ v(w)= \dfrac{\beta^{\alpha-1} w^{\alpha-2} e^{-\beta w}}{\Gamma(\alpha)}$$ and use the integration-by-parts formula $$\int_a^b u(w)v(w)dw = \left[u(w) \int v(w) dw\right]_a^b - \left[\int u'(w)\left(\int v(t)dt\right) dw \right]_a^b$$