$X$ ~ $Gamma(a, b)$. Find $E(X)$

Question

$X$ ~ $Gamma(a, b)$. Find E(X)

$$f_X(x) = \frac{b^{a}x^{a-1}e^{-bx}}{T(a)}$$

$E(X) = \int_{0}^{\infty}x f_X(x) dx = \frac{b^a}{T(a)} \int_{0}^{\infty}x^{a-1}e^{-bx}dx = \frac{b^a}{T(a)b^{a+1}}T(a+1) = \frac{b^{a}}{b^{a}b} \cdot \frac{a(a-1)!}{(a-1)!} = \frac{a}{b}$

Actually, I don't get it. How does $$\int_{0}^{\infty}x^{a-1}e^{-bx} dx = \frac{T(a+1)}{b^{a+1}}$$

attempt:

$$\int_{0}^{\infty} x^{a-1}e^{-bx} dx$$

let $u = bx \leftrightarrow x = \frac{u}{b}$, $du = b dx$

$$\frac{1}{b}\int_{0}^{\infty}\left(\frac{u}{b} \right)^{a-1}e^{-u} du = \frac{1}{b^{a}} \int_{0}^{\infty} u^{a-1}e^{-u} = \frac{T(a-1+1)}{b^a} = \frac{T(a)}{b^a}$$

What am I doing wrong?

Answer 1

When substituting in the density into $\int_{0}^{\infty} xf_X(x)dx$, you may have missed the added $x$ from the expectation, so it should be equal to $\frac{b^a}{\Gamma(a)}\int_{0}^{\infty} x^a e^{-bx}dx$. The rest of your algebra should work out fine with this change.