Oczekiwanie kwadratu Gamma

Jeśli rozkład gamma jest sparametryzowany za pomocą i β , to:$\alpha$$\beta$

$$E(\Gamma(\alpha, \beta)) = \frac{\alpha}{\beta}$$

Chciałbym obliczyć oczekiwaną kwadratową gamma, to znaczy:

$$E(\Gamma(\alpha, \beta)^2) = ?$$

Myślę, że to jest:

$$E(\Gamma(\alpha, \beta)^2) = \left(\frac{\alpha}{\beta}\right)^2 + \frac{\alpha}{\beta^2}$$

Czy ktoś wie, czy to ostatnie wyrażenie jest poprawne?

Oczekiwanie kwadratu dowolnej zmiennej losowej to jego wariancja plus oczekiwane kwadratowe, as

$\mathbb{D}^2(X)=\mathbb{E}([X-\mathbb{E}(X)]^2)=\mathbb{E}(X^2)-[\mathbb{E}(X)]^2 \Rightarrow \mathbb{E}(X^2) = \mathbb{D}^2(X)+[\mathbb{E}(X)]^2$

$\Gamma$$\alpha/\beta$ $\alpha/\beta^2$

$(\alpha/\beta)^2+\alpha/\beta^2$

To znaczy: masz rację.

$$f_X(x) = \frac{\beta^\alpha x^{\alpha-1} e^{-\beta x}}{\Gamma(\alpha)}, \quad x > 0.$$ $\alpha, \beta > 0$ $$\int_{x=0}^\infty f_X(x) \, dx = 1,$$ $$\int_{z=0}^\infty x^{z-1} e^{-z} \, dz = \Gamma(z).$$ $k$ $$\begin{align*} \mathrm{E}[X^k] &= \int_{x=0}^\infty x^k f_X(x) \, dx \\ &= \frac{1}{\Gamma(\alpha)} \int_{x=0}^\infty \beta^\alpha x^{\alpha+k-1} e^{-\beta x} \, dx \\ &= \frac{\Gamma(\alpha+k)}{\beta^k \Gamma(\alpha)} \int_{x=0}^\infty \frac{\beta^{\alpha+k} x^{\alpha+k-1} e^{-\beta x}}{\Gamma(\alpha+k)} \, dx \\ &= \frac{\Gamma(\alpha+k)}{\beta^k \Gamma(\alpha)}, \end{align*}$$ $1$$\alpha+k$$\beta$$k = 2$$\mathrm{E}[X^2] = \frac{\Gamma(\alpha+2)}{\beta^2 \Gamma(\alpha)} = \frac{(\alpha+1)\alpha}{\beta^2}.$ $$\begin{align*} M_X(t) = \mathrm{E}[e^{tX}] &= \int_{x=0}^\infty \frac{\beta^\alpha x^{\alpha-1} e^{-\beta x + tx}}{\Gamma(\alpha)} \, dx \\ &= \frac{\beta^\alpha}{(\beta-t)^\alpha} \int_{x=0}^\infty \frac{(\beta-t)^\alpha x^{\alpha-1} e^{-(\beta-t)x}}{\Gamma(\alpha)} \, dx \\ &= \biggl(\frac{\beta}{\beta-t}\biggr)^{\!\alpha}, \quad t < \beta, \end{align*}$$ $t$ $$M_X(t) = (1 - t/\beta)^{-\alpha},$$ $$\mathrm{E}[X^k] = \left[ \frac{d^k M_X(t)}{dt^k} \right]_{t=0} = \left[(1-t/\beta)^{-\alpha-k}\right]_{t=0} \prod_{j=0}^{k-1} \frac{\alpha+j}{\beta} = \frac{\Gamma(\alpha+k)}{\beta^k \Gamma(\alpha)}.$$