\displaystyle \mathrm{E}(g(X,Y))=\int_{-\infty}^\infty \int_{-\infty}^\infty g(x,y) f(x,y)\,\mathrm{d}x \,\mathrm{d}y.
Zmienne X i Y są niezależne, więc f(x,y) = f_X(x)f_Y(y).
Więc \displaystyle \mathrm{E}(g(X)h(Y))=\int_{-\infty}^\infty \int_{-\infty}^\infty g(x)h(y) f(x,y)\,\mathrm{d}x \,\mathrm{d}y = \int_{-\infty}^\infty \int_{-\infty}^\infty g(x) h(y) f_X(x)f_Y(y)\,\mathrm{d}x \,\mathrm{d}y =
\displaystyle \int_{-\infty}^\infty h(y) f_Y(y) \int_{-\infty}^\infty g(x) f_X(x)\,\mathrm{d}x \,\mathrm{d}y = \displaystyle \int_{-\infty}^\infty g(x) f_X(x)\,\mathrm{d}x \int_{-\infty}^\infty h(y) f_Y(y) \,\mathrm{d}y = \mathrm{E}(g(X))\mathrm{E}(h(Y))
\displaystyle M_x( t ) = (1-p+pe^t)^n
\displaystyle M_x '( t ) = n \cdot (1-p+pe^t)^{n-1} \cdot (1-p+pe^t)' = n \cdot (1-p+pe^t)^{n-1} \cdot pe^t
\displaystyle M_x ' '( t ) = np (e^t (1-p+pe^t)^{n-1} + e^t (n-1) (1-p+pe^t)^{n-2} p e^t )
\displaystyle E(X) = M_x'(0) = np
\displaystyle D^2(X) = E(X^2) - E(X)^2 = M_x' '(0) - n^2p^2 = np(1+(n-1)p) - n^2p^2 =
np + n^2p^2 - np^2 - n^2p^2 = np - np^2 = np(1-p) = npq
\displaystyle M_{aX+b}(t) = \int_{-\infty}^{\infty} e^{t(ax+b)} f(x) dx = e^{tb} \int_{-\infty}^{\infty} e^{tax} f(x) dx = e^{tb} M_{x}(at) = \mathrm{exp}\{tb + at\mu + \frac 1 2 a^2t^2\sigma^2\}
\displaystyle M_Y(t) = M_{\sum_i X_i}(t) = M_{X_1}(t)M_{X_2}(t)\cdots M_{X_n}(t) =
\mathrm{exp}\{t\mu_1 + \frac 1 2 t^2\sigma_1^2\}\mathrm{exp}\{t\mu_2 + \frac 1 2 t^2\sigma_2^2\}\cdots \mathrm{exp}\{t\mu_n + \frac 1 2 t^2\sigma_n^2\} =
\mathrm{exp}\{t(\mu_1 + \mu_2 + \dots + \mu_n) + \frac 1 2 t^2(\sigma_1^2 + \sigma_2^2 + \dots + \sigma_n^2)\}
Z iniekcji mgf-ów w rozkłady wynika, że Y \sim \mathcal{N}(\sum_i \mu_i, \sum_i \sigma_i^2).
\displaystyle \Gamma(p+1) = \int_0^{\infty} x^pe^{-x} dx = [x^p e^{-x}]|^\infty_0 + \int_0^{\infty} p x^{p-1} e^{-x} dx = p \Gamma(p)
\displaystyle M_x( t ) = (1- \frac t b )^{-p}
\displaystyle M_x'( t ) = -p(1-\frac t b)^{-p-1} (-\frac 1 b)
\displaystyle M_x''( t ) = \frac p b (-p-1) (1 - \frac t b)^{-p-2} (-\frac 1 b)
\displaystyle \mathrm{E}(X) = M_x'(0) = \frac p b
\displaystyle \mathrm{D}^2(X) = M_x' '(0) - \mathrm{E}(X)^2 = \frac {p(p+1)} {b^2} - \frac {p^2} {b^2} = \frac p {b^2}
Robimy podstawienie Z = X^2, dZ = 2XdX = 2\sqrt Z dX.
Pamiętajmy, że musimy pomnożyć przez czynnik 2,
bo zbiór wartości zmienia się z (-\infty, \infty) na [0, \infty).
\displaystyle f(z)dz = \frac 1 {\sqrt{2\pi}} e^{\frac 1 2 z} \frac 1 {2 \sqrt z} \cdot 2 dz. A więc \displaystyle f(z) = \frac 1 {\Gamma(1/2) 2^{1/2}} e^{z\cdot1/2} z^{-1/2}.
A zatem Z \sim \chi^2(1).
Mgf rozkładu chi-kwadrat z k stopniami swobody:
\displaystyle M_x(t) = (1-2\,t)^{-k/2}
\displaystyle Y \sim \sum^n_{i=1}\mathrm{X}^2(1)(0,1)
\displaystyle M_y(t) = \prod^n_{i=1}(1-2\,t)^{-1/2} = (1-2\,t)^{-n/2}
\displaystyle Y \sim \mathrm{X}^2(n)