M(t) = \exp \{ \lambda ( e^t - 1) \}.
m_1 = M^{(1)}(0) = \exp \{ \lambda ( e^0 - 1) \} \lambda ( e^0 ) = \lambda
m_2 = M^{(2)}(0) = \exp \{ \lambda ( e^0 - 1) \} \lambda ( e^0 ) + \exp \{ \lambda ( e^0 - 1) \} \lambda^2 e^0 e^0 = \lambda + \lambda^2
D^2(X) = m_2 - m_1^2 = \lambda.
M(t) = \exp \{ \mu t + \frac 1 2 \sigma^2 t^2 \}.
m_1 = \exp \{ \mu 0 + \frac 1 2 \sigma^2 0^2 \} (\mu + 0 \cdot \sigma^2) = \mu.
m_2 = \exp \{ \mu 0 + \frac 1 2 \sigma^2 0^2 \} (\sigma^2) + \exp \{ \mu 0 + \frac 1 2 \sigma^2 0^2 \} (\mu + 0 \cdot \sigma^2) (\mu + 0 \cdot \sigma^2) = \sigma^2 + \mu^2.
D^2(X) = m_2 - m_1^2 = \sigma^2.
X_k \sim \mathrm{Pois}(\lambda_k).
\displaystyle Y = \sum_k X_k.
M_Y(t) = M_{X_1}(t) M_{X_2}(t) \cdots M_{X_n}(t) = \exp\{\lambda_1(e^t-1)\}\exp\{\lambda_2(e^t-1)\}\cdots\exp\{\lambda_n(e^t-1)\} = \exp\{(e^t-1)(\lambda_1 + \lambda_2 + \dots + \lambda_n)\}
Więc \displaystyle Y \sim \mathrm{Pois}(\sum_k \lambda_k).
X_k \sim \mathrm{N}(\mu_k,\sigma_k^2).
\displaystyle Y = \sum_k X_k.
M_Y(t) = M_{X_1}(t) M_{X_2}(t) \cdots M_{X_n}(t) = \exp\{\mu_1t + \frac 12 \sigma_1^2t^2 \}\exp\{\mu_2t + \frac 12 \sigma_2^2t^2\}\cdots\exp\{\mu_nt + \frac 12 \sigma_n^2t^2\} = \exp\{(\mu_1+\mu_2+\dots+\mu_n)t + \frac 1 2 t^2 (\sigma_1^2 + \sigma_2^2 + \dots + \sigma_n^2)\}
Więc \displaystyle Y \sim \mathrm{N}(\sum_k \mu_k,\sum_k \sigma_k^2)
Znajdźmy najpierw mgf rozkładu.
Jeśli X \sim \mathrm{B}(n,p), to
\displaystyle M_X(t) = E(e^{tX}) = \sum_{X=0}^n e^{tX} \cdot {n \choose X} p^X (1-p)^{n-X} = \sum_{X=0}^n {n \choose X} (e^t p)^X (1-p)^{n-X} = (pe^t + (1-p))^n.
X_k \sim \mathrm{B}(n_k,p).
\displaystyle Y = \sum_k X_k.
M_Y(t) = M_{X_1}(t) M_{X_2}(t) \cdots M_{X_n}(t) = (pe^t + (1-p))^{n_1}(p \cdot e^t + (1-p))^{n_2}\cdots(p \cdot e^t + (1-p))^{n_m} = (p \cdot e^t + (1-p))^{n_1+n_2+\dots+n_m}
Więc \displaystyle Y \sim \mathrm{B}(\sum_k n_k,p)
W zadaniu wystarczy zrobić podstawienie za x=1-u.
B(p,q)= \displaystyle\int_0^{1}x^{p-1}(1-x)^{q-1}dx=| x=1-u;dx=du|=\int_0^{1}(1-u)^{p-1}u^{q-1}du=B(q,p)
B(p,q)= \displaystyle\int_0^{1}x^{p-1}(1-x)^{q-1}dx=| u'=x^{p-1};u=\frac {x^p} {p};v'=-(1-x)^{q-1};v=
\displaystyle \frac {(1-x)^q} {q}|=\frac 1 {p} x^p(1-x)^{q-1}|^1_0 + \frac 1 {p} \int_0^{1}x^{p}(q-1)(1-x)^{q-2}dx=
\displaystyle 0 + \frac {q-1} p(\int_0^{1}x^{p-1}(1-x)^{q-2}dx-\int_0^{1}x^{p-1}(1-x)^{q-1}dx)=
\displaystyle \frac {q-1} {p+q-1}\int_0^{1}x^{p-1}(1-x)^{q-2}dx=\frac {q-1} {p+q-1}B(p,q-1)
Można stwierdzić, że to jest szczególny przypadek zadania 11., ale można również rozwiązać bez niego:
\displaystyle \mathrm{B}(p,q) = \frac {q-1}{p+q-1} \mathrm{B}(p,q-1) = \frac {(q-1)(q-2)}{(p+q-1)(p+q-2)} \mathrm{B}(p,q-1) = \dots = \frac {(q-1)(q-2)\cdots 1}{(p+q-1)(p+q-2)\cdots(p-1)} \mathrm{B}(p,1) = \frac {\Gamma(p)\Gamma(q)}{\Gamma(p+q)}
\displaystyle \Gamma(p)\Gamma(q) = \int_0^{\infty}\int_0^{\infty}x^{p-1}y^{q-1}e^{-x-y}\;dxdy
Podstawmy x = s(1-t) oraz y = st. \displaystyle t = \frac y s = \frac {y(1-t)} x czyli \displaystyle t(1+\frac y x) = \frac y x. Więc \displaystyle t = \frac y {x+y} oraz s = x+y. Policzmy jakobian:
J = \left| \begin{array} {cc} \frac{ds}{dx} & \frac{ds}{dy} \cr \frac{dt}{dx} & \frac{dt}{dy} \end{array}\right| = \left| \begin{array} {cc} 1 & 1 \cr -\frac{y}{(x+y)^2} & \frac 1 {x+y}-\frac{y}{(x+y)^2} \end{array}\right| = \frac 1 {x+y}
a więc \displaystyle \Gamma(p)\Gamma(q) = \int_0^{1}\int_0^{\infty}s^{p-1}(1-t)^{p-1}s^{q-1}t^{q-1}e^{-s} \frac 1 s\;dsdt = \int_0^{1} (1-t)^{p-1}t^{q-1}\; dt \cdot \int_0^{\infty}s^{p-1}s^{q-1}e^{-s} \frac 1 s\;dt = \mathrm{B}(p,q) \Gamma(p+q)
X \sim \Gamma(b,p_1), Y \sim \Gamma(b,p_2).
Z = X+Y.
\displaystyle M_Z(t) = M_X(t)M_Y(t) = (1-\frac t b)^{-p_1}(1-\frac t b)^{-p_2} = (1-\frac t b)^{-(p_1+p_2)}.
A więc Z \sim \Gamma(b,p_1+p_2).