X_1 \sim \chi^2(k)
Y=X_1+X_2 \sim \chi^2(l).
X_1, X_2 - iid.
M_{X_1}(t)\cdot M_{X_2}(t) = M_Y(t).
(1-2\,t)^{-k/2} \cdot M_{X_2}(t) = (1-2\,t)^{-l/2}
A więc M_{X_2}(t) = (1-2\,t)^{-(l-k)/2}. Z iniekcji mgf-ów w rozkłady wynika, że X_2 \sim \chi^2(l-k).
X_i \sim \mathrm{Pois}(\lambda_i). X_i są parami iid.
Y = \sum_i X_i.
M_Y(t) = M_{X_1}(t)\cdots M_{X_n}(t) = \exp(\lambda_1 (e^t-1)) \cdots \exp(\lambda_n (e^t-1)) = \exp( (\lambda_1 + \dots + \lambda_n) (e^t-1))
A zatem Y \sim \mathrm{Pois}(\sum_i \lambda_i).
X_i są parami iid oraz E(X_i) = \mu, \; D^2(X_i) = \sigma^2.
\displaystyle E(\overline X) = E(\frac 1 n \sum_i X_i) = \frac 1 n \sum_i E(X_i) = \frac 1 n n \mu = \mu.
\displaystyle D^2(\overline X) = D^2(\frac 1 n \sum_i X_i) = \frac 1 {n^2} \sum_i D^2(X_i) = \frac 1 {n^2} n \sigma^2 = \frac {\sigma^2} n.
D_n = \left|\matrix{ 1 & -1 & -1 & \cdots & -1 \cr 1 & 1 & 0 & \cdots & 0 \cr 1 & 0 & 1 & \cdots & 0 \cr \vdots & \vdots & \vdots & \ddots & \vdots \cr 1 & 0 & 0 & \cdots & 1}\right| = (-1)^{n+1} \left|\matrix{ -1 & -1 & -1 & \cdots & -1 \cr 1 & 0 & 0 & \cdots & 0 \cr 0 & 1 & 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \ddots & \vdots \cr 0 & 0 & 0 & \cdots & 0}\right| + (-1)^{n+n} \left|\matrix{ 1 & -1 & \cdots & -1 \cr 1 & 1 & \cdots & 0 \cr \vdots & \vdots & \ddots & \vdots \cr 1 & 0 & \cdots & 1}\right| = (-1)(-1) + (-1)^{2n} D_{n-1} = D_{n-1} + 1 = n
y_1 = \overline x oraz y_k = x_k - \overline x.
y_1 - y_2 - \dots - y_n = \overline x - (x_2 + \dots + x_n) + (n-1) \overline x = n \overline x - (n\overline x - x_1) = x_1
y_1 + y_k = \overline x + x_k - \overline x = x_k. QED.
Niech Z = X+aY. Znajdźmy związek dz z dy. dZ = a\cdot dY.
\displaystyle F_Z(z) = P[Z \leq z] = P[X + aY \leq z] = \int_{-\infty}^{\infty} P[X + aY \leq z | X = x]f_X(x) \; dx = \int_{-\infty}^{\infty} P[x + aY \leq z]f_X(x) \; dx = \int_{-\infty}^{\infty} F_Y(\frac {z-x} a)f_X(x) \; dx
\displaystyle f_Z(z) = \frac d {dz} \int_{-\infty}^{\infty} F_Y(\frac {z-x} a)f_X(x) \; dx = \int_{-\infty}^{\infty} \frac {dF_Y(\frac {z-x} a)}{dz}f_X(x) \; dx = \int_{-\infty}^{\infty} \frac {dF_Y(\frac {z-x} a)}{a\cdot dy}f_X(x) \; dx = \int_{-\infty}^{\infty} \frac 1 a f_Y(\frac {z-x} a)f_X(x) \; dx
Rozwiązania zadań 6. i 7. to przypadki dla odpowiednio a=1 oraz a=-1.