====== Wnioskowanie Statystyczne - Lista 2. ======
===== Zadanie 1. =====

$X_1 \sim \chi^2(k)$

$Y=X_1+X_2 \sim \chi^2(l).$

$X_1, X_2$ - 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 $X_2 \sim \chi^2(l-k)$.

===== Zadanie 2. =====

$X_i \sim \mathrm{Pois}(\lambda_i).$ $X_i$ 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 $Y \sim \mathrm{Pois}(\sum_i \lambda_i).$

===== Zadanie 3. =====

$X_i$ są parami iid oraz $E(X_i) = \mu, \; D^2(X_i) = \sigma^2$.

$\displaystyle E(\overline X) = E(\frac 1 n \sum_i X_i) = \frac 1 n \sum_i E(X_i) = \frac 1 n n \mu = \mu$.

$\displaystyle D^2(\overline X) = D^2(\frac 1 n \sum_i X_i) = \frac 1 {n^2} \sum_i D^2(X_i) = \frac 1 {n^2} n \sigma^2 = \frac {\sigma^2} n$.

===== Zadanie 4. =====

$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$

===== Zadanie 5. =====

$y_1 = \overline x$ 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.

===== Zadanie 6. i 7. =====

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 $a=-1$.


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