====== Rachunek prawdopodobieństwa i statystyka - Lista 6. ======

===== Zadanie 1. =====

^ $x_i$ | -3 | -1 | 0 | 1 | 2 | 3 |
^ $p_i$ | 0.08 | 0.16 | 0.25 | 0.36 | 0.1 | 0.05 |

$Y = X^2$.

$\displaystyle P(Y = y_i) = \sum_{x_i: \; x_i^2 = y_i} p_i$, więc:

^ $y_i$ | 0 | 1 | 4 | 9 |
^ $q_i$ | 0.25 | 0.52 | 0.1 | 0.13 |

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

$\sin$ jest okresowy, więc $y_i$ jest tylko trzy.

$\displaystyle P(Y=-1) = \sum_{n: \; \sin (\pi n/2 ) = -1} 2^{-n} = \sum_k 2^{-4k-3} = \frac 1 8 \sum_k 2^{-4k} = \frac 1 8 \sum_k (\frac 1 {16})^{k} = \frac 1 8 \cdot \frac 1 {1 - \frac 1 {16}} = \frac {16} {120}$

$\displaystyle P(Y=1) = \sum_{n: \; \sin (\pi n/2 ) = 1} 2^{-n} = \sum_k 2^{-4k-1} = \frac 1 2 \sum_k 2^{-4k} = \frac {64} {120}$

$\displaystyle P(Y=0) =  \sum_{n: \; \sin (\pi n/2 ) = 0} 2^{-n} = \sum_k 2^{-4k-2} + \sum_k 2^{-4k-4} = \frac {32} {120} + \frac {8} {120}$

^ $y_i$ | -1 | 0 | 1 |
^ $P(Y=y_i)$ | $\frac {16} {120}$ | $\frac {40} {120}$ | $\frac {64} {120}$ |

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

$\displaystyle f(x) = \frac 1 {2 \sqrt {\pi}} \exp \{ - \frac 1 2 x^2 \}$.

$y = 3x+1$. Więc $x(y) = \frac 1 3 (y-1)$.

$g(y) = f(x(y)) |h'(y)|$, więc $\displaystyle g(y) = \frac 1 {6 \sqrt {\pi}} \exp \{ - \frac 1 {18} (y-1)^2 \}$

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

$f(x) = 3\exp \{ -3x \}$, dla $x > 0$.

$Y = X^2$. Więc $x(y) = y^{1/2}$.

$g(y) = 3 \exp \{ -3y^{1/2} \} \cdot | y^{-1/2} |$. Jeśli $y$ jest rzeczywiste, można zdjąć moduł.

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

$f(x) = \frac 1 {\sqrt {2 \pi}} \exp \{ - \frac 1 2 x^2 \}$.

$Y=X^2$. Więc $x(y) = y^{1/2}$.

$\displaystyle g(y) = \frac 2 {\sqrt {8 \pi y}} \exp \{ - \frac 1 2 y \} = \frac { (\frac 1 2)^{\frac 1 2} } {\Gamma(1/2)} \cdot y^{-1/2} \cdot \exp \{ - \frac 1 2 y \}$. 

A więc $Y \sim \Gamma(1/2, 1/2)$.

===== Zadanie 6. =====

$E(Z) = E( (X-aY)^2) = a^2E(Y^2) - 2aE(XY) + E(X^2) \geq 0$.

$\Delta = 4(E(XY) )^2 - 4(E(Y^2)E(X^2) ) \leq 0$.

$(E(XY) )^2 - E(Y^2)E(X^2) \leq 0$.

$|E(XY)| \leq \sqrt{E(Y^2)E(X^2)}$.

===== Zadanie 7. =====

$D^2(X+a) = E(X+a-E(X+a))^2 = E(X-EX)^2 = D^2(X)$

===== Zadanie 8. =====

$D^2(aX) = E(aX)^2 - (E(aX))^2 = a^2(EX^2-(EX)^2) = a^2D^2(X)$

===== Zadanie 9. =====

$\displaystyle E(Y) = E \bigg(\frac {X - EX} {\sqrt{D^2(X)}} \bigg) = \frac {E(X - EX)} {\sqrt{D^2(X)}} = \frac {EX - EX} {\sqrt{D^2(X)}} = 0$

$\displaystyle D^2(Y) = D^2 \bigg ( \frac {X-EX} {\sqrt{D^2(X)}} \bigg) = \frac {D^2(X-EX)} {D^2(X)} = \frac {D^2(X)} {D^2(X)} = 1$


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