| 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 |
\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} |
\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 \}
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.
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).
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)}.
D^2(X+a) = E(X+a-E(X+a))^2 = E(X-EX)^2 = D^2(X)
D^2(aX) = E(aX)^2 - (E(aX))^2 = a^2(EX^2-(EX)^2) = a^2D^2(X)
\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