====== Rachunek prawdopodobieństwa i statystyka - Lista 2. ======
{{:rachunek_prawdopodobienstwa_i_statystyka:10.lista02.pdf|Lista 2}}.
Dowód twierdzenia $Var(X) = E(X^2) - E^2(X)$ potrzebnego do większości zadań:
Trzy obserwacje:
*$E(a) = a$, gdy $a = const$,
*$E(E^2(X)) = E(X)$, bo $E^2(X)=const$,
*$E(f(X) + g(X)) = E(f(X)) + E(g(X))$ (oczywiste).
$Var(X) =^{def}
E(X - E(X))^2 =
E(X^2 - 2XE(X) + E^2(X)) =
E(X^2) - E(2XE(X)) + E(E^2(X)) =
E(X^2) - E(2)E(X)E^2(X) + E(E^2(X)) =
E(X^2) - 2E(X)E^2(X) + E^2(X) =
E(X^2) - E^2(X)$
===== Zadanie 1. =====
$E(X) = 2\cdot0,2 + 3\cdot0,4 + 4\cdot0,1 + 5\cdot0,3 = 3,5$.
$E(X^2) = 4\cdot0,2 + 9\cdot0,4 + 16\cdot0,1 + 25\cdot0,3 = 13,5$.
$D^2(X) = E(X^2) - E(X)^2 = 13,5 - 12,25 = 1.25$.
===== Zadanie 2. =====
| x | -2 | 3 | 5 |
| f(x) | 0,2 | 0,5 | 0,3 |
$EX = -2\cdot0,2 + 3\cdot0,5 + 5\cdot0,3 = -0,4 + 1,5 + 1,5 = 2,6$
$E(X^2) = 4\cdot0,2 + 9\cdot0,5 + 25\cdot0,3 = 12,8$.
$D^2(X) = E(X^2) - E(X)^2 = 12,8 - 6,7 = 6,1$.
===== Zadanie 3. =====
$\displaystyle E(X) = \sum_{k=0}^{n}{ k {n \choose k} p^k q^{n-k}}$\\
$ = \sum_{k=1}^n k \cdot \frac{n!}{k!(n-k)!} p^k(1-p)^{n-k} = \sum_{k=1}^n k \cdot \frac{n\cdot(n-1)!}{k\cdot(k-1)!(n-k)!} \cdot p \cdot p^{k-1}(1-p)^{n-k} = $
$\displaystyle = np \cdot \sum_{k=0}^{n-1}{ {n \choose k} p^k q^{n-k}} = np \cdot 1 = np $
$\displaystyle E(X^2) = \sum_{k=0}^n{ k^2 {n \choose k} p^k q^{n-k}} = np \cdot \sum_{k=0}^{n-1}{ (k+1) {n \choose k} p^k q^{n-k}} =$\\
$np \cdot (\sum_{k=0}^{n-1}{k {n \choose k} p^k q^{n-k}} + \sum_{k=0}^{n-1}{{n \choose k} p^k q^{n-k}}) = np \cdot ((n-1)p + 1) = np(np - p + 1)$
$D^2(X) = E(X^2) - E(X)^2 = np(np - p + 1) - n^2p^2 = -np^2 + np = np(1-p) = npq$
===== Zadanie 4. =====
$\displaystyle E(X) = \sum_{k=0}^{\infty}k\frac{\lambda^k}{k!}\,\mathrm{e}^{-\lambda} = \lambda\, \mathrm{e}^{-\lambda}\sum_{k=1}^{\infty}\frac{\lambda^{k-1}}{(k-1)!} = \lambda\, \mathrm{e}^{-\lambda}\sum_{j=0}^{\infty}\frac{\lambda^{j}}{j!} = \lambda$
$\displaystyle E(X^2) = \sum_{k=0}^{\infty}k^2\frac{\lambda^k}{k!}e^{-\lambda} = \sum_{k=0}^{\infty}(k(k-1) + k)\frac{\lambda^k}{k!}e^{-\lambda} = \sum_{k=0}^{\infty}k(k-1)\frac{\lambda^k}{k!}e^{-\lambda} + \sum_{k=0}^{\infty}k\frac{\lambda^k}{k!}e^{-\lambda} = \lambda^2 + \lambda$
$\displaystyle D^2(X) = E(X^2) - E(X)^2 = \lambda^2 + \lambda - \lambda^2 = \lambda$
===== Zadanie 5. =====
$p(x) = pq^{k-1}$
$\displaystyle E(X) = \sum_{k=1}^{\infty}{kpq^{k-1}} = p \sum_{k=1}^{\infty}{(q^k)'} = p (\sum_{k=1}^{\infty}{q^k})' = p (\frac q {1-q})' = \frac p {(1-q)^2} = \frac 1 p$
$\displaystyle E(X^2) = \sum_{k=1}^{\infty}{k^2pq^{k-1}} = p \sum_{k=1}^{\infty}{(-k+k(k+1) ) \cdot q^{k-1}} =$\\
$ -\frac 1 p + p(\sum_{k=1}^{\infty}{q^{k+1}})' ' = p(\frac {(1-p)^2} {p})' ' - \frac 1 p = p(\frac 2 {p^3}) - \frac 1 p = \frac {2-p}{p^2}$
$\displaystyle D^2(X) = E(X^2) - E(X)^2 = \frac {2-p}{p^2} - \frac 1 {p^2} = \frac {1-p}{p^2}$
===== Zadanie 6. =====
===== Zadanie 7. =====
$\displaystyle f(x) = \frac 1 {b-a} \text{, dla } a \leq x \leq b$
$\displaystyle E(X) = \int_a^b { \frac x {b-a} dx} = \frac 1 {b-a} ( \frac {b^2-a^2} 2 ) = \frac {b+a} 2$
$\displaystyle E(X^2) = \int_a^b { \frac {x^2} {b-a} dx} = \frac 1 {b-a} ( \frac {b^3-a^3} 3 ) = \frac {b^2+ab+a^2} 3$
$\displaystyle D^2(X) = E(X^2) - E(X)^2 = \frac {b^2+ab+a^2} 3 - \frac {b^2+2ab+a^2} 4 = \frac {b^2 -2ab + a^2} {12} = \frac {(b-a)^2} {12}$
===== Zadanie 8. =====
$\displaystyle f(x) = \lambda e^{-\lambda x} \text{, dla } x \geq 0$
$\displaystyle E(X) = \int_0^{\infty}{x\lambda e^{-\lambda x} dx} = \lim_{b -> \infty} \int_0^b{x\lambda e^{-\lambda x} dx} =
\lambda \lim_{b -> \infty} ([- x \cdot \frac {e^{-\lambda x}} {\lambda}]_0^b + \frac 1 \lambda \int_0^b{ e^{-\lambda x} dx}) = $
$\displaystyle = \lambda \lim_{b -> \infty} ([- x \cdot \frac {e^{-\lambda x}} {\lambda}]_0^b + [-\frac { e^{-\lambda x} } {\lambda^2}]_0^b ) =
\lambda \lim_{b -> \infty} (-b \cdot \frac { e^{-\lambda b} } {\lambda} - \frac { e^{-\lambda b} } {\lambda^2} + \frac 1 {\lambda^2} ) = \lambda (0 + 0 + \frac 1 {\lambda^2}) = \frac 1 {\lambda}$
$\displaystyle E(X^2) = \int_0^{\infty}{x^2\lambda e^{-\lambda x} dx} = \lim_{b -> \infty} \int_0^b{x^2\lambda e^{-\lambda x} dx} = \lambda \lim_{b -> \infty} ([- x^2 \cdot \frac {e^{-\lambda x}} {\lambda}]_0^b + \frac 2 {\lambda} \int_0^b{ xe^{-\lambda x} dx}) = $
$\displaystyle = \lim_{b -> \infty} ([- x^2 \cdot xe^{-\lambda x}]_0^b + 2 [-x \frac {e^{-\lambda x}} {\lambda}]_0^b + \frac 2 {\lambda} \int_0^b{ e^{-\lambda x} dx})
= \lim_{b -> \infty} ([- x^2 \cdot \frac {xe^{-\lambda x}} {\lambda}]_0^b + 2 [-x \frac {e^{-\lambda x}} {\lambda}]_0^b + \frac 2 {\lambda^2} [-e^{-\lambda x}]_0^b) =$
$ \displaystyle = \lim_{b -> \infty} (-b^2\frac {be^{-\lambda b}}{\lambda} -2b\frac{e^{-\lambda b}}{\lambda} - \frac 2 {\lambda^2} e^{-\lambda b} + \frac 2 {\lambda^2}) = (0 - 0 - 0 + \frac 2 {\lambda^2}) = \frac 2 {\lambda^2}$
I **nareszcie**:
$\displaystyle D^2(X) = E(X^2) - E(X)^2 = \frac 2 {\lambda^2} - \frac 1 {\lambda^2} = \frac 1 {\lambda^2}$
===== Zadanie 9. =====
===== Zadanie 10. =====