Rachunek prawdopodobieństwa i statystyka - Lista 2.

E(X) = 3,5. E(X^2) = 13,5.

D^2(X) = E(X^2) - E(X)^2 = 13,5 - 12,25 = 1.25.

E(X) = -1.2 + 0.3 + 1.5 = 0. E(X^2) = 2.4 + 0.3 + 1.5 = 4.2.

D^2(X) = E(X^2) - E(X)^2 = 4,2.

\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

\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

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}

Vide zadanie 5.

\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}

\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}