[1]
\(Pr\{X=k\}=\)「白玉を\(k-1\)回引いた後、赤玉を\(1\)回引く確率」
\(=\displaystyle \frac{1}{2}\cdot\frac{2}{3}\cdot \cdots \cdot \frac{k-1}{k}\cdot \frac{1}{k+1}\)
\(=\displaystyle \frac{1}{k(k+1)}\)
[2]
\(f(k) \geq 0\)かつ
\(\displaystyle \sum_{k=1}^\infty f(k) = \sum_{k=1}^\infty \frac{1}{k(k+1)}\)
\(\displaystyle = \sum_{k=1}^\infty \left( \frac{1}{k} -\frac{1}{k+1}\right)\)\(=1\)
より、\(f(k)\)は確率関数。
[3]
\(E[X] =\displaystyle \sum_{k=1}^\infty kf(k) \)
\(=\displaystyle \sum_{k=1}^\infty \frac{1}{k+1} \)
\(=\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots \)
\(\displaystyle > \int ^\infty_2 \frac{1}{x}dx\) (区分求積法の利用)
\(=\displaystyle [\ln x]^\infty_2 =\infty \)
よって平均値は存在せず、分散も存在しない。
[4]
\(P(k) \geq 0\)かつ
\(\displaystyle \sum_{k=1}^\infty P(k) = \sum_{k=1}^\infty \frac{18}{k(k+1)(k+2)(k+3)}\)
\(\displaystyle = 6\sum_{k=1}^\infty \left( \frac{1}{k(k+1)(k+2)} -\frac{1}{(k+1)(k+2)(k+3)}\right)\)\(=1\)
より、\(P(k)\)は確率関数。
[5]
\(E[Y]= \displaystyle\sum_{k=1}^\infty kP(k)\)
\(= \displaystyle\sum_{k=1}^\infty \frac{18}{(k+1)(k+2)(k+3)}\)
\(\displaystyle = 9\sum_{k=1}^\infty \left( \frac{1}{(k+1)(k+2)} -\frac{1}{(k+2)(k+3)}\right)\)
\(\displaystyle = \frac{3}{2}\)
[6]
\(E[Y(Y+1)]\)
\(E[Y(Y+1)]= \displaystyle\sum_{k=1}^\infty k(k+1)P(k)\)
\(= \displaystyle\sum_{k=1}^\infty \frac{18}{(k+2)(k+3)}\)
\(\displaystyle = 18\sum_{k=1}^\infty \left( \frac{1}{k+2} -\frac{1}{k+3}\right)\)
\(\displaystyle = 6\)
\(V[Y]\)
\(V[Y]=E[Y(Y+1)]-E[Y]-E[Y]^2\)
\(\displaystyle = 6-\frac{3}{2}-\left(\frac{3}{2}\right)^2\)
\(\displaystyle = \frac{9}{4}\)
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