记 \[ f(x)=\frac{x^n}{n!}e^{-x},\quad x\geq 0. \] 由于 \[ \int_0^\infty x^m e^{-x}\,dx=m!,\qquad m=0,1,2,\ldots, \] 可得 \[ \mathrm E X=\int_0^\infty x\frac{x^n}{n!}e^{-x}\,dx =\frac{(n+1)!}{n!}=n+1, \] \[ \mathrm E X^2=\int_0^\infty x^2\frac{x^n}{n!}e^{-x}\,dx =\frac{(n+2)!}{n!}=(n+1)(n+2). \] 因此 \[ \operatorname{Var}(X) =\mathrm E X^2-(\mathrm E X)^2 =(n+1)(n+2)-(n+1)^2=n+1. \] \((\mathrm I)\) 因为 \(X\geq 0\) 几乎处处，所以 \[ \{01\}\leq P\{|Y_n|\geq 1\} \leq \operatorname{Var}(Y_n) =\frac1n. \] 于是 \[ 1-\frac1n\leq P\{Y_n\leq 1\}\leq 1. \] 令 \(n\to\infty\)，由夹逼定理得 \[ \lim_{n\to\infty} P\left\{ \frac{\sum_{i=1}^n X_i-n(n+1)}{n\sqrt{n+1}}\leq 1 \right\} =1. \] \((\mathrm{III})\) 下面证明不存在这样的正实数 \(a\)。令 \[ Z_i=X_i^2,\qquad \overline Z_n=\frac1n\sum_{i=1}^n Z_i =\frac1n\sum_{i=1}^n X_i^2. \] 先计算 \(Z_i\) 的均值和方差。由前面结果， \[ \mathrm E Z_i=\mathrm E X_i^2=(n+1)(n+2). \] 又 \[ \mathrm E Z_i^2=\mathrm E X_i^4 =\int_0^\infty x^4\frac{x^n}{n!}e^{-x}\,dx =\frac{(n+4)!}{n!} =(n+1)(n+2)(n+3)(n+4). \] 因此 \[ \operatorname{Var}(Z_i) =\mathrm E Z_i^2-(\mathrm E Z_i)^2 \] \[ =(n+1)(n+2)(n+3)(n+4) -(n+1)^2(n+2)^2 \] \[ =(n+1)(n+2)\bigl[(n+3)(n+4)-(n+1)(n+2)\bigr] \] \[ =(n+1)(n+2)(4n+10). \] 于是 \[ \mathrm E \overline Z_n=(n+1)(n+2), \] 且由独立性， \[ \operatorname{Var}(\overline Z_n) =\frac1{n^2}\sum_{i=1}^n \operatorname{Var}(Z_i) =\frac{(n+1)(n+2)(4n+10)}{n}. \] 记 \[ \mu_n=(n+1)(n+2). \] 则由切比雪夫不等式， \[ P\left\{\left|\overline Z_n-\mu_n\right|\geq \frac{\mu_n}{2}\right\} \leq \frac{4\operatorname{Var}(\overline Z_n)}{\mu_n^2} \] \[ = \frac{4(n+1)(n+2)(4n+10)} {n(n+1)^2(n+2)^2} = \frac{4(4n+10)}{n(n+1)(n+2)} \longrightarrow 0. \] 因此 \[ P\left\{\overline Z_n\geq \frac{\mu_n}{2}\right\}\longrightarrow 1. \] 而 \[ \frac{\mu_n}{2}=\frac{(n+1)(n+2)}2\longrightarrow \infty. \] 所以对任意给定的正实数 \(a\) 以及任意 \(\varepsilon>0\)，当 \(n\) 充分大时， \[ a+\varepsilon<\frac{\mu_n}{2}. \] 于是 \[ \left\{\left|\overline Z_n-a\right|<\varepsilon\right\} \subset \{\overline Z_n0\) 都有 \[ \lim_{n\to\infty} P\left\{ \left| \frac1n\sum_{i=1}^n X_i^2-a \right|<\varepsilon \right\}=1. \]

随机刷题