随机变量的期望

解：

伯努力分布是两点分布，可知其概率质量函数为

\[
f(x)=\begin{cases}
p & \text{ 若 } x=1 \\
1-p & \text{ 若 } x=0
\end{cases}
\]

按照定义代入：

\[\begin{align*}
E(X) =& \sum_i^n x_if(x_i)\\
=& 1\times p+0\times (1-p) \\
=& p
\end{align*}\]

解：

按照定义：

\[\begin{align*}
E(X) =& \sum_i^n e^{x_i} f(x_i)\\
=& e^1 \times p+ e^0\times (1-p) \\
=& ep+1-p
\end{align*}\]

\[
f(x)=\begin{cases}
\frac{1}{b-a} & \text{如果} a<x<b\\
0 & \text{其他}
\end{cases}
\]

求E(X)。^[1]

解：

按照定义，
\[
\begin{align*}
E(X) =& \int_{-\infty}^{\infty}x f(x)\ \mathrm{d}x\\
=& \int_{-\infty}^{a}x 0\ \mathrm{d}x+ \int_{a}^{b}x f(x)\ \mathrm{d}x+\int_{b}^{\infty}x 0\ \mathrm{d}x \\
=& 0 + \int_{a}^{b}x \frac{1}{b-a}\ \mathrm{d}x + 0\\
=& \frac{x^2}{2(b-a)} \bigg|_a^b \\
=& \frac{b^2}{2(b-a)}-\frac{a^2}{2(b-a)} \quad \because a\neq b \\
=& \frac{1}{2}(a+b)
\end{align*}
\]

Mathematica命令：

求X的数学期望。^[2]

解：

按照定义，
\[
\begin{align*}
E(X) =& \int_{-\infty}^{\infty}x f(x)\ \mathrm{d}x\\
=& \int_{-\infty}^{0}x 0\ \mathrm{d}x+ \int_{0}^{\infty}\frac{3x}{(x+1)^4}\ \mathrm{d}x \\
=& \int_{0}^{\infty}\frac{3x}{(x+1)^4}\ \mathrm{d}x
\end{align*}
\]

发现 [latex]\int_{0}^{\infty}\frac{3x}{(x+1)^4}\ \mathrm{d}x[/latex] 不好求。我们先求[latex]\frac{3x}{(x+1)^4}[/latex]的不定积分。

\[\begin{align*}
\int \frac{3x}{(x+1)^4}\ \mathrm{d}x\ =& \text{使用积分常数法则}\qquad 3\int\frac{x}{(x+1)^4} \ \mathrm{d}x = 3\int\frac{x+1-1}{(x+1)^4} \ \mathrm{d}x\\
=& 3\int\frac{1}{(x+1)^3}-\frac{1}{(x+1)^4} \ \mathrm{d}x \qquad\text{使用和差法则}\\
=& 3\left[\int\frac{1}{(x+1)^3}\ \mathrm{d}x – \int\frac{1}{(x+1)^4} \ \mathrm{d}x \right] \qquad\text{令}f(x)=\frac{1}{x^3}\text{，令}g(x)=\frac{1}{x^4}\\
=& 3\left[\int f(x+1)\ \mathrm{d}x – \int g(x+1)\ \mathrm{d}x \right] \\
& \qquad\text{令}h(x)=x+1\text{，则}h'(x)=1\text{，令}i(x)=x+1\text{，则}i'(x)=1\\
=& 3\left[\int f(h(x))h'(x)\ \mathrm{d}x – \int g(i(x))i'(x)\ \mathrm{d}x \right] \\
& \qquad\text{符合换元积分格式，令}u=h(x)=x+1\text{，}v=i(x)=x+1\\
=& 3\left[\int f(u)\ \mathrm{d}u – \int g(v)\ \mathrm{d}v \right] = 3\left[\int \frac{1}{u^3}\ \mathrm{d}u – \int \frac{1}{v^4}\ \mathrm{d}v \right] \\
=& 3\left[-\frac{1}{2}\frac{1}{u^2}+ \frac{1}{3} \frac{1}{v^3} \right] = 3\left[\frac{1}{3(x+1)^3}-\frac{1}{2(x+1)^2} \right] \\
=& -\frac{3x+1}{2(x+1)^3}
\end{align*}\]

(如果对上面的积分法则和换元积分法不熟悉，可参考https://www.shuxuele.com/calculus/integration-by-substitution.html )

\[\begin{align*}
E(X) =& \int_{0}^{\infty}\frac{3x}{(x+1)^4}\ \mathrm{d}x \\
=& \left.\left( \int \frac{3x}{(x+1)^4}\ \mathrm{d}x \right) \right|_0^\infty \\
=& \left. -\frac{3x+1}{2(x+1)^3} \right|_0^\infty \\
=& \left(\lim_{x\to \infty} -\frac{3x+1}{2(x+1)^3}\right)+\frac{1}{2} \\
=& \frac{1}{2}
\end{align*}\]

所以，X的数学期望为[latex]\frac{1}{2}[/latex]。