随机变量的期望

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：

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 \(\int_{0}^{\infty}\frac{3x}{(x+1)^4}\ \mathrm{d}x\) 不好求。我们先求\(\frac{3x}{(x+1)^4}\)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\(\frac{1}{2}\)。