$$
E(x)=\int_{-\infty}^{\infty} N\left(x \mid \mu, \sigma^2\right) x d x=\mu
$$
证明
令$u=x-\mu$则有:
$$
\begin{aligned}
E(x) & =\int_{-\infty}^{\infty} N\left(x \mid \mu, \sigma^2\right) x d x \
& =\int_{-\infty}^{\infty} \frac{1}{\left(2 \pi \sigma^2\right)^{1 / 2}} \exp \left[-\frac{1}{2 \sigma^2}(x-\mu)^2\right] x d x \
& =\int_{-\infty}^{\infty} \frac{1}{\left(2 \pi \sigma^2\right)^{1 / 2}} \exp \left(-\frac{y^2}{2 \sigma^2}\right)(y+\mu) d y \
& =\frac{1}{\left(2 \pi \sigma^2\right)^{1 / 2}}\left[\int_{-\infty}^{\infty} \exp \left(-\frac{y^2}{2 \sigma^2}\right) y d y+\int_{-\infty}^{\infty} \exp \left[-\frac{y^2}{2 \sigma^2}\right) \mu d y\right]
\end{aligned}
$$
其中$\exp \left(-\frac{y^2}{2 \sigma^2}\right) y$满足奇函数的性质$f(x)=-f(-x)$因此这一项的积分为0所以:
$$
\begin{aligned}
E(x) & =\frac{1}{\left(2 \pi \sigma^2\right)^{1 / 2}} \int_{-\infty}^{\infty} \exp \left(-\frac{y^2}{2 \sigma^2}\right) \mu d y \
& =\frac{\mu}{\left(2 \pi \sigma^2\right)^{1 / 2}} \int_{-\infty}^{\infty} \exp \left(-\frac{y^2}{2 \sigma^2}\right) d y
\end{aligned}
$$
可以从《高斯函数概率密度函数的积分》得到
$$
I = \int_{-\infty}^{\infty} \exp\left(-\frac{x^2}{2}\right), dx \
$$
$$
I = \sqrt{2\pi}
$$
所以最终的结果为$E(x)=\mu$