服从正态分布的正弦函数、余弦函数期望。

如果X服从均值为$\mu$，方差为${\sigma }^2$的正态分布，计算sin(X)与cos(X)的数学期望。

利用特征函数(Characteristic Function)Wiki-Characteristic Function，我们知道$X\sim N(\mu ,{\sigma }^2)$的特征函数为：
$${\phi }_{X(t)}=E(e^{itX})=exp\left(i\mu t-\frac{{\sigma }^2{t}^2}{2}\right)=exp(-{\sigma }^2{t}^2/2)exp(i\mu t)$$
根据欧拉公式：
$$e^{ix}=\cos (x)+i\sin (x)$$

$$\begin{array}{rl}E(e^{itX})& =exp\left(i\mu t-\frac{{\sigma }^2{t}^2}{2}\right)\\ & =exp(-{\sigma }^2{t}^2/2)exp(i\mu t)\\ & =exp(-{\sigma }^2{t}^2/2)\left[\cos (\mu t)+i\sin (\mu t)\right]\\ & =exp(-{\sigma }^2{t}^2/2)\cos (\mu t)+exp(-{\sigma }^2{t}^2/2)\sin (\mu t)i\\ & =Real(E(e^{itX}))+iIm(E(e^{itX}))\end{array}$$

上述数学期望变为：
$$\begin{array}{rl}E(e^{itX})& =E\left(\cos (tX)+i\sin (tX)\right)\\ & =E(cos(tX))+iE(sin(tX))\\ & =Real(E(e^{itX}))+iIm(E(e^{itX}))\end{array}$$
对比上述的实数部分和虚数部分，得到：
$$\begin{gathered} E(\cos (tX))=exp(-{\sigma }^2{t}^2/2)\cos (\mu t) \\ E(\sin (tX))=exp(-{\sigma }^2{t}^2/2)sin(\mu t) \end{gathered}$$
最后，当t=1的时候：
$$\begin{gathered} E(\cos (X))=exp(-{\sigma }^2/2)\cos (\mu ) \\ E(\sin (X))=exp(-{\sigma }^2/2)sin(\mu ) \end{gathered}$$

参考资料：Mean and variance of Y=cos(bX) when X has a Gaussian distribution

应用：IPE位置编码中，对服从高斯分布正弦函数的数学期望计算：

# Code Source: 
# https://github.com/liuyuan-pal/NeRO/blob/3b4d421a646097e7d59557c5ea24f4281ab38ef1/network/field.py#L369-L378
def expected_sin(mean, var):"""Compute the mean of sin(x), x ~ N(mean, var)."""return torch.exp(-0.5 * var) * torch.sin(mean)  # large var -> small value.def IPE(mean,var,min_deg,max_deg):scales = 2**torch.arange(min_deg, max_deg)shape = mean.shape[:-1] + (-1,)scaled_mean = torch.reshape(mean[..., None, :] * scales[:, None], shape)scaled_var = torch.reshape(var[..., None, :] * scales[:, None]**2, shape)return expected_sin(torch.concat([scaled_mean, scaled_mean + 0.5 * np.pi], dim=-1), torch.concat([scaled_var] * 2, dim=-1))