# What is the expected value of cosine of a multivariate Gaussian?

Suppose $X \sim \mathcal{N}\left(\mu, \Sigma\right)$. How do I evaluate $\operatorname{E}\left[\cos \left(t^{T}X \right) \right]$ and $\operatorname{E}\left[\sin \left(t^{T}X\right) \right]$? Does this have to do with the characteristic function $\operatorname{E}\left[e^{it^{T}X} \right] =\exp \left\{i\mu^{T}t -\frac{1}{2}t^{T}\Sigma t\right\}$?

• Keep going... Think about the real and imaginary parts of your last equation! Jan 24, 2016 at 15:14
• Hint: $t^TX = \sum_i t_iX_i$ is a (univariate) Gaussian random variable $Y$ with mean $t^T\mu$ and variance $t^T\Sigma t$. So, can you figure out what $E[\cos Y]$ and $E[\sin Y]$ are? (See danimal's hint) Jan 24, 2016 at 15:17
• oh!! So $\operatorname{E}\left[e^{it^{T}X} \right]=\operatorname{E}\left[\cos \left(t^{T}X \right) \right]+i\operatorname{E}\left[\sin \left(t^{T}X\right) \right]$ and this is also equal to $e^{i\mu^{T}t -\frac{1}{2}t^{T}\Sigma t}$ which is $e^{i\mu^{T}t} e^{-\frac{1}{2}t^{T}\Sigma t} = \left(\cos \left(\mu^{T}t\right) +i\sin \left(\mu^{T}t\right) \right)e^{-\frac{1}{2}t^{T}\Sigma t}$. Now I only have to match the real and imaginary part! Epiphany! Thank you! Jan 24, 2016 at 15:21

I will post an answer to my own question. Throughout the answer, the Euler's formula is extensively used: $$e^{ix} = \cos x + i \sin x$$ Plugging in $x = t^{T}X$, \begin{align*} e^{it^{T}X} &= \cos \left(t^{T}X\right) + i \sin \left(t^{T}X\right) \\ \mathbb{E}\left[e^{it^{T}X} \right] &= \mathbb{E} \left[\cos \left(t^{T}X\right) \right] + i \mathbb{E}\left[\sin \left(t^{T}X\right) \right] \\ &= \exp\left\{i\mu^{T}t-\frac{1}{2}t^{T}\Sigma t \right\} \left(\text{characteristic function} \right)\\ &= \exp\left\{i\mu^{T}t \right\}\exp\left\{-\frac{1}{2}t^{T}\Sigma t \right\}\\ &= \left(\cos \left(\mu^{T}t\right) + i\sin\left(\mu^{T}t\right) \right)\exp \left\{-\frac{1}{2}t^{T}\Sigma t \right\} \end{align*} Therefore, the comparing the real parts and imaginary parts, \begin{align*}\mathbb{E}\left[\cos \left(t^{T}X\right) \right] &= \cos \left(\mu^{T}t\right)\exp \left\{-\frac{1}{2}t^{T}\Sigma t \right\}\\ \mathbb{E}\left[\sin \left(t^{T}X\right) \right] &= \sin \left(\mu^{T}t\right)\exp\left\{ -\frac{1}{2}t^{T}\Sigma t\right\}. \end{align*}