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My attempt to calculate the characteristic function $\mathbb{E}\left(e^{it^T X}\right)$ of multivariate normal distributed $X \sim \mathcal{N}_d(\mu,\Sigma)$, finding a lecture note that writes as follows, but I am confused about the $\overset{?}{=}$ step.

\begin{align*} \phi_X(t) &= \mathbb{E}\left(e^{it^T X}\right) \\ &= \int_{-\infty}^{\infty} e^{it^T X} f_X(x) dx \\ &= \int_{-\infty}^{\infty} e^{it^T X} \frac{1}{(2\pi)^{d/2}|\Sigma|^{1/2}} e^{-\frac{1}{2}(x-\mu)^T \Sigma^{-1} (x-\mu)} dx \\ &= \int_{-\infty}^{\infty} \frac{1}{(2\pi)^{d/2}|\Sigma|^{1/2}} \exp{\left(-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)+it^T x\right)} dx \\ &\overset{?}{=} \exp{\left(it^T \mu - \frac{1}{2} t^T \Sigma t\right)} \\ &= e^{it^T \mu} e^{-\frac{1}{2}t^T\Sigma t} \end{align*}

Can you show me more details about that step?

Jason
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Analysis Newbie
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1 Answers1

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What they do is completing the square:

\begin{align}-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)+it^T x & = -\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)+it^T (x - \mu) + it^T \mu \\ & = -\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)+it^T (x - \mu) +\frac{1}{2} t^T \Sigma t - \frac{1}{2} t^T \Sigma t\\ & \;\;\;\; +\ it^T \mu \\ & = -\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)+it^T \Sigma \Sigma^{-1} (x - \mu) +\frac{1}{2} t^T \Sigma \Sigma^{-1}\Sigma t \\ & \;\;\;\; - \frac{1}{2} t^T \Sigma t +\ it^T \mu \\ & = -\frac{1}{2}(x-\mu-i\Sigma t)^T\Sigma^{-1}(x-\mu-i\Sigma t) - \frac{1}{2} t^T \Sigma t +\ it^T \mu \end{align}

Then, the two last terms factorize out with the exponential, while the first gives rise to a reparametrized Gaussian. So that integral is 1.

Raskolnikov
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    Could you please elaborate how you came from the second last to last line? $ \begin{align} -\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)+it^T \Sigma \Sigma^{-1} (x - \mu) +\frac{1}{2} t^T \Sigma \Sigma^{-1}\Sigma t - \frac{1}{2} t^T \Sigma t +\ it^T \mu = \ \ -\frac{1}{2}(x-\mu-i\Sigma t)^T\Sigma^{-1}(x-\mu-i\Sigma t) - \frac{1}{2} t^T \Sigma t +\ it^T \mu \end{align} $ – wklm Oct 15 '22 at 19:17
  • The second term is equal to $\frac{1}{2}it^T \Sigma \Sigma^{-1} (x - \mu) + \frac{1}{2}i(x - \mu)^T \Sigma^{-1} \Sigma t$. – Raskolnikov Oct 15 '22 at 20:35
  • @Raskolnikov, Why? And what does it give to us? Even if I put that equation, which is equal to the second term, as you said, I get to terms, one of which contains ((x−μ)^T−it^TΣ) and another contains (i(x−μ)^T−t^TΣ), so, anyways I cant get, how did you get over it – Timur Apr 14 '23 at 05:16
  • @Timur: I'm not sure what you are struggling with. Note that $$\left[it^T \Sigma \Sigma^{-1} (x - \mu)\right]^T = it^T \Sigma \Sigma^{-1} (x - \mu)) = (x - \mu)^T \Sigma^{-1} i\Sigma t ; .$$ The first equality is correct because a number (1x1 matrix) transposed is equal to itself. The second is just mechanically applying the transposition rule to a product. – Raskolnikov Apr 14 '23 at 06:05