average of exp of linear function How to prove that $<e^{iA}>=e^{-<A^2>/2}$ ?, where average is taken w.r.t. a gaussian distribution (of any number of variables, let two (x,y)) and A is a linear function (without constant term, i.e, for two variables it will be lin combination of x and y)?
I tried to expand out exp in LHS in terms of cousin and sine then sine terms will vanish but cosine will remain and that will not be equal to the term that we'll get by expanding out RHS!
I got LHS= $1-\dfrac{<A^2>}{2!}+\dfrac{<A^4>}{4!}-\dfrac{<A^6>}{3!}+\ldots$ , and RHS=$1-\dfrac{<A^2>/2}{1!}+\dfrac{(<A^2>/2)^2}{2!}-\dfrac{(<A^2>/2)^3}{3!}+\ldots$, which, at least manifestally, doesn't seems to be equal!
 A: $$\displaystyle \frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty e^{i(a+bx)}e^{-x^2} \,dx=  e^{ia-b^2/4}$$
$$\displaystyle \exp\left\{-\frac{1}{2\sqrt{\pi}}\int_{-\infty}^\infty (a+bx)^2 e^{-x^2} \,dx\right\}= e^{-a^2/2-b^2/4}$$
Not equal.
UPDATE
With $a=0$ they are equal.
In two dimensions:
$$\displaystyle \frac{1}{{\pi}}\int_{-\infty}^\infty \, \int_{-\infty}^\infty e^{i(\alpha x +\beta y)}e^{-x^2} e^{-y^2} \,dx\, dy=  e^{-(\alpha^2/4 + \beta^2/4)}$$
$$\displaystyle \exp\left\{-\frac{1}{2 {\pi}}\int_{-\infty}^\infty \, \int_{-\infty}^\infty (\alpha x + \beta y)^2 e^{-x^2} e^{-y^2} \,dx \, dy \right\}= e^{-(\alpha^2/4+\beta^2/4)}$$
Yes, they are equal.  You can easily extend this to more than two dimensions.
Note:  We've made use of $\int_{-\infty}^\infty x e^{-x^2} \, dx=0 \quad \Longrightarrow \quad \int_{-\infty}^\infty \int_{-\infty}^\infty xy\, e^{-(x^2+y^2)} \, dx\, dy=0.$
A: I assume yours is the case with mean zero.  First let's recall that for $s\in \mathbb{R},$ the characteristic function of a univariate normal random variable, $Y\sim N(\mu_Y,\sigma^2_Y)$, is $$E[\exp\left\{isY\right\}]=\exp\left\{is\mu_Y-\frac{1}{2}\sigma^2_Ys^2\right\}.$$
In the multivariate case, for $X:=(X_1,...X_n)'\sim N(\mu,\Sigma)$, note that for $t\in\mathbb{R}^n,$
$$A:=t'X\sim N(t'\mu,t'\Sigma t),$$
and applying the univariate characteristic function above (taking $s=1,Y=A$) tells us
$$E[\exp\left\{iA\right\}]=\exp\left\{it'\mu-\frac{1}{2}t'\Sigma t \right\}.$$
Now, in the case where $\mu=0$, then $E[A^2]=\text{Var}(A)=t'\Sigma t,$ so we can equivalently write
$$E[\exp\left\{iA\right\}]=\exp\left\{-\frac{1}{2}E[A^2] \right\}.$$
