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Let $X_1,X_2,\dots$ be i.i.d. r.v. and $X_i\sim \mathcal{N}(0,1)$. Further let $S_n=X_1+\dots+X_n$, $\phi \in \mathbb{R}$

How do I calculate $\mathbb{E}(e^{\phi S_n})$?

The integral looks extremly complicated

$$\mathbb{E}(e^{\phi S_n})=\int\limits_{-\infty}^\infty \cdots\int\limits_{-\infty}^\infty e^{\phi (x_1 +\dots x_n)}\frac{1}{(2\pi)^{n-1}} e^{-\frac{1}{2} (x_1^2+\cdots x_n^2)} \mathrm{d}x_1\dots\mathrm{d}x_n$$

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  • $\begingroup$ What is $\phi$? $\endgroup$ – Jack Jan 10 '17 at 16:57
  • $\begingroup$ If $\phi$ is just a constant, use independence and the fact that $m_{X+Y}(t) = m_X(t) m_Y(t)$ for independent rvs $X$ and $Y$. $\endgroup$ – Nigel Overmars Jan 10 '17 at 16:58
  • $\begingroup$ @Jack $\phi$ is a constant. $\endgroup$ – MarcE Jan 10 '17 at 17:40
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Assuming $\phi$ is meant to be a constant, this integral is fairly easy.

Making the substitutions $x_i = v_i+\phi$ the integral (or at least the correct integral for the expectation; I believe your normalization is not in place) becomes the product of $n$ integrals, each of which is of the form $$ \frac1{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac12 y_i^2 + \frac12 \phi^2} $$ each of these is just the normalization integral for the unit normal distribution, so the answer will be $$\Bbb E(e^{\phi S_n}) = e^{\frac{n}{2}\phi^2} $$

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