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Let $\Omega, F, P)$ be the classic setting. I saw that if $f$ is a function which satisfies some assumptions then the integral with respect to the brownian motion is Gaussian. Ie $\int_{0}^{t} f_u dB_u \sim normal$

1)How we can see this gaussianity?

Ok generally, the integral of a progressively measurable process $f$ is a limit of the integrals $I(f^{n}) = \int_{0}^{t} f^{n}_u dB_u$ where $f^{n}$ are simple processes and these integrals are gaussian by definition. In fact we have also $L^{2}\Omega)$ convergence which implies weak one. So the distribution functions of $I(f^{n})$ random variables converge uniformly( for all $x \in R$) to the distribution function of $I(f)$. But why this implies gaussianity of $I(f)$. Ok I get that the distribution function of $I(f)$ should be continuous but why gaussian? Actually this is a simple question in probability theory but I'm just thinking about it )))

Thanks

noob-mathematician
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    The distributional limit normally distributed random variables is (possibly degenerate) normal. See http://math.stackexchange.com/questions/232540/the-limit-of-a-convergent-gaussian-random-variable-sequence-is-still-a-gaussian – Chris Janjigian Oct 29 '14 at 18:05

1 Answers1

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The following result can be found as Lemma A.3 of these lecture notes:

Let $X_n$ be a sequence of normally distributed random variables with mean zero and variances $\sigma_n^2 \in [0,\infty)$. Suppose $X_n \to X$ in distribution. Then $\sigma_n^2$ converges to some $\sigma^2 \in [0,\infty)$, and $X$ is normally distributed with mean zero and variance $\sigma^2$.

One can prove this pretty easily by looking at Fourier transforms.

In particular, this holds if $X_n \to X$ almost surely, or in probability, or in $L^p$ for any $p \in [1,\infty]$.

Nate Eldredge
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