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It's well known that $X_t:=\int_0^tB_\tau d\tau$ where $\{B_\tau\}$ is a 1D standard Brownian motion is distributed as $N(0, t^3/3)$. Is there any "immediate" way to see this fact?

The easiest one I can get: discretise $X_t$ into Riemannian sum, and break each $B_{\tau_i}$ into independent increments over tiny intervals, then we see the sum is just a linear sum of independent normal distributions, then take the limit and use convergence in distribution to conclude.

I wouldn't say this is hard, but this is in no way trivial or immediate either. Can we somehow see this fact without any effort or whatsoever? Thanks.

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    $\begingroup$ Writing the integral as limit of Riemann sums is ths natural easiest way to prove this. $\endgroup$ – Kavi Rama Murthy Jan 16 at 8:26
  • $\begingroup$ @KaviRamaMurthy Ok, but I'm still expecting something easier, like a "continuous" argument (cleverly determine its characterisitc function, so to speak). Discretisation looks non-tidy, albeit natural. $\endgroup$ – Vim Jan 16 at 8:29
  • $\begingroup$ Why do you have to breat $B_{\tau_i}$ into tiny intervals...? The random vector $(B_{t_1},\ldots,B_{t_n})$ is Gaussian for any choice of $t_i$ (as $B$ is a Gaussian process), and hence $$\sum_{i=1}^n c_i B_{t_i}$$ is Gaussian for any choice of constants $c_i$. $\endgroup$ – saz Jan 16 at 8:57
  • $\begingroup$ @saz you're definitely correct. It's just that estimating the variance of independent sums is easier. $\endgroup$ – Vim Jan 16 at 12:53
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    $\begingroup$ @Vim Well, the expectation as well as the variance can be calculated using Fubini's theorem... that's not really a problem. From my point of view, the key point is to show that $X_t$ is Gaussian. By the way, I don't think that it possible to get rid of the discretisation - simply because the integral is defined as the limit of the discretisations. Obviously, you can apply some "big" result and hide the discretisation there... but that doesn't make the proof "easier", in my oppinion. $\endgroup$ – saz Jan 16 at 14:03
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There are three things which need to be done:

  1. Check that $X_t$ is Gaussian.
  2. Compute its expectation.
  3. Compute its variance.

1. $X_t$ is Gaussian: It seems to me that the most natural (and also direct) way to prove this is an approximation by Riemann sums. Noting that

$$X_t^{(n)} := \sum_{j=1}^n \frac{1}{n} B_{t j/n} \tag{1}$$

is Gaussian for each $t>0$ (because $(B_t)_{t \geq 0}$ is a Gaussian process), we find that $$X_t = \lim_{n \to \infty} X_t^{(n)}$$ is Gaussian as pointwise limit of Gaussian random variables. For an alternative reasoning see the very end of my answer.

2. Compute $\mathbb{E}(X_t)$: Since $\mathbb{E}(B_s)=0$ for all $s \geq 0$, it follows that each $X_t^{(n)}$ (defined in $(1)$) has expectation zero, and hence its limit $X_t = \lim_n X_t^{(n)}$ has expectation zero. Alternatively, we can apply Fubini's theorem:

$$\mathbb{E}(X_t) = \mathbb{E} \left( \int_0^t B_s \, ds \right) = \int_0^t \underbrace{\mathbb{E}(B_s)}_{=0} \, ds =0.$$

3. Compute $\text{var}(X_t)$: Since we already know that $\mathbb{E}(X_t)=0$, we clearly have $\text{var}(X_t) = \mathbb{E}(X_t^2)$. As

$$X_t^2 = \int_0^t \int_0^t B_s B_r \, ds \, dr$$

it follows from Fubini's theorem that

$$\mathbb{E}(X_t^2) = \int_0^t \int_0^t \mathbb{E}(B_s B_r) \, ds \, dr. \tag{2}$$

By symmetry, we thus get

$$\mathbb{E}(X_t^2) = 2 \int_0^t \int_0^r \underbrace{\mathbb{E}(B_s B_r)}_{\min\{s,r\}=s} \, ds \, dr = 2 \int_0^t \int_0^r s \, ds = \frac{t^3}{3}$$

If you don't like symmetrization, then note that (2) implies

$$\mathbb{E}(X_t^2) = \int_0^t \int_0^s \underbrace{\mathbb{E}(B_s B_r)}_{\min\{s,r\}=s} \, ds \, dr + \int_0^t \int_s^t \underbrace{\mathbb{E}(B_s B_r)}_{\min\{s,r\}=r} \, ds \, dr$$

and each of the integrals can be calculated explicitly using standard calculus.


Let me close this answer with a result which combines all the three steps into one.

Proposition: Let $(L_t)_{t \geq 0}$ be a Lévy process with characteristic exponent $\psi$, i.e. $$\mathbb{E}\exp(i \xi L_t) = \exp(-t \psi(\xi)), \qquad t \geq 0, \xi \in \mathbb{R}. \tag{3}$$ Then the characteristic function of $$X_t := \int_0^t L_s \, ds$$ equals $$\phi(\xi) = \exp \left(- \int_0^t \psi(\xi s) \, ds \right), \qquad \xi \in \mathbb{R}.$$

If you are not familar with Lévy processes (that it, stochastic processes with independent and stationary increments), then you can just think of a Brownian motion; in this case $\psi$ is given by $\psi(\xi) = \xi^2/2$. Applying the proposition, we thus find that the characteristic function of $X_t = \int_0^t B_s \, ds$ equals $$\exp \left(- \frac{t^3}{3} \frac{\xi^2}{2} \right)$$ which is the characteristic function of $N(0,t^3/3)$, and so $X_t \sim N(0,t^3/3)$.

Proof of the proposition: For fixed $n \in \mathbb{N}$ and $t>0$ set $t_j := t j/n$ for $j=1,\ldots,n$, and set

$$\phi_n(\xi) := \mathbb{E} \exp \left( i \xi \frac{1}{n} \sum_{j=1}^n L_{t_j} \right).$$

Denote by $\mathcal{F}_t := \sigma(L_s; s \leq t)$ the canonical filtration of $(L_t)_{t \geq 0}$. Using the tower property of the conditional expectation, we find

$$\begin{align*} \phi_n(\xi) &= \mathbb{E} \bigg\{ \mathbb{E} \bigg[ \exp \left( i \xi \frac{1}{n} \sum_{j=1}^n L_{t_j} \right) \mid \mathcal{F}_{t_{n-1}} \bigg] \bigg\} \\ &= \mathbb{E} \bigg\{ \exp \left( i \xi \frac{1}{n} \sum_{j=1}^{n-1} L_{t_j} \right) \mathbb{E} \bigg[ \exp \left( i \xi \frac{1}{n} L_{t_n} \right) \mid \mathcal{F}_{t_{n-1}} \bigg] \bigg\} \tag{4} \end{align*}$$

Since $(L_t)_{t \geq 0}$ has independent increments, we have

$$\begin{align*} \mathbb{E} \bigg[ \exp \left( i \xi \frac{1}{n} L_{t_n} \right) \mid \mathcal{F}_{t_{n-1}} \bigg] &=\exp(i \xi/n L_{t_{n-1}}) \mathbb{E} \bigg[ \exp \left( i \xi \frac{1}{n} (L_{t_n}-L_{t_{n-1}}) \right) \mid \mathcal{F}_{t_{n-1}} \bigg] \\ &= \exp(i \xi/n L_{t_{n-1}}) \mathbb{E}\exp\left( i \xi \frac{1}{n} (L_{t_n}-L_{t_{n-1}}) \right). \end{align*}$$

Using that $(L_t)_{t \geq 0}$ has stationary increments, i.e. $L_{t_n}-L_{t_{n-1}} \sim L_{t_n-t_{n-1}}=L_{1/n}$, and using $(3)$ we thus get

$$ \mathbb{E} \bigg[ \exp \left( i \xi \frac{1}{n} L_{t_n} \right) \mid \mathcal{F}_{t_{n-1}} \bigg] = \exp(i \xi/n L_{t_{n-1}}) \exp \left(- \frac{1}{n} \psi \left( \frac{\xi}{n} \right) \right).$$

Plugging this into $(4)$, we obtain that

$$\phi_n(\xi) = \mathbb{E} \bigg\{ \exp \left( i \xi \frac{1}{n} \sum_{j=1}^{n-2} L_{t_j} + i \xi \frac{2}{n} L_{t_{n-1}} \right) \exp \left(- \frac{1}{n} \psi \left( \frac{\xi}{n} \right) \right) .$$

Iterating this reasoning (i.e. next conditioning on $\mathcal{F}_{t_{n-2}}$, then on $\mathcal{F}_{t_{n-3}}$, ...) we conclude that

$$\phi_n(\xi) = \exp \left( - \frac{1}{n} \sum_{j=1}^n \psi \left( \frac{j}{n} \xi \right) \right).\tag{5}$$

Finally, we note that

$$X_t \stackrel{\text{def}}{=} \int_0^t L_s \, ds = \lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^n L_{tj/n},$$

and thus, by letting $n \to \infty$ in (5), we get

$$\mathbb{E}\exp(i \xi X_t) = \exp \left(- \int_0^t \psi(s \xi) \, ds \right).$$

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  • $\begingroup$ Thanks for such a detailed answer! $\endgroup$ – Vim Jan 17 at 0:39
  • $\begingroup$ @Vim You're welcome. $\endgroup$ – saz Jan 17 at 7:30

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