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It's well known that $X_t:=int_0^tB_tau dtau$ 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.

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

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

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

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