In this post we will define the following results:

Stochastic integral Result Variance
$\int_0^t \mathrm{d}B_s$ $B_t$ $t$
$\int_0^t s \mathrm{d}B_s$ $tB_t - \mathrm{ibm}(t, \omega)$ $\tfrac{1}{3}t^3$
$\int_0^t B_s \mathrm{d}B_s$ $\tfrac{1}{2}B_t^2 - \tfrac{1}{2}t$ $\tfrac{1}{2}t^2$
$\int_0^t B_s^2 \mathrm{d}B_s$ $\tfrac{1}{3}B_t^3 - \mathrm{ibm}_t$ $t^3$
$\int_0^t B_s^k\mathrm{d}B_s$ $\tfrac{1}{k+1}B_t^{k+1} - k \int_0^t B_s^{k-1} \mathrm{d}s$
$\int_0^t \exp\left(B_s - \tfrac{s}{2}\right)\mathrm{d}B_s$ $\exp(B_t - \tfrac{t}{2}) - 1$ $e^t - 1$

Introduction

Recall that the integrated Brownian motion (IBM), which is defined as

$$\begin{aligned} \mathrm{ibm}(t, \omega) = \int_0^t B_s \mathrm{d}s, \end{aligned}$$

is a random variable which follows the normal distribution with zero mean and variance $t^3/3$. We will use the notation $\mathrm{ibm}_t$ hereafter.

This MSE thread is worth reading.

Hereafter, we will give a few examples of stochastic integrals of the Itô type which are given by

$$I(t, \omega) {}={} \int_0^t f(t, \omega) \mathrm{d}B_t,$$

where $f$ is an Itô-integrable function. We shall also use the shorthand notation $I_t(\omega) = I(t, \omega)$.

Let us recall some important results from stochastic calculus:

Theorem 1 (Itô isometry). Let $f:(t_1, t_2)\times \Omega \to {\rm I\!R}$ be an Itô-itegral function on $(t_1, t_2)$. Then

$${\rm I\!E}\left[\left(\int_{t_1}^{t_2}f(t, \omega){\rm d}B_t\right)^2\right] {}={} {\rm I\!E}\left[\int_{t_1}^{t_2}f^2(t, \omega){\rm d}t\right].$$

Theorem 2 (Itô's lemma). Let $X_t$ be an Itô process with

$${\rm d}X_t = u{\rm d}t + v{\rm d}B_t.$$

Let $g:[0, \infty) \times {\rm I\!R} \to {\rm I\!R}$ be twice continuously differentiable on $[0, \infty) \times {\rm I\!R} $. Then, the process $Y_t = g(t, X_t)$ is an Itô process with

$${\rm d}Y_t = \frac{\partial g}{\partial t}(t, X_t){\rm d}t + \frac{\partial g}{\partial x}(t, X_t){\rm d}X_t + \tfrac{1}{2} \frac{\partial^2 g}{\partial x^2}(t, X_t)({\rm d}X_t)^2,$$

where $({\rm d}X_t)^2 = ({\rm d}X_t) \cdot ({\rm d}X_t)$ is computed according to the rules

$${\rm d}t \cdot {\rm d}t = {\rm d}t\cdot {\rm d}B_t = {\rm d}B_t \cdot {\rm d}t = 0, \ {\rm d}B_t \cdot {\rm d}B_t = {\rm d}t.$$



Stochastic Integrals

Stochastic integral No 1

Consider the stochastic integral

$$\begin{aligned}I_1(t, \omega) = \int_0^t s \mathrm{d}B_s = t B_t - \mathrm{ibm}_t, \end{aligned}$$

where $\mathrm{ibm}_t$ denotes the integrated Brownian motion given above.

The expected value of $I_1$ is ${\rm I\!E}[I_1] = {\rm I\!E}[\int_0^t s \mathrm{d}\mathrm{B}_s] = 0$ and its variance is given by

$$\begin{aligned} \mathrm{Var}[I_1] = {\rm I\!E}[I_1^2] {}={}& {\rm I\!E}\left[\left(\int_0^t s \mathrm{d}B_s\right)^2\right]\\{}={}& {\rm I\!E}\left[\int_0^t s^2 \mathrm{d}s\right] {}={}\frac{t^3}{3}.\end{aligned}$$

We have used the Itô isometry.

Note that the above computation is easier than trying to do

$${\rm I\!E}[I_1^2] {}={} {\rm I\!E}[(tB_t - \mathrm{ibm}_t)^2].$$


Stochastic integral No 2

Define

$$\begin{aligned}I_2 = \int_0^t B_s \mathrm{d}B_s\end{aligned}$$

Using Itô's lemma, we obtain

$$\begin{aligned}I_2 = \tfrac{1}{2}B_t^2 - \tfrac{1}{2}t\end{aligned}$$

and ${\rm I\!E}[I_2] = 0$, while the variance is

$$\begin{aligned} \mathrm{Var}[I_2] = {\rm I\!E}[I_2^2] {}={}& {\rm I\!E}[(\tfrac{1}{2}B_t^2 - \tfrac{1}{2}t)^2] \\ {}={}& \tfrac{1}{4}{\rm I\!E}[(B_t^2 - t)^2] \\ {}={}& \tfrac{1}{4}{\rm I\!E}[B_t^4 + t^2 - 2tB_t^2]\end{aligned}$$

and using the fact that ${\rm I\!E}[B_t^2] = t$ and ${\rm I\!E}[B_t^4] = 3t^2$, it is ${\rm I\!E}[I_2^2] = \tfrac{1}{2}t^2$.


Stochastic integral No 3

Define the following stochastic integral which involves the squared Brownian motion

$$\begin{aligned}I_3 = \int_0^t B_s^2 \mathrm{d}B_s\end{aligned}$$

This can be computed by applying Itô's formula to $f(t, x) = \tfrac{1}{3}x^3$ and gives

$$\begin{aligned}I_3 = \tfrac{1}{3}B_t^3 -\mathrm{ibm}_t,\end{aligned}$$

where $\mathrm{ibm}_t$ is the integrated Brownian motion.

The expectation of $I_3$ is clearly ${\rm I\!E}[\int_0^t B_s^2 \mathrm{d}B_s] = 0$ and its variance is

$$\begin{aligned}\mathrm{Var}[I_3] = {\rm I\!E}[I_3^2] {}={}& {\rm I\!E}[(\tfrac{1}{3}B_t^3 -I_t)^2]\\ {}={}& \tfrac{1}{9}{\rm I\!E}[B_t^6] + {\rm I\!E}[I_t^2] - \tfrac{2}{3} {\rm I\!E}[B_t^3 I_t],\end{aligned}$$

and ${\rm I\!E}[B_t^6] = 15 t^3$, ${\rm I\!E}[I_t^2] = t^3/3$ and

$$\begin{aligned}{\rm I\!E}[B_t^3 I_t] = {\rm I\!E} \left[ B_t^3 \int_0^t B_s \mathrm{d}s\right] = {\rm I\!E} \left[ \int_0^t B_t^3 B_s \mathrm{d}s\right] = \int_0^t {\rm I\!E}[B_t^3 B_s]\mathrm{d}s,\end{aligned}$$

Now using Isserlis's Theorem and the fact that $0 \leq s \leq t$ and ${\rm I\!E}[B_t B_s] = s$ and ${\rm I\!E}[B_t^2] = t$ we have ${\rm I\!E}[B_t^3 B_s] = 3ts$, therefore

$${\rm I\!E}[B_t^3 I_t] = \int_0^t {\rm I\!E}[B_t^3 B_s]\mathrm{d}s = \int_0^t 3ts \mathrm{d}s = \tfrac{3}{2}t^3.$$

Overall,

$$\begin{aligned} \mathrm{Var}[I_3] = t^3.\end{aligned}$$

We can generalise the above result using the formula

$$\begin{aligned}\int_0^t B_s^k \mathrm{d}B_s = \tfrac{1}{k+1}B_t^{k+1} - k \int_0^t B_s^{k-1}\mathrm{d}s\end{aligned}$$

and proceeding as above.


Stochastic integral No 4

Consider the stochastic integral

$$\begin{aligned}I_4 = \int_0^t \exp (B_s - \tfrac{1}{2}s) \mathrm{d}B_s\end{aligned}$$

Using Itô’s formula, we may show that this is equal to

$$\begin{aligned}I_4 = \exp (B_t - \tfrac{1}{2}t) - 1 = e^{B_t}e^{-\tfrac{1}{2}t} - 1.\end{aligned}$$

Since $B_t$ follows the normal distribution with zero mean and variance $t$, the random variable $e^{B_t}$ follows the log-normal distribution.

The expectation of $I_4$ is ${\rm I\!E}[\int_0^t \exp (B_s - \tfrac{1}{2}s) \mathrm{d}B_s] = 0$ and its variance is

$$\begin{aligned}\mathrm{Var}[I_4] {}={} {\rm I\!E}[I_4^2] {}={}& {\rm I\!E}[(\exp (B_t - \tfrac{1}{2}t) - 1)^2]\\ {}={}& {\rm I\!E}[\exp (2B_t - t) - 2\exp (B_t - \tfrac{1}{2}t) + 1]\end{aligned}$$

where ${\rm I\!E}[\exp (2B_t - t)] = e^{-t}{\rm I\!E}[\exp (2B_t)]$. Since $2B_t \sim \mathcal{N}(0, 4t)$, ${\rm I\!E}[e^{2B_t}] = e^{\frac{4t}{2}} = e^{2t}$, so ${\rm I\!E}[\exp (2B_t - t)] = e^t$.

Similarly, ${\rm I\!E}[2\exp (B_t - \tfrac{1}{2}t)] = 2e^{-\frac{t}{2}}{\rm I\!E}[e^{B_t}] = 2$.

Overall,

$$\begin{aligned}\mathrm{Var}[I_4] {}={} e^t - 1.\end{aligned}$$