Ito Calculus – quantdev.blog

Exercises

Exercise 1 (A strange martingale) Let $(B_t,t\geq 0)$ be a standard Brownian motion. Consider the process:

\[ M_t = \frac{1}{\sqrt{1-t}}\exp\left(\frac{-B_t^2}{2(1-t)}\right), \quad \text{ for }0 \leq t < 1 \]

Show that $M_t$ can be represented by:

\[ M_t = 1 + \int_0^t \frac{-B_s M_s}{1-s}dB_s, \quad \text{ for } 0 \leq t \leq 1 \]

Deduce from the previous question that $(M_s,s \leq t)$ is a martingale for $t < 1$ and for the Brownian filtration.
Show that $\mathbb{E}[M_t] = 1$ for all $t < 1$.
Prove that $\lim_{t \to 1^-} M_t = 0$ almost surely.

Solution.

Let $f(t,x) = \frac{1}{\sqrt{1-t}}e^{-\frac{x^2}{2(1-t)}}$. We have:

\[ \begin{align*} \frac{\partial f}{\partial t} &= \frac{\sqrt{1-t}\cdot e^{-\frac{x^2}{2(1-t)}}\cdot \frac{x^2}{2(1-t)^2}\cdot (-1) - e^{-\frac{x^2}{2(1-t)}}\cdot \frac{1}{2\sqrt{1-t}}\cdot(-1)}{(1-t)}\\ &=\frac{e^{-\frac{x^2}{2(1-t)}}}{(1-t)}\left( -\frac{x^2}{2(1-t)^{3/2}}+\frac{1-t}{2(1-t)^{3/2}}\right)\\ &= \frac{e^{-\frac{x^2}{2(1-t)}}((1-t) - x^2)}{2(1-t)^{5/2}} \end{align*} \]

Also, the first and second derivatives with respect to the space variable $x$ are:

\[ \begin{align*} \frac{\partial f}{\partial x} &= \frac{1}{\sqrt{1-t}}\exp\left(-\frac{x^2}{2(1-t)}\right)\left(-\frac{x}{(1-t)}\right)\\ &= -\frac{x}{(1-t)^{3/2}} \exp\left(-\frac{x^2}{2(1-t)}\right) \end{align*} \]

\[ \begin{align*} \frac{\partial^2 f}{\partial x^2} &= - \frac{1}{(1-t)^{3/2}}\left\{\exp\left(-\frac{x^2}{2(1-t)}\right) + x\exp\left(-\frac{x^2}{2(1-t)}\right)\left(-\frac{x}{(1-t)}\right)\right\}\\ &= - \frac{1}{(1-t)^{3/2}}\exp\left(-\frac{x^2}{2(1-t)}\right) \left\{1 - \frac{x^2}{(1-t)}\right\}\\ &= \frac{(x^2 - (1-t))}{(1-t)^{5/2}}\exp\left(-\frac{x^2}{2(1-t)}\right) \end{align*} \]

By Ito’s formula:

\[ \begin{align*} df(t,B_t) &= (f_t + \frac{1}{2}f_{xx})dt + f_x dB_t\\ dM_t &= -\frac{B_t}{(1-t)} \frac{e^{-\frac{B_t^2}{2(1-t)}}}{\sqrt{1-t}}\\ &= -\frac{B_t M_t}{(1-t)} \end{align*} \]

where in the second step, we used the fact that $f_t + \frac{1}{2}f_{xx} = 0$, so the $dt$ term is zero.

In the integral form:

\[ \begin{align*} M_t - M_0 &= \int \frac{- B_s M_s}{1 - s}dB_s\\ M_t &= 1 + \int \frac{- B_s M_s}{1 - s}dB_s \end{align*} \]

The Ito integral $\int_0^t \frac{-B_s M_s}{1-s}dB_s$ is a continuous martingale with respect to the brownian filtration. Hence $(M_t,t \geq 0)$ is a martingale.
The expectation of an Ito integral is zero. Hence, $\mathbb{E}[M_t] = 1$ for all $t < 1$.
Using Markov’s inequality followed by (concave) Jensen’s inequality, because we know that $\mathbb{E}[B_t^2]=t$, $\mathbb{E}[B_t^4]=3t^2$ and $Var(B_t^2) = 2t^2$ for any $\epsilon > 0$:

--- title: "Ito Calculus" author: "Quasar" date: "2024-07-12" categories: [Stochastic Calculus] image: "image.jpg" toc: true toc-depth: 3 --- ## Exercises ::: {#exr-a-strange-martingale} (A strange martingale) Let $(B_t,t\geq 0)$ be a standard Brownian motion. Consider the process: $$ M_t = \frac{1}{\sqrt{1-t}}\exp\left(\frac{-B_t^2}{2(1-t)}\right), \quad \text{ for }0 \leq t < 1 $$ (a) Show that $M_t$ can be represented by: $$ M_t = 1 + \int_0^t \frac{-B_s M_s}{1-s}dB_s, \quad \text{ for } 0 \leq t \leq 1 $$ (b) Deduce from the previous question that $(M_s,s \leq t)$ is a martingale for $t < 1$ and for the Brownian filtration. (c) Show that $\mathbb{E}[M_t] = 1$ for all $t < 1$. (d) Prove that $\lim_{t \to 1^-} M_t = 0$ almost surely. ::: *Solution*. Let $f(t,x) = \frac{1}{\sqrt{1-t}}e^{-\frac{x^2}{2(1-t)}}$. We have: $$ \begin{align*} \frac{\partial f}{\partial t} &= \frac{\sqrt{1-t}\cdot e^{-\frac{x^2}{2(1-t)}}\cdot \frac{x^2}{2(1-t)^2}\cdot (-1) - e^{-\frac{x^2}{2(1-t)}}\cdot \frac{1}{2\sqrt{1-t}}\cdot(-1)}{(1-t)}\\ &=\frac{e^{-\frac{x^2}{2(1-t)}}}{(1-t)}\left( -\frac{x^2}{2(1-t)^{3/2}}+\frac{1-t}{2(1-t)^{3/2}}\right)\\ &= \frac{e^{-\frac{x^2}{2(1-t)}}((1-t) - x^2)}{2(1-t)^{5/2}} \end{align*} $$ Also, the first and second derivatives with respect to the space variable $x$ are: $$ \begin{align*} \frac{\partial f}{\partial x} &= \frac{1}{\sqrt{1-t}}\exp\left(-\frac{x^2}{2(1-t)}\right)\left(-\frac{x}{(1-t)}\right)\\ &= -\frac{x}{(1-t)^{3/2}} \exp\left(-\frac{x^2}{2(1-t)}\right) \end{align*} $$ $$ \begin{align*} \frac{\partial^2 f}{\partial x^2} &= - \frac{1}{(1-t)^{3/2}}\left\{\exp\left(-\frac{x^2}{2(1-t)}\right) + x\exp\left(-\frac{x^2}{2(1-t)}\right)\left(-\frac{x}{(1-t)}\right)\right\}\\ &= - \frac{1}{(1-t)^{3/2}}\exp\left(-\frac{x^2}{2(1-t)}\right) \left\{1 - \frac{x^2}{(1-t)}\right\}\\ &= \frac{(x^2 - (1-t))}{(1-t)^{5/2}}\exp\left(-\frac{x^2}{2(1-t)}\right) \end{align*} $$ By Ito's formula: $$ \begin{align*} df(t,B_t) &= (f_t + \frac{1}{2}f_{xx})dt + f_x dB_t\\ dM_t &= -\frac{B_t}{(1-t)} \frac{e^{-\frac{B_t^2}{2(1-t)}}}{\sqrt{1-t}}\\ &= -\frac{B_t M_t}{(1-t)} \end{align*} $$ where in the second step, we used the fact that $f_t + \frac{1}{2}f_{xx} = 0$, so the $dt$ term is zero. In the integral form: $$ \begin{align*} M_t - M_0 &= \int \frac{- B_s M_s}{1 - s}dB_s\\ M_t &= 1 + \int \frac{- B_s M_s}{1 - s}dB_s \end{align*} $$ (b) The Ito integral $\int_0^t \frac{-B_s M_s}{1-s}dB_s$ is a continuous martingale with respect to the brownian filtration. Hence $(M_t,t \geq 0)$ is a martingale. (c) The expectation of an Ito integral is zero. Hence, $\mathbb{E}[M_t] = 1$ for all $t < 1$. (d) Using Markov's inequality followed by (concave) Jensen's inequality, because we know that $\mathbb{E}[B_t^2]=t$, $\mathbb{E}[B_t^4]=3t^2$ and $Var(B_t^2) = 2t^2$ for any $\epsilon > 0$: