## Introduction

These notes are mostly based on the book Stochastic Calculus for Finance vol. II, Chapter 4. I give a few propositions and focus on exercises of Shreve by make use of the Ito-Doeblin formula. The use of Ito-Doeblin formula is almost purely practical to solve continuous-time stochastic models. My treatment is slightly different from the Shreve since I emphasize on the differential forms of the formulas.

First of all, stochastic calculus is different from the ordinary calculus and looks a bit scary. But the difference lies on a very simple fact: Continuous functions have zero quadratic variation whereas the stochastic processes have nonzero. The difference results in additional terms in the differentiation and we show that without knowing deep theory, one can solve SDEs. This note focus on the practical use of the formula, not to the theory behind of it.

## Ito-Doeblin Formula

Brownian motion case. We want to differentiate functions of Brownian motion: $f(W(t))$. Think of Brownian motion as a random walk with infinitesmall steps. For each $t$, $W(t)$ is a random variable and distributed normally. Note that, we'll give a formula for general processes in the next section.

Proposition 1.(Ito-Doeblin for Brownian motion) $d(f(W(t))$ can be written as, \begin{align} df(W(t)) = f' (W(t)) dW(t) + \frac{1}{2} f'' (W(t))dt \end{align}

You probably noticed the extra second term comes from the properties of stochastic processes. By integrating this equation one can obtain, \begin{align*} f(W(t)) - f(W(0)) = \int_0^t f'(W(u))dW(u) + \frac{1}{2} \int_0^t f''(W(u))du \end{align*} Although the precise form is this integral form, we'll use differential form for practical purposes. Note that, first integral is a stochastic integral and second is a Lebesgue integral. However, even if you do not know about stochastic integration, you'll see that, Ito-Doeblin formula will work for you. Now suppose we have a function of both time and Brownian motion, i.e., we have $f(t,W(t))$. What is the extension of Ito-Doeblin formula for this case? Answer is in the following proposition.

Proposition 2. $d(f(t,W(t))$ can be written as, \begin{align} df(t,W(t)) = f_t(t,W(t))dt + f_x(t,W(t))dW(t) + \frac{1}{2} f_{xx}(t,W(t))dt \end{align}

If we integrate this equation from $0$ to $T$, we get similarly, \begin{align*} f(T,W(T)) &= f(0,W(0)) + \int_0^T f_t(W(t))dt + \int_0^T f_x(W(t))dW(t) \nonumber \\ &+ \frac{1}{2} \int_0^T f_{xx}(W(t))dt \end{align*} You probably wonder about the meaning of $f_t(t,W(t))$ and $f_x(t,W(t))$. These mean differentiation with respect to first and second variable respectively. When doing differentation wrt to second variable $W(t)$, which is a stochastic process, we take it as an ordinary variable. The variation comes from the stochasticity is captured by the last term. Besides, there is no need to different procedure for differentiation. In the exercises, you'll see this is very easy indeed.

Ito-Doeblin for Ito Processes. We do not define what the Ito process is. You can think of an Ito process as a general stochastic process without jumps. We denote it with $X(t)$ and give the Ito-Doeblin formula as in the following.

Proposition 3. $d(f(t,X(t))$ can be written as, \begin{align} df(t,X(t)) &= f_t(t,X(t))dt + f_x(t,X(t))dX(t) \nonumber \\ &+\frac{1}{2} f_{xx}(t,X(t))dX(t) dX(t) \end{align}

This is similar to the case of Brownian motion except the last term. We omit the integral form since it is straightforward.

## Examples

Before starting examples, we present a result from the theory of stochastic analysis.

Proposition 4. The following things hold, \begin{align*} dt dt &= 0 \\ dW(t) dt &= 0 \\ dW(t) dW(t) &= dt \end{align*}

The proof of this proposition is simple given the information that $W(t)$ is a standart Normal random variable. However, we omit it.

Example 1. (Solving the generalized geometric Brownian motion equation) This is Exercise 4.5 in Shreve, vol. II. Let $S(t)$ be a positive stochastic process that satisfies the generalised geometric Brownian motion differential equation, \begin{align} dS(t) = \alpha(t) S(t) dt + \sigma(t) S(t) dW(t) \end{align} where $\alpha(t)$ and $\sigma(t)$ are adapted to the filtration $\cF(t), t\geq 0$, associated with the Brownian motion $W(t),t\geq 0$. First of all, we aim to compute $d \log S(t)$ using Ito-Doeblin formula so that we would like to have a formula for $d \log S(t)$ that does not involve $S(t)$. Then we'll integrate it. Notice that $f(S(t)) = \log S(t)$ and we seek for $df(S(t))$. Then, we use the formula as follows. \begin{align}\label{BaslangicFormula} d\log(S(t)) = \frac{dS(t)}{S(t)} - \frac{1}{S^2(t)} dS(t) dS(t) \end{align} Notice that while computing $f_x$ and $f_{xx}$ part in the Ito-Doeblin formula, we take $S(t)$ as a ordinary variable and did ordinary differentiation wrt $S(t)$. In the above equation, consider the term $dS(t) dS(t)$. To handle this term, we use the last proposition. We have explicit expression of $dS(t) = \alpha(t) S(t) dt + \sigma(t) S(t) dW(t)$. So, \begin{align*} dS(t) dS(t) = (\alpha(t) S(t) dt + \sigma(t) S(t) dW(t))(\alpha(t) S(t) dt + \sigma(t) S(t) dW(t)) \end{align*} Notice because of the last proposition, from this expression, we only have $dS(t) dS(t) = \sigma^2(t) S^2(t) dt$. So, by putting also $dS(t)$, \eqref{BaslangicFormula} becomes, \begin{align} d\log(S(t)) = \sigma(t) dW(t) + (\alpha(t) - \sigma^2(t)) dt \end{align} Now we turn to the integral form: \begin{align} \log(S(t)) = \log(S(0)) + \int_0^t \sigma(s) dW(s) + \int_0^t (\alpha(s) - \sigma^2(s)) ds \end{align} If we exponentiate, \begin{align} S(t) = S(0) \exp\left[ \int_0^t \sigma(s) dW(s) + \int_0^t (\alpha(s) - \sigma^2(s)) ds \right] \end{align} We solved the SDE.

Example 2. (Solving the Vasicek Equation) This is Exercise 4.8 in Shreve, vol. II. The Vasicek interest rate SDE is, \begin{align}\label{Vasicek} dR(t) = (\alpha - \beta R(t)) dt + \sigma dW(t) \end{align} where $\alpha,\beta$ and $\sigma$ are positive constants. We solve this equation in two steps: (1) We compute $d(e^{\beta t} R(t))$ by using Ito-Doeblin formula. (2) We integrate it and solve it for $R(t)$. To compute $d(e^{\beta t} R(t))$, we use Ito-Doeblin. Think of $d(e^{\beta t} R(t))$ as $df(t,R(t)$. Then we obtain, \begin{align}\label{expVasicek} d(e^{\beta t} R(t)) = \beta e^{\beta t} R(t) dt + e^{\beta t} dR(t) \end{align} We rearrange \eqref{Vasicek} to obtain an expression for $R(t)$: \begin{align*} R(t) = \frac{\alpha}{\beta} dt + \frac{\sigma}{\beta} dW(t) - \frac{1}{\beta} dR(t) \end{align*} We put this into \eqref{expVasicek}, \begin{align}\label{expVasicek2} d(e^{\beta t} R(t)) &= e^{\beta t}(\alpha dt + \sigma dW(t) - dR(t)) + e^{\beta t} dR(t)\nonumber \\ &= e^{\beta t}(\alpha dt + \sigma dW(t)) \end{align} Hence, we integrate the both sides of the equation as the following, \begin{align*} e^{\beta t} R(t) = R(0) + \int_0^t e^{\beta u} \alpha du + \int_0^t e^{\beta u} \sigma dW(u) \end{align*} Hence, we obtain the solution $R(t)$ as, \begin{align} R(t) = R(0) e^{-\beta t} + \frac{\alpha}{\beta} (1 - e^{-\beta t}) + \sigma \int_0^t e^{-\beta (t-u)} dW(u) \end{align}