# Lecture 24 (March 25, 2022)

### Ito process as a stochastic differential equation

Recall that the Ito’s process is

$dx_t = \mu(x_t,t) dt + \sigma(x_t,t)dw_t$

where $w_t$ is the standard BM with $w_0 = 0$ . If $G(x,t)$ is twice differentiable

Ito formula:

$dG(x_t,t) = \left(\frac{\partial G}{\partial x}\mu + \frac{\partial G}{\partial t} + \frac{1}{2}\frac{\partial^2 G}{\partial x^2}\sigma^2\right)dt + \sigma\frac{\partial G}{\partial x} dw_t$

For example:

$dx_t = \mu dt + \sigma dw_t$

If $y_t = G(x,t) = x_t^2$, then

$\frac{\partial G}{\partial x} = 2x \quad \frac{\partial^2 G}{\partial x^2} = 2 \quad \frac{\partial G}{\partial t} = 0$

\begin{aligned} dy_t &= (2x_t \mu + \frac{1}{2}\times 2\sigma^2) dt + 2x_t\sigma dw_t \\ &=(2x_t\mu + \sigma^2)dt + 2x_t\sigma dw_t \end{aligned}

Another example If $\mu = 0$ and $\sigma = 1$, then $dx_t = dw_t$. This leads to that

$dw_t^2 = dt + 2 w_t dw_t$

The term $dt$ which should not appear in classical differential calculus, is from the quadratic variation of $w_t$.

Another example: Geometric Brownian motion. Consider stock price at time $t$, it satisfies that

$dP_t = \mu P_t dt + \sigma P_t dt$

As an Ito process, $\mu(P_t, t) = \mu P_t$ and $\sigma(P_t,t) = \sigma P_t$. We can get the formula for log prices $\log P_t$. Let $G(x,t) = \log(x)$, then

$\frac{\partial G}{\partial x} = \frac{1}{x}\quad \frac{\partial^2 G}{\partial x^2} = -\frac{1}{x^2} \quad \frac{\partial G}{\partial t} = 0$

$d\log P_t = \left( \mu - \frac{\sigma^2}{2}\right) dt + \sigma dw_t$

Summary: If the price is a geometric BM, then the log price follows a generalized BM with drift $\mu-\sigma^2/2$ and volatility $\sigma$.

This shows that the change of log-prices

$\log P_{t_2} - \log P_{t_1} \sim \mathcal{N} \left( (\mu-\frac{\sigma^2}{2})(t_2-t_1), \sigma^2 (t_2-t_1) \right)$

If we have data $p_0, p_1,\dots p_n$ at equally spaced time interval $\Delta$, we want to estimate $\mu$ and $\sigma$. Note first that the log return follows the distribution

$r_t = \log P_t - \log P_{t-1} \sim \mathcal{N}((\mu-\sigma^2/2)\Delta, \sigma^2\Delta)$

then we let

$\bar{r}_n = \frac{1}{n}\sum_{t=1}^n r_t \quad s_n = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (r_t-\bar{r}_n)^2}$

By law of large numbers

$\bar{r}_n \rightarrow \mathbb{E}[r_t] = (\mu -\frac{\sigma^2}{2})\Delta \quad s_n \rightarrow \sqrt{\text{Var}(r_t)} = \sigma\sqrt{\Delta}$

method of moments:

$\hat{\sigma} = \frac{s_n}{\sqrt{\Delta}}, \quad \hat\mu = \frac{\bar{r}_n}{\Delta} + \frac{\hat{\sigma}^2}{2} = \frac{\bar{r}_n}{\Delta} + \frac{s_n^2}{2\Delta}$

Def [log-normed distribution] $X$ is log-normed distribution if $Y = \log(X)$ follows a normed distribution.

If $Y\sim \mathcal{N}(\mu,\sigma^2)$, then $X\sim \text{log-normed} (\mu,\sigma^2)$ s.t.

$\mathbb{E} X = \exp (\mu + \sigma^2/2) \quad \text{Var}(X) = \exp(2\mu+\sigma^2)[\exp(\sigma^2)-1]$

$t_1 < t_2$, then

$p_{t_2}/p_{t_1} \sim \text{log-normal} \left((\mu-\frac{\sigma^2}{2})(t_2-t_1), \sigma^2(t_2-t_1) \right)$

Then

$\mathbb{E}\left[\frac{P_{t_2}}{P_{t_1}}\right] = \exp \left[(\mu-\frac{\sigma^2}{2})(t_2-t_1) + \frac{\sigma^2(t_2-t_1)}{2} \right] = \exp(\mu(t_2-t_1))$ $\text{Var}\left(\frac{P_{t_2}}{P_{t_1}}\right) = \exp\left[ 2\mu(t_2-t_1) \right]\left\{ \sigma^2(t_2-t_1) -1 \right\}$

This shows that

$\mathbb{E}[P_{t_2}|P_{t_1}] = P_{t_1} \mathbb{E} \left[\frac{P_{t_2}}{P_{t_1}} \right] = P_{t_1}\exp(\mu(t_2-t_1))$

and

$\text{Var}[P_{t_2}|P_{t_1}] = P_{t_1}^2 \text{Var}\left(\frac{P_{t_2}}{P_{t_1}} \right) = P_{t_1}^2\exp(2\mu(t_2-t_1))\left\{\exp(\sigma^2(t_2-t_1)) - 1\right\}$

Summary:

\begin{aligned} &dP_t = \mu P_t dt + \sigma P_t dw_t \\ &d\log P_t = (\mu-\frac{\sigma^2}{2}) dt + \sigma dw_t \end{aligned}

This shows that

$\log P_t - \log P_0 = (\mu-\frac{\sigma^2}{2})t + \sigma(w_t - w_0)$

which gives

$P_t = P_0 \exp \left( (\mu-\frac{\sigma^2}{2})t + \sigma w_t\right)$

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