02 - Stochastic Differential Equation - Part II

$V(S)$ when S is a SDE - Itô III

Let $V$ be a function of $S$ where $S$ satisfies the stochastic differential equation $dS=\mu Sdt+\sigma SdX(t)$ Note how in this case $\mu$ and $\sigma$ are constants. In more elaborate models, both can be time-dependent variables and be stochastic themselves.

Using a one-dimension Taylor Series expansion, we can write $$V(S+dS)\approx V(S)+\frac{dV}{dS}dS+\frac{1}{2}\frac{d^{2}V}{d{S}^2}d{S}^2$$

To express $d{S}^2$? $$d{S}^2=(dS{)}^2={\mu }^2{S}^{2}d{t}^2+2\mu \sigma S^{2}dtdX(t)+{\sigma }^2{S}^{2}dX(t{)}^2$$

$$\begin{array}{}\text{(1)}& d{S}^2={\sigma }^2{S}^{2}dX(t{)}^2={\sigma }^2{S}^{2}dt\end{array}$$

Tip

We could generalize this a bit further Let’s have a function $$d{G}_t=A(t,X_t)dt+B(t,X_t)d{W}_t$$

Then $$\mathbb{E}[d{G}_t]=\mathbb{E}[Adt]+\mathbb{E}[Bd{W}_t]=A\mathbb{E}[dt]+B\mathbb{E}[d{W}_t]$$

Recall that $\mathbb{E}[d{W}_t]=0$, hence $$\mathbb{E}[d{G}_t]=A\mathbb{E}[dt]$$ Considering the variance, we can write $$\mathbb{V}ar[d{G}_t]=\mathbb{V}ar[Adt]+\mathbb{V}ar[Bd{W}_t]=A^2\mathbb{V}ar[dt]+B^2\mathbb{V}ar[d{W}_t]=B^{2}dt$$ Recall that $\mathbb{V}ar[dt]=0$, hence $$\mathbb{V}ar[d{G}_t]=B^2\mathbb{V}ar[d{W}_t]$$

Going back to our expansion and considering $dV=V(S+dS)-V(S)$: $$dV=\frac{dV}{dS}(\mu Sdt+\sigma SdX(t))+\frac{1}{2}\frac{d^{2}V}{d{S}^2}({\sigma }^2{S}^{2}dt)$$ $$\begin{array}{}\text{(2)}& dV=(\mu S\frac{dV}{dS}+\frac{1}{2}{\sigma }^2{S}^2\frac{d^{2}V}{d{S}^2})\cdot dt+\left(\sigma S\frac{dV}{dS}\right)\cdot dX(t)\end{array}$$

Example

Let $V(S)=log(S)$ with S satisfies the usual SDE: $dS=\mu Sdt+\sigma Sd{X}_t$. We can then use the above SDE form.

$$\frac{dV}{dS}=\frac{1}{S}$$ $$\frac{d^{2}V}{d{S}^2}=-\frac{1}{S^2}$$

Using above Equation 2: $$dV=(\mu S\frac{1}{S}+\frac{1}{2}{\sigma }^2{S}^2\frac{-1}{S^2})\cdot dt+\left(\sigma S\frac{1}{S}\right)\cdot dX(t)$$ $$dV=(\mu -\frac{1}{2}{\sigma }^2)\cdot dt+\sigma \cdot dX(t)$$ Using the integral form:

$${\int }_0^{t}d(logS)={\int }_0^t\mu -\frac{1}{2}{\sigma }^{2}d\tau +{\int }_0^t\sigma dX(\tau )$$ $$log(S_t)-log(S_0)=\mu t-\frac{1}{2}{\sigma }^{2}t+\sigma (X_t-X_0)$$ $$log\left(\frac{S_t}{S_0}\right)=\mu t-\frac{1}{2}{\sigma }^{2}t+\sigma (X_t-X_0)$$ $$S_t=S_0\cdot e^{\mu t-\frac{1}{2}{\sigma }^{2}t+\sigma (X_t-X_0)}$$

Using $X_0=0$ and $X_t=\varphi \sqrt{t}$: $$S_t=S_0\cdot e^{\mu t-\frac{1}{2}{\sigma }^{2}t+\sigma \varphi \sqrt{t}}$$

Another example with interest rate

Vasicek model

This model developed in 1978 by Vasicek is about interest rate.

The basic SDE takes this form $$\begin{array}{}\text{(3)}& dr=(\eta -\gamma r)dt+\sigma dX\end{array}$$ In this model, $\eta$, $\gamma$ and $\sigma$ are all constant.

• $\gamma$ is the speed of reversion to the (long term) mean rate. It’s the rate of reversion.
• we demote $\overline{r}$ the mean interest rate such that $\overline{r}=\frac{\eta }{\gamma }$

$$\begin{array}{}\text{(4)}& dr=\gamma (\overline{r}-r)dt+\sigma dX\end{array}$$

If we let $u=r-\overline{r}$, then $du=dr$ because we consider $\overline{r}$ as a constant. Hence $$du=-\gamma udt+\sigma dX$$ $$du+\gamma udt=\sigma dX$$
$$e^{\gamma t}du+\gamma u{e}^{\gamma t}dt=e^{\gamma t}\sigma dX$$ $$d(u{e}^{\gamma t})=\sigma e^{\gamma t}dX$$ $${\int }_0^{t}d(u_s{e}^{\gamma s})=\sigma {\int }_0^t{e}^{\gamma s}d{X}_s$$ $$u(t)e^{\gamma t}-u(0)=\sigma {\int }_0^t{e}^{\gamma s}d{X}_s$$ $$u(t)=u(0)e^{-\gamma t}+\sigma {\int }_0^t{e}^{\gamma (s-t)}d{X}_s$$

$V(t,S)$ when S is a SDE - Itô IV

This time $V$ is a function of both time $t$ and $S$ which satisfies the usual SDE: $dS=\mu Sdt+\sigma Sd{X}_t$

We can also do a Taylor Series expansion on this. And recall Equation 1 when dealing with $d{S}^2$

$$V(S+dS,t+dt)\approx V(S,t)+\frac{\partial V}{\partial S}dS+\frac{\partial V}{\partial t}dt+\frac{1}{2}\frac{{\partial }^{2}V}{\partial S^2}d{S}^2$$ $$V(S+dS,t+dt)-V(S,t)\approx \frac{\partial V}{\partial S}dS+\frac{\partial V}{\partial t}dt+\frac{1}{2}\frac{{\partial }^{2}V}{\partial S^2}{\sigma }^2{S}^{2}dt$$

$$dV=(\frac{\partial V}{\partial t}+\frac{1}{2}\frac{{\partial }^{2}V}{\partial S^2}{\sigma }^2{S}^2)dt+\frac{\partial V}{\partial S}dS$$

$$dV=(\frac{\partial V}{\partial t}+\frac{1}{2}\frac{{\partial }^{2}V}{\partial S^2}{\sigma }^2{S}^2)dt+\frac{\partial V}{\partial S}(\mu Sdt+\sigma Sd{X}_t)$$ $$\begin{array}{}\text{(5)}& dV=(\frac{\partial V}{\partial t}+\frac{1}{2}\frac{{\partial }^{2}V}{\partial S^2}{\sigma }^2{S}^2+\mu S\frac{\partial V}{\partial S})dt+\sigma S\frac{\partial V}{\partial S}d{X}_t\end{array}$$

Transition probabilty function as ODE

For background on the transition probability function, check this post

Recall from the above linked post that $y,t$ are current state and $y^{\prime },t^{\prime }$ are future values.
The transition probabilty function was expressed as a partial differential equation of the form $\frac{\partial P}{\partial t^{\prime }}=C^2\frac{{\partial }^{2}P}{\partial y^{\prime 2}}$ which was the FKE.

In the case of our usual Stochastic Differential Equation, which we write under a more general form like $$dY=A(y,t)dt+B(y,t)d{W}_t$$ We can find that the transition probability function satisfies the following Partial differential equation: $$\begin{array}{}\text{(6)}& \frac{\partial P}{\partial t^{\prime }}=\frac{1}{2}\frac{{\partial }^2(B(y^{\prime },t^{\prime }{)}^{2}p)}{\partial y^{\prime 2}}-\frac{\partial (A(y^{\prime },t^{\prime })p)}{\partial y^{\prime }}\end{array}$$

In our financial model, we’ll apply Equation 6 with the usual SDE $dS=\mu Sdt+\sigma Sd{W}_t$. Remember that $\mu$ and $\sigma$ are constant here.

$$\frac{\partial P}{\partial t^{\prime }}=\frac{1}{2}\frac{{\partial }^2({\sigma }^2{S}^{\prime 2}p)}{\partial S^{\prime 2}}-\frac{\partial (\mu S^{\prime }p)}{\partial S^{\prime }}$$

The solution to this partial differential equation has been developed in the trinomial post:

$$\begin{array}{}\text{(7)}& p(S,t;S^{\prime },t^{\prime })=\frac{1}{\sigma S^{\prime }\sqrt{2\pi (t^{\prime }-t)}}{e}^{\frac{-log\left(\frac{S}{S^{\prime }}\right)+(\mu +\frac{1}{2}{\sigma }^2)}{2{\sigma }^2(t^{\prime }-t)}}\end{array}$$

Steady-state

In some case, there are situation (random-walk) with a long term mean reversal - we say that they have a steady state distribution. This means that in the long run, the $p(y,t;y^{\prime },t^{\prime })$ doesn’t depend of the starting point $y,t$; the probability becomes time independent. Think of situations such as interest rate and volatility.

In the case of a steady state situation, $\frac{\partial P}{\partial t^{\prime }}=0$ since the process becomes time independent in the long run. And the probability (now written $p_{} (y’) $)satisfiestheordinarydifferentialequation:$$\frac{1}{2}\frac{d(B(y^{\prime }{)}^2{p}_{\mathrm{\infty }})}{d{y}^{\prime 2}}-\frac{d(A(y^{\prime })p_{\mathrm{\infty }})}{d{y}^{\prime }}=0$$ {#eq-FKE2} It isn’t anymore a partial differential equation as the time component vanishes (aka steady-state, long-term reversal)

Vasicek revisited

Recall from above the Vasicek model $$dr=\gamma (\overline{r}-r)dt+\sigma d{W}_t$$ Using ?@eq-FKE2, we can write the steady state distribution $p_{\mathrm{\infty }}{r}^{\prime }$ following ordinary differential equation. $$\frac{1}{2}{\sigma }^2\frac{d^2{p}_{\mathrm{\infty }}}{d{r}^{\prime 2}}-\gamma \frac{d((\overline{r}-r)p_{\mathrm{\infty }})}{d{r}^{\prime }}=0$$ Integrating both sides, we get: $$\frac{1}{2}{\sigma }^2\frac{d{p}_{\mathrm{\infty }}}{d{r}^{\prime }}+\gamma (r-\overline{r})p_{\mathrm{\infty }}=K$$ K being a constant. $$\frac{1}{2}{\sigma }^2\frac{d{p}_{\mathrm{\infty }}}{d{r}^{\prime }}=-\gamma (r-\overline{r})p_{\mathrm{\infty }}+K$$ Letting the constant be 0 (need explanation here) Letting the $\mathrm{\infty }$ just for convenience purposes.

$$\frac{1}{p}dp=\frac{-2\gamma }{{\sigma }^2}(r-\overline{r})d{r}^{\prime }$$ Integrating both sides, $$\int \frac{1}{p}dp=\frac{-2\gamma }{{\sigma }^2}\int (r-\overline{r})d{r}^{\prime }$$ $$log(p)=\frac{-2\gamma }{{\sigma }^2}\frac{1}{2}(r-\overline{r}{)}^2+K$$ Using a normalizing constant, as we inverse the log $$\begin{array}{}\text{(8)}& p(r)=A\cdot e^{-\frac{\gamma }{{\sigma }^2}(r-\overline{r}{)}^2}\end{array}$$

We know that ${\int }_{\mathbb{R}}p(r)dr=1$, hence $$A{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{\gamma }{{\sigma }^2}(r-\overline{r}{)}^2}dr=1$$ We can integrate this using substitution $u=\frac{\sqrt{\gamma }(r-\overline{r})}{\sigma }$ with $\frac{du}{dr}=\frac{\sqrt{\gamma }}{\sigma }$ $$A{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-u^2}\frac{\sigma }{\sqrt{\gamma }}du=1$$ $$A\frac{\sigma }{\sqrt{\gamma }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-u^2}du=1$$ $$A\frac{\sigma }{\sqrt{\gamma }}\sqrt{\pi }=1$$ $$A=\frac{\sqrt{\gamma }}{\sigma \sqrt{\pi }}=\frac{1}{\sigma }\sqrt{\frac{\gamma }{\pi }}$$

Putting it all back together in Equation 8: $$p_{\mathrm{\infty }}(r)=A\cdot e^{-\frac{\gamma }{{\sigma }^2}(r-\overline{r}{)}^2}=\frac{1}{\sigma }\sqrt{\frac{\gamma }{\pi }}{e}^{-\frac{\gamma }{{\sigma }^2}(r-\overline{r}{)}^2}$$ This means: in our case of a steady state stochastic process, the variable $r$ follows a normal distribution with mean $\text{bar{r}}}$ and standard deviation $\frac{\sigma }{\sqrt{2\cdot \gamma }}$

$V(t,S_1,S_2)$ Modeling correlated random walks

We have now 2 assets

$$\{\begin{array}{l}d{S}_1={\mu }_1{S}_{1}dt+{\sigma }_1{S}_{1}d{X}_1\\ d{S}_2={\mu }_2{S}_{2}dt+{\sigma }_2{S}_{2}d{X}_2\end{array}$$

Using a Taylor series expansion $$V(t+dt,S_1+d{S}_1,S_2+d{S}_2)\approx V(t,S1,S2)+\frac{\partial V}{\partial t}dt+\frac{\partial V}{\partial S_1}d{S}_1+\frac{\partial V}{\partial S_2}d{S}_2+\frac{1}{2}\frac{{\partial }^{2}V}{\partial S_1^2}d{S}_1^2+\frac{1}{2}\frac{{\partial }^{2}V}{\partial S_2^2}d{S}_2^2+\frac{{\partial }^{2}V}{\partial S_1\partial S_2}d{S}_{1}d{S}_2$$ $$\begin{array}{}\text{(9)}& dV=\frac{\partial V}{\partial t}dt+\frac{\partial V}{\partial S_1}d{S}_1+\frac{\partial V}{\partial S_2}d{S}_2+\frac{1}{2}\frac{{\partial }^{2}V}{\partial S_1^2}d{S}_1^2+\frac{1}{2}\frac{{\partial }^{2}V}{\partial S_2^2}d{S}_2^2+\frac{{\partial }^{2}V}{\partial S_1\partial S_2}d{S}_{1}d{S}_2\end{array}$$

Now we can write down some results we’ll plug into that expansion

• $d{S}_i={\mu }_i{S}_{i}dt+{\sigma }_i{S}_{i}d{X}_i$
• Using Equation 1, we write $d{S}_i^2={\sigma }_i^2{S}_i^{2}dt$
• Letting $\mathbb{E}[{\varphi }_1{\varphi }_2]=\rho$, we write $\mathbb{E}[d{X}_{1}d{X}_2]=\mathbb{E}[{\varphi }_1\sqrt{dt}{\varphi }_2\sqrt{dt}]=\rho dt$
• Recall that all terms less than dt = 0. Hence in the product $d{S}_{1}d{S}_2=({\mu }_1{S}_{1}dt+{\sigma }_1{S}_{1}d{X}_1)({\mu }_2{S}_{2}dt+{\sigma }_2{S}_{2}d{X}_2)$, the only term that we are considering is the outter product ${\sigma }_1{S}_{1}d{X}_1{\sigma }_2{S}_{2}d{X}_2$. Finally, re-using above result, $d{S}_{1}d{S}_2=\rho {\sigma }_1{\sigma }_2{S}_1{S}_{2}dt$.

Re-writing Equation 9 by substituting $d{S}_1$ and $d{S}_2$, and factoring the terms in dt and those in $d{X}_t$, we get: $$dV=(\frac{\partial V}{\partial t}+{\mu }_1{S}_1\frac{\partial V}{\partial S_1}+{\mu }_2{S}_2\frac{\partial V}{\partial S_2}+{\sigma }_1^2{S}_1^2\frac{{\partial }^{2}V}{\partial S_1^2}+{\sigma }_2^2{S}_2^2\frac{{\partial }^{2}V}{\partial S_2^2}+\rho {\sigma }_1{\sigma }_2{S}_1{S}_2\frac{{\partial }^{2}V}{\partial S_1\partial S_2})dt+({\sigma }_1{S}_1\frac{\partial V}{\partial S_1}d{X}_1+{\sigma }_2{S}_2\frac{\partial V}{\partial S_2}d{X}_2)$$

Itô