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Consider the stochastic differential equation

$$dS=\left(\mu S+\frac{\Lambda-S}{\omega}\right)dt+\sigma S dW_t.$$

such that $W_t$ is a Wiener process and $\mu,\sigma,\Lambda,\omega\in\mathbb{R}^+$. As $\omega\rightarrow 0$, does $S_t\rightarrow\Lambda$ for all $\mu$ (or is $S\sim\Lambda$)? If the expected value of $S_t$ was given by drift, $\frac{dS}{dt}=\frac{\Lambda-S_t}{\omega}+\mu$, then $\mathbb{E}[S_t]=c_0e^{-t/\omega}+\Lambda+\mu\omega$ and $\mathbb{E}[S_t]\rightarrow\Lambda$ as $\omega\rightarrow0$. Any help would be much appreciated.

Response to Kurt G.

Suppose

\begin{align} S_t&=\Phi_t\Bigg(S_0-\frac{\Lambda}{\omega}\int_0^t\frac{1}{\Phi_s}\,ds\Bigg),\\\ \Phi_t&:=e^{\sigma W_t\,-\,\sigma^2t/2\,+\,\alpha t}.\ \end{align}

If $W_t=0$ for all $t$ (to make our calculation much simpler), then

\begin{align} \int_0^t \frac{1}{\exp\left(\alpha s-\frac{\sigma^2 s}{2}\right)} ds=-\frac{2\left(e^{\frac{1}{2}t\left(\sigma^{2}-2\alpha\right)}-1\right)}{2\alpha-\sigma^{2}}. \end{align}

Hence \begin{equation} \begin{split} S&=e^{\alpha t-\frac{\sigma^2t}{2}}\left(S_0-\frac{\Lambda}{\omega}\left(-\frac{2\left(e^{\frac{1}{2}t\left(\sigma^2-2\alpha\right)}-1\right)}{2\alpha-\sigma^2}\right)\right),\\ &=e^{\alpha t-\frac{t\sigma^2}{2}}\left(S_0+\frac{2\Lambda\left(e^{\frac{t\left(\sigma^2-2\alpha\right)}{2}}-1\right)}{\omega\left(2\alpha-\sigma^2\right)}\right),\\ &=e^{\alpha t-\frac{t\sigma^2}{2}}S_0+\frac{2\Lambda\left(1- e^{\frac{2\alpha t-t\sigma^2}{2}}\right)}{\omega(2\alpha-o^2)}. \end{split} \end{equation}

So, as $\omega\rightarrow 0$, $S\rightarrow -\Lambda$. This seems counterintuitive since $S$ should be increasing when $S<\Lambda$ and decreasing when $S>\Lambda$. So, where is the mistake?

UNOwen
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1 Answers1

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For $\Lambda=0$ the answer is fairly simple because then the SDE is the familiar Black-Scholes SDE $$ \frac{dS_t}{S_t}=\alpha\,dt+\sigma\,dW_t\,,\quad\quad\alpha=\mu-\frac{1}{\omega}\,, $$ which has the solution $$ S_t=S_0\,e^{\sigma W_t-\,\sigma^2t/2\,+\,\alpha\, t}\,. $$ Clearly $E[S_t]=S_0\,e^{\alpha t}$ which converges to $\Lambda=0$ when when $\omega\to 0$ because $\alpha\to-\infty\,.$ To figure out what happens pathwise we can look at $$ E[S_t^2]=S_0^2\,e^{\sigma^2t\,+\,2\,\alpha\,t}\,. $$ Obviously, for every fixed $t$ this squared $L_2$-norm converges to zero when $\omega\to 0\,.$ Therefore, $S_t$ converges almost surely to $\Lambda=0\,.$

In general, maybe the following helps: The explicit solution to your SDE is \begin{align} S_t&=\Phi_t\Bigg(S_0+\frac{\Lambda}{\omega}\int_0^t\frac{1}{\Phi_s}\,ds\Bigg)\,,\\ \Phi_t&:=e^{\sigma W_t\,-\,\sigma^2t/2\,+\,\alpha t}\,. \end{align} (see Wikipedia). This is easily verified. Because $\Phi_t$ is identical to $S_t$ when $\Lambda$ is zero we know from above that $\Phi_t$ converges almost surely to zero when $\omega\to 0\,.$

To figure out what $S_t$ is doing when $\omega\to 0$ observe that $$\tag{1} \Phi_t\int_0^t\frac{1}{\Phi_s}\,ds=e^{\sigma W_t\,-\,\sigma^2t/2}\int_0^te^{-\sigma W_s\,+\,\sigma^2s/2\,+\,\alpha(t-s)}\,ds\,. $$ Since $e^{\alpha(t-s)}$ converges to zero when $\omega\to 0$ I think it follows from the dominated convergence theorem that the expresson (1) converges to zero. It remains to show that $$ \frac{\Phi_t}{\omega}\int_0^t\frac{1}{\Phi_s}\to 1\,. $$

Kurt G.
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  • I think setting $\mu = \Lambda = 0$ might be simplifying a bit too much. Then the constant solution $S_t = 0$ satisfies the SDE, but that isn't the case in general. The constant process $S_t = \Lambda$ doesn't satisfy the SDE, so it would be interesting if $\lim_{\omega \rightarrow \infty} S_t = \Lambda$ a.s. – user6247850 Sep 03 '21 at 18:37
  • Thanks for your response, the SDE also produces the Black-Scholes formula when $\omega\rightarrow\infty$. Does a more general solution exist when $\omega\rightarrow 0$ (i.e. when $\mu$ and $\Lambda$ are not necessarily $0$)? Note that I missed the term $Su$ in the original question but this doesn't change your case, when $\mu=\Lambda=0$. – UNOwen Sep 04 '21 at 03:59
  • @Kurt G. Thank you for your generalisation, which was very helpful. Please see the recent edit to my original question. My calculations show that the general form yields $S\rightarrow -\Lambda$ as $\omega\rightarrow 0$ when taking $W_t=0$. There might be a minus sign missing somewhere (either in the general solution or my working/SDE). – UNOwen Sep 04 '21 at 14:21
  • @UNOwen . Sure. I had a wrong minus sign. Will just fix it. – Kurt G. Sep 05 '21 at 07:59
  • Great, so that all works then. I assume setting $W_t=0$ in my calculation of the integral yields the expectance of $S_t$ (i.e. $\mathbb{E}(x)$)? Thanks for all your help. – UNOwen Sep 05 '21 at 09:02