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I didn't study stochastic process in a systematic way, but I need to use it in financial analysis. Here's my question.

I know the solution of the SDE $dX_t = \mu X_tdt + \sigma X_t dW_t$, given that $\mu$ and $\sigma$ are constants, and now I was asked to solve the following SDE

$dX_t = \mu(t)X_tdt + \sigma X_t dW_t$, given that $\mu(t)$ is a growth function

My attempt

So I simply follow the procedure as that when $\mu$ is constant. Let $y=f(x)= \ln X$ (actually I don't know why $\ln X$ is chosen, could anyone explain it to me?) and apply Ito's lemma. We have $$dy=(\mu (t) - \frac{\sigma ^2}{2})dt + \sigma dW_t$$ Differentiating both sides, the solution is given by $$y_t =y_0 + \int_0^t \mu (s)s ds - \frac{\sigma ^2}{2}t + \sigma W_t $$ Since $X=e^y$, we have $$X_t = e^{y_0 + \int_0^t \mu (s)s ds - \frac{\sigma ^2}{2}t + \sigma W_t} =X_0 e^{ \int_0^t \mu (s)s ds - \frac{\sigma ^2}{2}t + \sigma W_t}$$ Is my try correct?

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  • $\begingroup$ If you know Ito's lemma, you can actually check if your solution is correct by just applying on the expression you have $X_t$. $\endgroup$ – Calculon Oct 5 '16 at 11:01
  • $\begingroup$ There are different ways to think about why you should consider $\ln(X_t)$. One way to think about it is to note that the ODE $dX_t=\mu X_t dt$ has an exponential solution, and that therefore the "first order noise term" in Ito's formula will also behave nicely under this substitution (since that term is $\sigma X_t dW_t$). Taking the logarithm would then remove the exponential which would simplify our lives. But as usual you have to handle the "second order noise term" (in this case $\frac{\sigma^2}{2} t$) in order to finally get the right answer. $\endgroup$ – Ian Oct 5 '16 at 11:11
  • $\begingroup$ Now to understand why this trick still works in your case, you can either just grind it out, or you can think about comparing to the method of integrating factors for first order linear ODE, which relates the solution to $y'+p(t)y=g(t)$ to $e^{\int p(s) ds}$. $\endgroup$ – Ian Oct 5 '16 at 11:13
  • $\begingroup$ @Calculon Thank you for the reply. Actually my doubt lies mainly on the term $int_0^t \mu(s)s ds$. Since I am not familiar with stochastic integral, I don't know whether it is legitimate to do that. $\endgroup$ – IDontKnowMath Oct 5 '16 at 11:25
  • $\begingroup$ @IDontKnowMath Note that $\int_0^t\mu(s)ds$ is just an ordinary (Riemann/Lebesgue) integral if $\mu$ is a function of time only. $\endgroup$ – Calculon Oct 5 '16 at 11:29
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The reason $\ln(x)$ is chosen is this:

(Note: where I say $B_{t}$ I mean Brownian motion, which you denoted in your question as $W_{t}$)

The SDE you provided is one of the few we can explicitly solve. I'll talk about Geometric Brownian Motion (GBM) $dX_{t} = \mu X_{t} \,dt + \sigma X_{t} \,dB_{t}$, but as you mentioned in your question, your case is the same when $\mu$ becomes a function of $t$.

You can "multiply" the stochastic differential equation (SDE) in its differential form by $\frac{1}{X_{t}}$ to get $$\frac{1}{X_{t}}dX_{t} = \frac{1}{X_{t}}\mu X_{t} \,dt + \frac{1}{X_{t}}\sigma X_{t} \,dB_{t} $$ and this simplifies to $$ \frac{1}{X_{t}}dX_{t} = \mu \,dt + \sigma \,dB_{t}.$$

Notice that the right hand side no longer depends on $X_{t}$. Now recall Ito's formula for a $C^{2,1}$ function $f(t,x)$ ($C^{2,1}$ means $f$ is twice differentiable in $x$ and once differentiable in $t$). Ito's formula tells us if $X_{t}$ satisfies the previous SDE, then $f(t,X_{t})$ will satisfy:

$$d(f(t,X_{t})) = \frac{\partial f}{\partial t}(t,X_{t}) \,dt + \frac{\partial f}{\partial x}(t,X{t}) \,dX_{t} + \frac{1}{2} \frac{\partial^{2} f}{\partial x^{2}}(t,X_{t}) \,d[X]_{t}$$ where $[X]_{t}$ is the quadratic variation process of $X_{t}$. Notice that in the SDE given to us by Ito's formula, one of the terms on the right hand size is $\frac{\partial f}{\partial x}(t,X{t}) \,dX_{t}$. This almost looks like our $\frac{1}{X_{t}} \,dX_{t}$, which was the left hand side of our original SDE. If we can choose $f(t,X_{t})$ wisely such that these are equal (i.e., such that $\frac{\partial f}{\partial x}(t,X{t})= \frac{1}{X_{t}}$, then we can solve this special SDE.

Hopefully you see that we should have $f(t,x) = \ln(x)$ for the above equality to hold. Okay, so since $f$ doesn't depend on $t$, let's call it $f(x)$ to save space. Substituting this $f$ into Ito's formula above gives:

$$d(\ln(X_{t})) = 0 \,dt + \frac{1}{X_{t}} \,dX_{t} - \frac{1}{2} \frac{1}{X_{t}^2} \,d[X]_{t}$$

Hmm... on the right hand side there is conveniently a $\frac{1}{X_{t}} \,dX_{t}$ (which was the whole point! That's why we choose $f$ as we did) and we have an SDE for this already. We have that $\frac{1}{X_{t}}dX_{t} = \mu \,dt + \sigma \,dB_{t}$.

Okay, so solving for $\frac{1}{X_{t}}\,dX_{t}$ in the Ito's formula SDE gives

$$\frac{1}{X_{t}} \,dX_{t} =d(\ln(X_{t})) + \frac{1}{2} \frac{1}{X_{t}^2} \,d[X]_{t} $$

and this allows us to set the right hand side of the above equal to $\mu \,dt + \sigma \,dB_{t}$. So we have $$d(\ln(X_{t})) + \frac{1}{2} \frac{1}{X_{t}^2} \,d[X]_{t} = \mu \,dt + \sigma \,dB_{t}. $$

What is $d[X]_{t}$? If you do the computation, you get $\sigma^{2} X_{t}^{2} \,dt$, so that our equation becomes $$d(\ln(X_{t})) + \frac{1}{2} \sigma^{2} \,dt = \mu \,dt + \sigma \,dB_{t}. $$

This simplifies to $$\ln(X_{t}) = \ln(X_{0}) + \int \limits_{0}^{t}(\mu - \frac{1}{2} \sigma^{2}) \,dt + \int \limits_{0}^{t}\sigma \,dB_{t} $$

so that $X_{t} = X_{0}e^{\int \limits_{0}^{t}(\mu - \frac{1}{2} \sigma^{2}) \,dt + \int \limits_{0}^{t}\sigma \,dB_{t}}$.

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  • $\begingroup$ Well explained. I finally understand the whole thing. Thanks!!! $\endgroup$ – IDontKnowMath Oct 5 '16 at 11:41
  • $\begingroup$ @IDontKnowMath I'm glad! :) $\endgroup$ – layman Oct 5 '16 at 11:58

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