Ito's Lemma Simple Application

Question

I need help applying Ito's Lemma to show a given result.

$B_t$ is standard Brownian motion

$dS_t = 0.4 S_tdt + 0.5 S_tdB_t$

I need to find $dlog(S_t)$

I am told it is $(0.4-1/8)dt + 0.5 dB_t$

From what I see because it has the $S_t$ term in it I have to apply a Taylor expansion instead of Ito's Lemma: Using the Taylor expansion on wiki: https://upload.wikimedia.org/math/7/6/7/767a4f324c2c43f86a2ecab83fe9de36.png $dlog(S_t) = \frac{dS_tdt}{S_tdt} + \frac{dS_t}{S_t} -\frac{dS_t^2}{2S_t^2} = (0.8-1/8) dt + dB_t$

Where have I gone wrong? I can see that if the df/dt = 0 I get the correct result but log(S_t) is a function of t..

Thanks

Answer 1

Ito's formula tells you that

$$f(t,X_t)=f(0,X_0)+\int_0^t \frac{\partial f}{\partial s}(s,X_s) ds + \int_0^t \frac{\partial f}{\partial x}(s,X_s) dX_s + \int_0^t \frac{1}{2} \frac{\partial^2 f}{\partial x^2}(s,X_s) (dX_s)^2$$

where $(dX_s)^2$ is a notational shorthand for the quadratic variation of $X_s$. If $X_t$ is a 1D diffusion process satisfying $dX_t=b(X_t) dt + \sigma(X_t) dW_t$ then $(dX_t)^2=\sigma(X_t)^2 dt$. (My habit is to use $W$ instead of $B$; it is the same.) So overall, if $X_t$ is a 1D diffusion process as above, you have

$$f(t,X_t)=f(0,X_0)+\int_0^t \frac{\partial f}{\partial s}(s,X_s) ds + \int_0^t \frac{\partial f}{\partial x}(s,X_s) b(X_s) ds + \int_0^t \frac{\partial f}{\partial x}(s,X_s) \sigma(X_s) dW_s + \int_0^t \frac{1}{2} \frac{\partial^2 f}{\partial x^2}(s,X_s) \sigma(X_s)^2 ds.$$

In your case this reads:

$$\log(S_t)=\log(S_0)+0+\int_0^t \frac{1}{S_s} (0.4S_s) ds + \int_0^t \frac{1}{S_s} (0.5S_s) dW_s + \int_0^t \frac{1}{2} \frac{-1}{S_s^2} (0.5S_s)^2 ds \\ = \log(S_0) + 0.4t + 0.5W_t - \frac{t}{8}.$$

It is not so clear to me where your arithmetic errors are, but I've obtained the result you anticipated, so maybe this helps. One thing I can see is that you got tripped up on the partial derivative with respect to $t$; in your case you have $f(t,x)=\log(x)$ so the partial derivative with respect to $t$ is just zero. What you are actually computing is essentially the total derivative with respect to $t$, which is not zero.

Answer 2

If you have an Ito process $$dX_t=\mu_tdt+\sigma_tdB_t$$ then Ito's lemma states that $$df(X_t)=f'(X_t)dX_t+\frac{1}{2}f''(X_t)\sigma_t^2dt$$

Here $\mu_t=0.4S_t$ and $\sigma_t=0.5S_t$. We also have $f'(x)=\frac{1}{x}$ and $f''(x)=-\frac{1}{x^2}$.

So $d(\log S_t)=\frac{dS_t}{S_t}-\frac{1}{2}\frac{(0.5S_t)^2}{S_t^2}dt=0.4dt+0.5dB_t-0.125dt$