Why does the following differential formula hold?

Question

I am currently studying Stochastic Calculus for Continuous Time Finance models. I have stumbled upon the equation $$ d ln(S_t) = \frac{dS_t}{S_t} - \frac{1}{2S_t^2}(d(S(t))^2 $$ Why does that hold? I really do not get it. I thought that $$ dln(S_t) = \frac{1}{S_t} $$

I am not sure if this is relevant, but I have $$ dS_t = r_t S_t dt + \sigma_t S_t d\tilde{W}_t $$

Ito's lemma. Your last formula is not relevant. – Kurt G. Dec 28 '22 at 17:21 — Kurt G., Dec 28 '22 at 17:21
Thank you for clarifying Kurt G. – Zebraboard Dec 28 '22 at 17:39 — Zebraboard, Dec 28 '22 at 17:39

score 1 · Accepted Answer · answered Dec 28 '22 at 17:30

Thanks to Kurt G. for clarifying. It is now clear to me. For some reason, I did not even think of that. But, we use Ito's Lemma with $f = ln(S_t)$, so that $$ f_t = 0 \\ f_{x} = \frac{1}{S_t} dS_t \\ f_{xx} = -\frac{1}{S_t^2} (d(S(t))^2 $$ Adding these terms we get $$ dln(S_t) = \frac{dS_t}{S_t} - \frac{1}{S_t^2} (d(S_t))^2 $$ In this particular situation, with $=+\tilde{W}_t$, we then get $$ d lnS_t= r_t dt + \sigma_t d\tilde{W}_t - \frac{1}{2S_t^2}(d(S(t))^2 $$ $$ d lnS_t = r_tdt + \sigma_t d\tilde{W}_t - \frac{1}{2}\sigma^2dt $$ where $(d(S_t))^2$ is computed using $dtdt = dtW_t = 0$ and $dW_tdW_t = dt$

Why does the following differential formula hold?

1 Answers1