How Itô’s Formula Connects to the Black–Scholes PDE

# Is the Black Scholes PDE actually immediate from Ito's lemma?


# Is the Black Scholes PDE actually immediate from Ito's lemma?












Ito's lemma replaces $dS^2$ by $vol^2*dt$, however it is repeatedly mentioned that the lemma manifests in the integral form and the differential form below is merely a short hand for the integral form:

$dC-C_{s}dS=C_{t}dt+0.5C_{ss}vol^*dt$ ---- (1)

(1) is true only if we integrate both sides. Simple counter example is Ito's lemma written as: $dW^2-2WdW=dt$ where LHS is stochastic and RHS is deterministic. If we integrate both sides here as well, it becomes accurate.

Now from (1) we create the B.S. P.D.E -

$C_{t}+0.5C_{ss}vol=r(C-C_{s}S)$

as if (1) was differentially true (i.e true at each step $dt$ - what we call "locally risk free hence grows at risk free rate"), rather than only true in it's integral form.

Now obviously if it is differentially true (i.e. true at every infinitesmal step dt) then it would obviously be true in integral form, which is all we care about. But it seems to be an unaddressed point. Any thoughts what I am missing? Is there a clear link, without assumptions that leads from (1) to the PDE?

## Answer by Kurt G. (score 1)

https://quant.stackexchange.com/a/79348

- The derivations of the Black & Scholes PDE being accurate or not has absolutely nothing to do with whether we write the Ito formula for the call price, in differential form, $$\tag1 dC=C_t\,dt+C_s\,dS_t+\frac12 C_{ss}S_t^2\sigma^2\,dt\,, $$ or equivalently, in integral form, $$\tag2 C(T,S_T)-C(0,S_0)=\int_0^T C_t(t,S_t)\,dt+\int_0^TC_s(t,S_t)\,dS_t+\frac12\int_0^TC_{ss}(t,S_t)\,S_t^2\,\sigma^2\,dt\,. $$ As we know: (1) is just a short notation for (2). The factor $S_t^2$ is missing in your formulas.



- The level of rigor in BS' original paper regarding the hedging portfolio being "instantaneously riskless" has been amply discussed during the decades. See this post to get started with this. In short: today we have a better way of deriving BS' PDE which remains correct as it was all the time.

- My favourite way of deriving the PDE you can find here.