I'm a math master student. I'm trying to follow a course-note that unfortunately has chasms to fill for this specific derivation. Rigour unfortunately has been thrown to the gutter.

Let $G$ denote a payoff function of some option, and suppose $G$ depends only on the final value of the underlying stock, so $G(S_T)$, where $T$ denotes the maturity. Let $V_t := F(t, S_t)$ denote the price process of the option. Lastly, let $f_x$ denote the partial of $f$ wrt $x$. Then, below is the Black-scholes PDE (as defined in my notes): \begin{equation}\label{eq1} \begin{aligned} F_t(t, s) + (r-q)sF_s(t,s) + \frac{1}{2}\sigma^2 s^2 F_{ss}(t,s) - rF(t, s) &= 0 \\ F(T, s) &= G(s). \end{aligned} \end{equation} In particular $G$ also doubles as the boundary condition in the Feynman-Kac formula. To make everything more concrete, let me state Ito's lemma for SDEs and the Feynman-Kac formula. Firstly though let the SDE be $dX_t = \mu(t, X_t)dt + \sigma(t, X_t)dW_t$, with initial condition $X_s=x$ for some $s \leq t$.

Ito's lemma (sde): $F(t, S_t) - F(s, S_s) = \int_s^t\left[\sigma(u, S_u)F_s(u, S_u)\right]dW_u + \int_s^t\left[ F_t(u, S_u) + \mu(u, S_u)F_s(u, S_u) + \frac{1}{2}\sigma(u, S_u)F_{ss}(u, S_u) \right]du$

Feynman-Kac formula: Let $B(s) := F(T,s)$ be the boundary conditions. Now suppose $$F_t(t, s) + \mu(t, s)F_s(t, s) + \frac{1}{2}\sigma^2(t, s)F_{ss}(t, s) = 0.$$ In particular this implies that the second integral in the Ito's lemma is 0, and thus we obtain that ($s \leq t$): $$ F(s, S_s) = \mathbb{E}\left[ B(S_T) |\mathcal{F}_s \right] $$

How is PDE "proved" in my lecture notes? First note that under the risk-neutral principle, $V_t = F(t, S_t) = \exp(-r(T-t))\mathbb{E}\left[G(S_T)|\mathcal{F}_t\right]$.

Then we suppose a function $H(t,s)$ that is a solution to: $$ \begin{aligned} H_t(t, s) + (r-q)sH_s(t, s) + \frac{1}{2}\sigma^2s^2H_{ss}(t, s) &= 0 \\ H(T, s) &= \exp(-rT)G(s), \end{aligned} $$ where the RHS of the second line is the boundary condition $B(s)$. Note, to compare with the Feynman-Kac formula, in this case $\mu(t, s) = (r-q)s$ and $\tilde{\sigma}(t, s) = \sigma s$.

Then by Feynman-Kac $$H(t, S_t) = \mathbb{E}\left[ B(s) | \mathcal{F}_t \right] = \exp(-rT)\mathbb{E}\left[ G(s) | \mathcal{F}_t \right].$$ Multiply both sides by $\exp(rt)$ to obtain $$ \exp(rt)H(t, S_t) = \exp(-r(T-t))\mathbb{E}\left[G(S_T)|\mathcal{F}_t\right] = F(t, S_t). $$

Finally the notes simply say take the partials of $F$ to obtain the initial PDE, but I can't make sense of this step.

I see components all over the place, but I can't assemble them together. We have the functions $\mu$ and $\tilde{\sigma}$, we have an expression for $F$ dependant on some $H$ satisfying the Feynman-Kac conditions, why do we need to create the PDE for $F$ then?

I feel like my question is slightly incoherent (and notation may be slightly messy), but that reflects the confused state I'm in now.

  • $\begingroup$ I'm not even looking for a full blown solution here. The general idea on what to do to obtain the original PDE is a massive help. $\endgroup$ Commented Jan 25 at 23:23

2 Answers 2


First, I suggest looking at the first derivation in this this link from Fabrice Rouah. While there are a few derivations of the BS PDE (and several in that file), this is one a lot of people will be familiar with, and it helps with the economic intuition.

To try to answer fully in the context of what you've provided, I think the BS PDE in your first setup is wrong; in particular, it looks like it's missing a term of the form $rF(t, s)dt$

With these, $F$ is not in the form of the Feynman-Kac formula, but $H$ is. Applying Ito to $F = e^{-rt}H$ will result in the BS PDE given above, as well as the solution being expressed as an expectation times an exponential.

Hope this helps close the gap.

  • 1
    $\begingroup$ Ah yes I missed the $-rF(t,s)$. I'll try to digest what you've written. $\endgroup$ Commented Jan 26 at 12:42

Sometimes I wonder how did I get my math degree.

"Just take the partials and substitute to the original to get the original PDE". Quite literally.

Recall we have $F(t, s) = \exp(rt)H(t, s)$, which implies: $$ \begin{aligned} F_t(t,s) &= r\exp(rt)H(t,s) + exp(rt)H_t(t,s) \\ F_s(t,s) &= \exp(rt)H_s(t,s) \\ F_{ss}(t,s) &= \exp(rt)H_{ss}(t,s) \end{aligned} $$

Now consider lhs of the original PDE: $$ \begin{aligned} F_t(t,s) + \mu F_s(t,s) + \frac{\tilde{\sigma}^2}{2} F_s(t,s) - rF(t,s). \end{aligned} $$ Simply substituting the terms will result in the equation above being 0. Furthermore, $F(T,s) = \exp(rT)H(T,s) = G(s)$ indeed. Thus the Black-scholes PDE is derived.


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