Feynman-Kac and Option Price Convexity

Worked Solution

How to Think About It: This problem ties together the three pillars of derivatives pricing: the PDE approach (Black-Scholes PDE), the probabilistic approach (risk-neutral expectation), and qualitative behavior of option prices (convexity implies positive vega). Before writing formulas, think about it economically. A call payoff is a convex function of the stock price. A convex function of a random variable benefits from more randomness -- that is Jensen's inequality in one sentence. More volatility means more spread in the terminal stock distribution, and convexity turns extra spread into extra value. That is why vega is positive.

Quick Sanity Checks: Convexity of the call payoff $(S - K)^+$ is obvious -- it is the max of a linear function and zero. The price of a call should increase in $\sigma$ because higher vol fattens both tails, but the convex payoff captures upside and is floored at zero on the downside. At $\sigma = 0$ the call price is just the discounted intrinsic value; as $\sigma \to \infty$ the call price approaches $S$ (the stock itself). So the price is increasing and concave in $\sigma$.

Derivation:

Part (i): Feynman-Kac Representation

The Feynman-Kac theorem states:

$V(S, t) = e^{-r(T-t)} \, \mathbb{E}^{\mathbb{Q}}\!\left[g(S_T) \,\big|\, S_t = S\right]$

where $S_T = S \exp\!\left(\left(r - \tfrac{\sigma^2}{2}\right)(T - t) + \sigma\, W_{T-t}^{\mathbb{Q}}\right)$.

This expectation equals the solution to the Black-Scholes PDE:

$\frac{\partial V}{\partial t} + r S \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} - r V = 0$

with terminal condition $V(S, T) = g(S)$, provided:

• $V \in C^{2,1}$ (twice continuously differentiable in $S$, once in $t$) on $(0, \infty) \times [0, T)$.
• $g$ satisfies a polynomial growth condition: $|g(S)| \leq C(1 + S^p)$ for some constants $C, p > 0$.
• $V$ itself satisfies a polynomial growth bound: $|V(S, t)| \leq C(1 + S^p)$ uniformly on $[0, T]$. This ensures the discounted price process is a true martingale (not just a local martingale).
• The underlying diffusion coefficients ($rS$ drift and $\sigma S$ volatility) are Lipschitz and satisfy linear growth conditions, which GBM satisfies.

These conditions ensure the stochastic integral in the Ito expansion of $e^{-rt}V(S_t, t)$ is a true martingale, so the conditional expectation formula holds.

Part (ii): Convexity Preservation

We want to show: if $g$ is convex, then for any $t < T$, the map $S \mapsto V(S, t)$ is convex.

Fix $t < T$, $S_1, S_2 > 0$, and $\lambda \in [0, 1]$. Let $S_3 = \lambda S_1 + (1 - \lambda) S_2$.

Under GBM, the terminal stock price starting from $S$ is:

$S_T = S \cdot \exp\!\left(\left(r - \tfrac{\sigma^2}{2}\right)(T - t) + \sigma W_{T-t}\right) = S \cdot M$

where $M = \exp\!\left(\left(r - \tfrac{\sigma^2}{2}\right)(T - t) + \sigma W_{T-t}\right)$ is a positive random variable that does not depend on $S$. This is the key property: GBM is linear in its initial condition.

Starting from $S_3$:

$S_T^{(3)} = S_3 \cdot M = \lambda S_1 M + (1 - \lambda) S_2 M = \lambda S_T^{(1)} + (1 - \lambda) S_T^{(2)}$

where we couple the three processes by using the same Brownian path $M$ (same $\omega$).

Since $g$ is convex:

$g(S_T^{(3)}) = g\!\left(\lambda S_T^{(1)} + (1 - \lambda) S_T^{(2)}\right) \leq \lambda \, g(S_T^{(1)}) + (1 - \lambda)\, g(S_T^{(2)})$

Taking $\mathbb{E}^{\mathbb{Q}}$ of both sides (expectation preserves inequalities):

$\mathbb{E}^{\mathbb{Q}}[g(S_T^{(3)})] \leq \lambda \, \mathbb{E}^{\mathbb{Q}}[g(S_T^{(1)})] + (1 - \lambda) \, \mathbb{E}^{\mathbb{Q}}[g(S_T^{(2)})]$

Multiplying by $e^{-r(T-t)}$:

$V(S_3, t) \leq \lambda \, V(S_1, t) + (1 - \lambda) \, V(S_2, t)$

This is exactly convexity of $V(\cdot, t)$. $\square$

Note: the proof hinges on GBM being multiplicative -- $S_T = S \cdot M$ is linear in $S$ for a fixed path. This would fail if the SDE had additive noise.

Part (iii): Call Price Increasing in Volatility

Let $C(\sigma)$ denote the call price as a function of volatility. We want to show $\frac{\partial C}{\partial \sigma} \geq 0$ (positive vega).

The call payoff $g(S) = (S - K)^+$ is convex, so by Part (ii), $V(S, t)$ is convex in $S$.

Convexity of $V$ in $S$ means $\frac{\partial^2 V}{\partial S^2} \geq 0$ (i.e., gamma $\Gamma \geq 0$).

Now differentiate the Black-Scholes PDE with respect to $\sigma$. Let $V_{\sigma} = \frac{\partial V}{\partial \sigma}$ (this is vega). Differentiating:

$\frac{\partial V_{\sigma}}{\partial t} + rS \frac{\partial V_{\sigma}}{\partial S} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V_{\sigma}}{\partial S^2} - r V_{\sigma} = -\sigma S^2 \frac{\partial^2 V}{\partial S^2}$

The right-hand side is $-\sigma S^2 \Gamma \leq 0$ (since $\Gamma \geq 0$).

The terminal condition is $V_{\sigma}(S, T) = 0$ (the payoff does not depend on $\sigma$).

This is a backward PDE for $V_{\sigma}$ with a non-positive source term and zero terminal condition. By the Feynman-Kac representation applied to this PDE:

$V_{\sigma}(S, t) = \mathbb{E}^{\mathbb{Q}}\!\left[\int_t^T e^{-r(u-t)} \sigma S_u^2 \, \Gamma(S_u, u) \, du \,\bigg|\, S_t = S\right]$

Since $\sigma > 0$, $S_u > 0$, and $\Gamma \geq 0$, the integrand is non-negative, so $V_{\sigma} \geq 0$.

Therefore the call price is increasing in volatility. $\square$

Practical Interpretation: This result is fundamental to options trading. Positive vega for convex payoffs means any portfolio with net convexity benefits from volatility increases. This is why traders say "long gamma = long vol." It also explains why implied volatility is a well-defined concept for calls and puts: the Black-Scholes price is strictly monotone in $\sigma$ (for non-degenerate cases), so you can always invert the market price to get a unique implied vol.

Answer: (i) $V(S,t) = e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[g(S_T) | S_t = S]$, equivalent to the BS PDE under polynomial growth, $C^{2,1}$ smoothness, and true martingale conditions. (ii) Convexity of $g$ plus linearity of GBM in its initial condition ($S_T = S \cdot M$) gives convexity of $V(\cdot, t)$ via Jensen's inequality applied pathwise. (iii) Convexity implies $\Gamma \geq 0$; differentiating the PDE in $\sigma$ gives vega as an integral of $\Gamma$ against positive weights, so vega $\geq 0$.