# Pricing function $P(S,t)$ is convex in $S$ for all $t$

I am now reading Alternative Characterization of American Put Options by Carr et all (available at http://www.math.nyu.edu/research/carrp/papers/pdf/amerput7.pdf). There is a theorem called 'Main Decomposition of the American Put'.

Theorem 1 (Main Decomposition of the American Put) On the continuation region $\mathcal{C}$, the American put value, $P_0$, can be decomposed into the corresponding European put price, $p_0$, and the early exercise premium, $e_0$: $$P_0=p_0+e_0$$ where $$e_0=rK \int_{0}^{T} \exp{(-rt)} N\bigg( \frac{\ln{(B_t / S_0)}-e_2 t}{\sigma \sqrt{t}} \bigg)dt,$$ $$e_2=r-\frac{\sigma^2}{2}, \,$$ and $$N(x)=\int_{0}^{x} \frac{\exp{(-z^2/2)}}{\sqrt{2\pi}}dz$$ is the standard normal distribution function.

The proof in the appendix starts with: We wish to prove that: $$P_0=p_0+rK \int_{0}^{T} \exp{(-rt)} N\bigg( \frac{\ln{(B_t / S_0)}-e_2 t}{\sigma \sqrt{t}} \bigg)dt.$$ Let $Z_t \equiv \exp{(−rt)}P_t$ be the discounted put price, defined in the region $D \equiv \{(S, t) : S ∈ [0, \infty), t ∈ [0, T]\}$. In this region, the pricing function $P(S, t)$ is convex in $S$ for all $t$, continuously differentiable in $t$ for all $S$, and a.e. twice continuously differentiable in $S$ for all $t$.

My question is regarding the statement: "the pricing function $P(S, t)$ is convex in $S$ for all $t$". Is it assumed or can we prove it?

I read the definition of convex function from http://mathworld.wolfram.com/ConvexFunction.html:

A convex function is a continuous function whose value at the midpoint of every interval in its domain does not exceed the arithmetic mean of its values at the ends of the interval. More generally, a function $f(x)$ is convex on an interval $[a,b]$ if for any two points $x_1$ and $x_2$ in $[a,b]$ and any $\lambda$ where $0< \lambda <1$, $$f[\lambda x_1 + (1- \lambda x_2)] \leq \lambda f(x_1)+ (1- \lambda) f(x_2)$$

I also have read a question in https://math.stackexchange.com/questions/112063/price-of-a-european-call-option-is-a-convex-function-of-strike-price-k but I am not sure if it can be applied to my question because

(1). I assume the $P(S,t)$ in my question to be the American put value instead of European one,

(2). the question in the link is about convex function of strike price while my question is about convex function in $S$ in all $t$ (or are they the same?), and

(3). the convex function definition I got seems different.

Can anyone help me to explain why $P(S, t)$ is convex in $S$ for all $t$? Thank you.

• Since P is continuous, perhaps it suffices to show that the second derivative of P(S,t) with respect to S is positive. Mar 15, 2016 at 23:32
• @AlexC, Thank you for the suggestion, but I don't think the explicit form of $P(S,t)$ is known yet. Mar 16, 2016 at 2:13

The Carr et al. (1992) paper you are referring to assumes that the underlying asset follows a geometric Brownian motion (GBM).

Within a wider setting including the GBM case, it was shown by El Karoui et al. (1998) as well as Hobson (1998) that the valuation function of an American plain vanilla option is a convex function of the asset price.

References

El Karoui, Nicole, Monique Jeanblanc-Picque and Steven E. Shreve (1998) “Robustness of the Black and Scholes Formula”, Mathematical Finance, Vol. 8, No. 2, pp. 93-126

Hobson, David G. (1998) “Volatility MIsspecification, Option Pricing and Superreplication via Coupling”, Annals of Applied Probability, Vol. 8, No. 1, pp. 193-205

The price of an American option is the Bermuda option in the pointwise limit in $$S$$ as the maximal exercising interval approaches zero. See the proof in this answer.

The Bermuda option at any exercising time can be evaluated inductively via the dynamic programming principle as the maximum of the payoff and the risk-neutral expected value, i.e, the European option price, at the current exercise time of the Bermuda option price at the next exercise time which is inductively assumed to be convex in $$S$$. The European option price at time $$t$$ of a payoff function convex in $$S_T$$ at the exercise time $$T$$ is convex in $$S_t$$ at time $$t$$. The maximum of convex functions is again convex. The dominant convergence theorem guarantees the pointwise limit of a sequence of convex functions is again convex. Therefore the American option is convex in $$S_{t_0}$$ at the present valuation time $$t_0$$.