# Convexity of an American put option

Is the price of an American put on an underlying without dividend convex with respect to the strike?

It is indeed. The price of an American option is the Bermuda option in the limit that the exercising interval approaches zero. 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 of the Bermuda option price at the next exercise time. The latter is inductively assumed to be whilst the former is convex in the strike. 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 strike. As a matter of fact the same deduction applies to an option where the principle of dynamic programming is applicable any the payoff function is convex with respect to an arbitrary random variable .

My other answer is more straightforward without resorting to the convergence of the Bermuda options. However, the Bermuda option approach can also be used to prove the convexity of the American option with respect to a variable dependent on the sample, such as $$S_t$$, while the other direct method is inapplicable here.

We will show the convergence of the Bermuda option price to its associated American option price.

Let $$A$$ be the price at time $$0$$ of an American option with a continuous payoff function $$g(S)$$ on the underlying $$S$$ expiring at time $$1$$, i.e. $$A=\sup_{\tau\in\mathbb F[0,1]}\mathbf E g(S_\tau),$$ where $$\mathbb FS$$ stands for the set of all stopping times taking value in set S. Let $$(T_n)_{n=1}^\infty$$ be a sequence of sets where $$T_n:=\{0,t_1,t_2,\cdots,t_{n-1},t_n=1\}$$ with $$0 and $$\max_{0\le i\le n-1}(t_{i+1}-t_i)\to 0$$ as $$n\to\infty$$. The associated Bermuda option price at time $$0$$ is $$B_n=\sup_{\tau\in\mathbb FT_n}\mathbf Eg(S_\tau).$$

Lemma: $$\lim_{n\to\infty} B_n=A.$$

Proof: Fix an arbitrary stopping time $$\tau\in\mathbb F[0,1]$$ and $$\epsilon>0$$.

Define simple function $$\tau_{T_n}:=\sum_{i=0}^{n-1} t_i\mathbf 1_{[t_i,t_{i+1})}.$$ $$\tau_{T_n}\to \tau$$ almost surely as $$n\to\infty$$. Since $$g$$ is continuous, by the dominated convergence theorem, $$\mathbf E|g(S_{\tau_{T_n}})-g(S_{\tau})|\to0$$ as $$n\to\infty$$. $$\exists N(\tau,\epsilon)\ni$$ $$\mathbf Eg(S_{\tau_{T_n}})>\mathbf Eg(S_\tau)-\epsilon \tag1$$ $$\forall n>N(\tau,\epsilon)$$. For such $$n$$ $$B_n=\sup_{\tau\in\mathbb FT_n}\mathbf Eg(S_\tau)\ge \mathbf Eg(S_{\tau_{T_n}}) \tag2.$$ There are infinitely many $$k>N(\tau,\epsilon)$$, that $$\liminf_{n\to\infty} B_n\ge B_k \tag3$$ Combining $$(1), (2)$$ and $$(3)$$, we have $$\liminf_{n\to\infty} B_n>\mathbf Eg(S_{\tau})-\epsilon.$$

As $$\tau$$ and $$\epsilon$$ are arbitrary $$\liminf_{n\to\infty} B_n\ge \sup_{\tau\in\mathbb F[0,1]}\mathbf E g(S_{\tau}). \tag4$$

On the other hand, it is obvious that $$\sup_{\tau\in\mathbb F[0,1]}\mathbf E g(S_{\tau})\ge\sup_{\tau\in\mathbb FT_n}\mathbf Eg(S_\tau)=:B_n.$$ That leads to $$A=\sup_{\tau\in\mathbb F[0,1]}\mathbf E g(S_{\tau})\ge\liminf_{n\to\infty} B_n. \tag5$$

Finally, combining $$(4)$$ and $$(5)$$, we obtain the desired result. $$\quad\quad\square$$

• Is there any reference to support your second sentence 'The price of an American option is the Bermuda option in the limit that the exercising interval approaches zero'? – Idonknow Jun 6 at 1:35
• @Idonknow: I have added the proof to that proposition. – Hans Jun 8 at 3:17
• @Idonknow: I have put up a much more straightforward proof in a new answer. Check it out. – Hans Jun 9 at 16:11
• yeah I saw it. Nice and elegant! – Idonknow Jun 9 at 16:21

Here is a much more straightforward proof of the convexity of the American option with respect to a parameter, if it is independent of time and sample, than my previous one, though I am happy to have made the connection amongst the dynamics programming principle, the discrete time process and the continuous time process there.

Let $$g(t,\omega,x)$$ be the discounted payoff function dependent on sample $$\omega$$ at time $$t$$ that is convex in parameter (e.g. strike) $$x$$ that is independent of $$(t,\omega)$$, and $$\mathcal T$$ the set of all stopping time. The price of an American option $$A(x) := \sup_{\tau\in\mathcal T}\mathbf Eg(\tau,\omega,x)$$. For the sake of notational brevity, we shall drop $$\omega$$ from the following derivation.

$$\forall\tau\in\mathcal T,\,\lambda\in[0,1]$$, by convexity $$g(\tau,\lambda x+(1-\lambda)y)\le \lambda g(\tau,x)+(1-\lambda)g(\tau,y).$$ Then \begin{align} \mathbf Eg(\tau,\lambda x+(1-\lambda)y) &\le \lambda\mathbf E g(\tau,x)+(1-\lambda)\mathbf Eg(\tau,y) \\ &\le \lambda\sup_{\tau\in\mathcal T}\mathbf E g(\tau,x)+(1-\lambda)\sup_{\tau\in\mathcal T}\mathbf Eg(\tau,y) \\ &=\lambda A(x)+(1-\lambda) A(y). \end{align} Thus $$A(\lambda x+(1-\lambda)y)=\sup_{\tau\in\mathcal T}\mathbf Eg(\tau,S_\tau,\lambda x+(1-\lambda)y)\le \lambda A(x)+(1-\lambda)A(y),$$ or that $$A(x)$$ is convex in $$x$$.

• This is indeed more straightforward. – Gordon Jun 8 at 23:04
• Did you miss an expectation in defining the American Option price? – Arshdeep Singh Duggal 2 days ago
• @ArshdeepSinghDuggal: Yes, indeed. Thanks for catching the typo. Corrected. Please check it out. – Hans 2 days ago

Let $$\mathscr{T}$$ be the set of stopping times with values in $$[0, T]$$. Note that, for any $$\tau \in \mathscr{T}$$, $$\lambda_1\ge 0$$, $$\lambda_2 \ge 0$$, and $$\lambda_1+\lambda_2 =1$$, \begin{align*} &\ \max(\lambda_1 K_1+\lambda_2 K_2 -S_{\tau}, 0) \\ =&\ \max\big(\lambda_1 (K_1-S_{\tau})+\lambda_2 (K_2 -S_{\tau}), 0\big)\\ =&\ \lambda_1\max(K_1-S_{\tau}, 0)\\ +&\max\big(\lambda_1 (K_1-S_{\tau})-\lambda_1\max(K_1-S_{\tau}, 0)+\lambda_2 (K_2 -S_{\tau}), -\lambda_1\max(K_1-S_{\tau}, 0)\big)\\ =&\ \lambda_1\max(K_1-S_{\tau}, 0)\\ +&\max\big(\lambda_1 (K_1-S_{\tau})-\lambda_1\max(K_1-S_{\tau}, 0)+\lambda_2 (K_2 -S_{\tau}), -\lambda_1\max(K_1-S_{\tau}, 0)\big)\pmb{1}_{K_1 < S_{\tau}}\\ +&\max\big(\lambda_1 (K_1-S_{\tau})-\lambda_1\max(K_1-S_{\tau}, 0)+\lambda_2 (K_2 -S_{\tau}), -\lambda_1\max(K_1-S_{\tau}, 0)\big)\pmb{1}_{K_1 \ge S_{\tau}}\\ =&\ \lambda_1\max(K_1-S_{\tau}, 0)\\ +&\max\big(\lambda_1 (K_1-S_{\tau})+\lambda_2 (K_2 -S_{\tau}), 0\big)\pmb{1}_{K_1 < S_{\tau}}\\ +&\max\big(\lambda_2 (K_2 -S_{\tau}), -\lambda_1\max(K_1-S_{\tau}, 0)\big)\pmb{1}_{K_1 \ge S_{\tau}}\\ \le&\ \lambda_1\max(K_1-S_{\tau}, 0) + \max\big(\lambda_2 (K_2 -S_{\tau}), 0\big)\pmb{1}_{K_1 < S_{\tau}} +\max\big(\lambda_2 (K_2 -S_{\tau}), 0\big)\pmb{1}_{K_1 \ge S_{\tau}}\\ =&\ \lambda_1\max(K_1-S_{\tau}, 0) + \lambda_2\max(K_2-S_{\tau}, 0). \end{align*} Let $$B_t$$ be the money-market account value at time $$t$$. Then, from Appendix 1 of Methods of Mathematical Finance, there exists a stopping time $$\tau*\in \mathscr{T}$$ such that \begin{align*} E\left(\frac{\max(\lambda_1 K_1+\lambda_2 K_2 -S_{\tau*}, 0)}{B_{\tau*}} \right) = \sup_{\tau\in \mathscr{T}} E\left(\frac{\max(\lambda_1 K_1+\lambda_2 K_2 -S_{\tau}, 0)}{B_{\tau}} \right). \end{align*} Therefore, \begin{align*} &\ \sup_{\tau\in \mathscr{T}} E\left(\frac{\max(\lambda_1 K_1+\lambda_2 K_2 -S_{\tau}, 0)}{B_{\tau}} \right) \\ =&\ E\left(\frac{\max(\lambda_1 K_1+\lambda_2 K_2 -S_{\tau*}, 0)}{B_{\tau*}} \right)\\ \le&\ \lambda_1 E\left(\frac{\max(K_1 -S_{\tau*}, 0)}{B_{\tau*}} \right) + \lambda_2 E\left(\frac{\max(K_2 -S_{\tau*}, 0)}{B_{\tau*}} \right)\\ \le&\ \lambda_1 \sup_{\tau\in \mathscr{T}} E\left(\frac{\max(K_1 -S_{\tau}, 0)}{B_{\tau}} \right) + \lambda_2 \sup_{\tau\in \mathscr{T}} E\left(\frac{\max(K_2 -S_{\tau}, 0)}{B_{\tau}} \right). \end{align*} That is, the American put is convex with respect to the strike.

• Great stuff. Do you think this will extend to a rates Bermudan, where the underlying is different at every exercise date? – Arshdeep Singh Duggal Jun 19 at 10:13
• Likely. what are the specific underlings, CMS? – Gordon Jun 19 at 12:08
• Yes, usual co-terminal swap rates. – Arshdeep Singh Duggal Jun 19 at 13:37
• You may ask this as a separate question, as some other details may also needed. – Gordon Jun 19 at 14:22
• @Gordon For the long algebra sequence, after the first equality, isn't it sufficient to just state that $\max(\cdot,0)$ function is convex? – ir7 2 days ago