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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 random variable of concern. 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 and the payoff function is convex with respect to the value of an arbitrary random variable .

My other answer is more straightforward without resorting to the convergence of the Bermuda options, but only applicable to a deterministic parameter; whilst the following Bermuda option approach can also be used to prove the convexity of the American option with respect to a random variable, such as $S_t$.

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 $0<t_1<t_2<\cdots<t_{n-1}<1$ 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. $\quad\quad\square$

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 random variable of concern. 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 and the payoff function is convex with respect to the value of an arbitrary random variable .

My other answer is more straightforward without resorting to the convergence of the Bermuda options, but only applicable to a deterministic parameter; whilst the following Bermuda option approach can also be used to prove the convexity of the American option with respect to a random variable, such as $S_t$.

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 $0<t_1<t_2<\cdots<t_{n-1}<1$ 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 $S_t$ is almost surely continuous with respect to $t$ and $g$ 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. $\quad\quad\square$

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 and the payoff function is convex with respect to the value of an arbitrary random variable .

My other answer is more straightforward without resorting to the convergence of the Bermuda options, but only applicable to a deterministic parameter; whilst the following Bermuda option approach can also be used to prove the convexity of the American option with respect to a random variable, such as $S_t$.

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 $0<t_1<t_2<\cdots<t_{n-1}<1$ 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. $\quad\quad\square$

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 random variable of concern. 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 and the payoff function is convex with respect to the value of an arbitrary random variable .

My other answer is more straightforward without resorting to the convergence of the Bermuda options, but only applicable to a deterministic parameter; whilst the following Bermuda option approach can also be used to prove the convexity of the American option with respect to a random variable, such as $S_t$.

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 $0<t_1<t_2<\cdots<t_{n-1}<1$ 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. $\quad\quad\square$

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 and the payoff function is convex with respect to the value of 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 random variable, such as $S_t$, while the 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 $0<t_1<t_2<\cdots<t_{n-1}<1$ 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. $\quad\quad\square$

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 and the payoff function is convex with respect to the value of an arbitrary random variable .

My other answer is more straightforward without resorting to the convergence of the Bermuda options, but only applicable to a deterministic parameter; whilst the following Bermuda option approach can also be used to prove the convexity of the American option with respect to a random variable, such as $S_t$.

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 $0<t_1<t_2<\cdots<t_{n-1}<1$ 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. $\quad\quad\square$