# ECSE 506: Stochastic Control and Decision Theory

Example: Call options

An investor has a call option to buy one share of a stock at a fixed price $$p$$ dollars and has $$T$$ days to exercise this option. For simplicity, we assume that the investor makes a decision at the beginning of each day.

The investory may decide not to exercise the option but if he does exercise the option when the stock price is $$x$$, he effectively gets $$(x-p)$$.

Assume that the price of the stoch varies with independent increments, i.e., the price on day $$t+1$$ is $X_{t+1} = X_t + W_t$ where $$\{W_t\}_{t \ge 1}$$ is an i.i.d. process with mean $$\mu$$.

Let $$\tau$$ denote the stopping time when the investor exercises his option. Then the optimization problem is to maximize $\EXP\big[ (X_{\tau} - p )\cdot \IND\{\tau \le T \} \big].$

Since this is an optimal stopping problem with perfect state observation, the optimal strategy is given by the solution of the following dynamic program

Dynamic program \begin{align*} V_{T}(x) &= \max\{ x-p, 0 \} \\ V_{t}(x) &= \max\{ x-p, \EXP[ V_{t+1}(x + W) \}. \end{align*}

# 1 Qualitative properties of the value function

Lemma
1. For all $$t$$, $$V_t(x)$$ is increasing in $$x$$.
2. For all $$t$$, $$V_t(x) - x$$ is decreasing in $$x$$.
3. For all $$x$$, $$V_t(x) \ge V_{t+1}(x)$$.

#### Proof

The first property follows immediately from monotonicity of terminal reward and the monotonicity of the dynamics. From Assignment 2, to show the third property, we need to show that $$V_{T-1}(x) \ge V_T(x)$$. Observe that $V_{T-1}(x) = \max\{x - p, \EXP[V_{T}(x + W) \} \ge \max\{ x - p, 0 \} = V_T(x).$

Now we prove the second property using backward induction. At $$t=T$$, $V_T(x) - x = \max\{ -p, -x \}$ which is decreasing in $$x$$. This forms the basis of induction. Now assume that $$V_{t+1}(x) - x$$ is decreasing in $$x$$. Then, \begin{align*} V_t(x) - x &= \max\{ -p, \EXP[ V_{t+1}(x+W) ] - x \} \\ &= \max\{ -p, \EXP[ V_{t+1}(x+W) - (x + W) ] + \EXP[W] \}. \end{align*} By the induction hypothesis the second term is decreasing in $$x$$. The minimum of a constant and a decreasing function is decreasing in $$x$$. Thus, $$V_t(x) - x$$ is decreasing in $$x$$. This completes the induction step.

Lemma

At any time $$t$$, if it is optimal to sell when the stock price is $$x^\circ$$, then it is optimal to sell at all $$x \ge x^\circ$$.

#### Proof

Since it is optimal to sell at $$x^\circ$$, we must have $$$\label{eq:p1} x^\circ - p \ge \EXP[V_{t+1}(x^\circ + W) ]$$$ Since $$V_{t}(x) - x$$ is decreasing in $$x$$, we have that for any $$x \ge x^\circ$$, $$$\label{eq:p2} \EXP[ V_{t+1}(x + W) - x ] \le \EXP[ V_{t+1}(x^\circ + W) - x^\circ ] \le -p$$$ where the last inequality follows from \eqref{eq:p1}. Eq \eqref{eq:p2} implies that $\EXP[ V_{t+1}(x+W) ] \le x - p.$ Thus, the stopping action is also optimal at $$x$$.

Theorem

The optimal strategy is of the threshold type. In particular, there exist numbers $$s_1 \ge s_2 \ge \cdots \ge s_T$$ such that it is optimal to exercise the option at time $$t$$ if and only if $$x_t \ge s_t$$.

#### Proof

Let $$S_t = \{x : g_t(x) = 1\}$$. The previous Lemma shows that $$S_t$$ is of the form $$[s_t, \infty)$$, where $$s_t = \min \{ x : g_t(x) = 1\}$$, where we assume that $$s_t = \infty$$ is it is not optimal to stop in any state. Thus proves the threshold property.

To show that the thresholds are decreasing with time, it suffices to show that $$S_t \subseteq S_{t+1}$$. Suppose $$x \in S_t$$. Then, $$$\label{eq:p3} x - p \ge \EXP[ V_{t+1}(x + W) ] \ge \EXP[ V_{t+2}(x + W) ],$$$ where the first inequality follows because $$x \in S_t$$ and the second inequality follows because $$V_{t+1}(x) \ge V_{t+2}(x)$$. Eq \eqref{eq:p3} implies that $$x \in S_{t+1}$$. Hence, $$S_t \subseteq S_{t+1}$$ and, therefore, the optimal thresholds are decreasing.

# References

The above model for pricing options was introduced by Taylor (1967).

Taylor, H.M. 1967. Evaluating a call option and optimal timing strategy in the stock market. Management Science 14, 1, 111–120. Available at: http://www.jstor.org/stable/2628546.

This entry was last updated on 31 Mar 2020 and posted in MDP and tagged optimal stopping, structural results, threshold strategy, finance.