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A $(K_1, t_1, K_2, t_2)$ double call option is one that can be exercised either at time $t_1$ with strike price $K_1$ or at time $t_2$ ($t_2 > t_1$) with strike price $K_2$. Argue that you would never exercise at time $t_1$ if $K_1 > e^{-r(t_2-t_1)}K_2$.

This was exercise 5.26 on p. 90 of Sheldon M. Ross's "An Elementary Introduction to Mathematical Finance", 3rd edition, Cambridge University Press, 2011.

I'd appreciate help in solving this problem. My foremost problem is formalizing the phrase "you would never exercise at time $t_1$". Does it mean "You would never exercise at time $t_1$ and then just hold the underlying asset until time $t_2$"? What if I exercise at time $t_1$, but, rather than holding the underlying asset until time $t_2$, I partly invest it in something and partly deposit it in a savings account? Is there no such investment/savings strategy that would make it more advantageous for me to exercise at time $t_1$?


My attempted solution

I will show that, under a certain extra assumption (namely $(*)$ below), if the inequality $K_1 > e^{-r(t_2-t_1)}K_2$ holds, then the expected profit when exercising at $t_1$ is no greater than the expected profit when exercising at $t_2$ (i.e. $(**)$ below).

But will this solve the problem? Well, it depends on two factors:

(a) Is this inequality of expected profits (i.e. $(**)$ below) what is meant by the phrase "you would never exercise at time $t_1$"? If not, what mathematical formula does this phrase translate into?

(b) Is the extra assumption that I am going to make (i.e. $(*)$ below) reasonable? If not, how can the argument be modified to avoid relying on this assumption?

We shall start by introducing a couple symbols to denote the double call option as well as the underlying asset that the double call option derives from. Denote the underlying asset by $S$ and denote the option by $X$. $S$ and $X$ are stochastic processes $S = (S_t)_{t \in [0, \infty)}, X = (X_t)_{t \in [0, \infty)}$, where, for every $t \in [0, \infty)$, $S_t$ is the underlying's payoff at time $t$ and $X_t$ is the double call option's payoff at time $t$. As always the payoffs $S_0, X_0$ at the present time $t = 0$ are taken to be constant.

Furthermore, denote the market's fixed interest rate by $r$ (we assume $r \in [0, \infty)$), and denote the profit process associated with $X$ by $P$, in other words, for every $t \in [0, \infty)$, $$ P_t = e^{-rt}X_t - X_0 $$

We shall now make the extra assumption, to which we alluded above, that $$ e^{-rt_1}S_{t_1} \sim e^{-rt_2}S_{t_2} \tag{*} $$ i.e. the distribution of the underlying's present value at time $t_1$ is the same as its distribution at time $t_2$.

We may now state our main result: $$ E(P_{t_1}) \leq E(P_{t_2}) \tag{**} $$ where the expectation is calculated w.r.t. the actual market probability measure (as opposed to the EMM probability measure, if such exists, though note that the argument to follow does not make any assumption regarding the existence or absence of arbitrage in the market. In fact, as far as mathematical correctness is concerned, the following chain of equalities and inequalities holds no matter what probability measure the expectations are calculated with respect to, as long as it's the same one throughout).

$$ \begin{align} E\left(P_{t_1}\right) &= E\left(e^{-rt}X_{t_1} - X_0\right) \\ &= E\left(e^{-rt_1}(S_{t_1} - K_1)_+ - X_0\right) \\ &= E\left((e^{-rt_1}S_{t_1} - e^{-rt_1}K_1)_+ - X_0\right)\\ &= E\left((e^{-rt_2}S_{t_2} - e^{-rt_1}K_1)_+ - X_0\right)\\ &\leq E\left((e^{-rt_2}S_{t_2} - e^{-rt_2}K_2)_+ - X_0\right)\\ &= E\left(P_{t_2}\right) \end{align} $$ where the inequality is due to the fact that, by assumption, $K_1 > e^{-r(t_2-t_1)}K_2$, which is equivalent, after rearrangement of terms, to $$ K_1e^{-rt_1} > K_2e^{-rt_2} $$

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    $\begingroup$ Is there a typo at $e^{r(t_1-t_1)}K_2$ ? $\endgroup$ Commented Aug 7, 2015 at 15:54
  • $\begingroup$ @calculus: Yes, indeed. Corrected. Thanks. $\endgroup$
    – Evan Aad
    Commented Aug 7, 2015 at 15:55
  • $\begingroup$ What have you tried? Can you handle, say, the case when $t_1$ is $0$? Also you have problems with your discount factor. $\endgroup$
    – lulu
    Commented Aug 7, 2015 at 15:56
  • $\begingroup$ @lulu: I'm not sure I understand what I'm supposed to do. Am I supposed to show that the profit when exercising at time $t_1$ is never greater than the profit when exercising at time $t_2$? $\endgroup$
    – Evan Aad
    Commented Aug 7, 2015 at 15:58
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    $\begingroup$ Well, try that. Scenario A: I exercise my call on $t_1$ and hold the asset til $t_2$ vs. Scenario B: I don't exercise the call early, I invest the cash, and wait until $t_2$. $\endgroup$
    – lulu
    Commented Aug 7, 2015 at 16:02

3 Answers 3

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There are a few missing (but standard) assumptions underlying this. As a technical point, you have to assume that the market is liquid and arbitrage free. in particular, you have to assume that there is a rational market available into which you could sell the unexercised option at any time. Or, similarly, a market wherein you can short the asset liquidly without incurring special financing costs. Also, and more critically, you have to assume that the asset "carries at the riskless rate". The claim may be false otherwise (if, say, the asset is like oil, which incurs non-trivial storage costs, or if the asset finances at a rate other than the riskless rate, or if the asset releases cash, like a bond or a dividend paying stock).

Note: to tie these assumptions into the approach begun in the question, the assumption that the asset carries at the riskless rate is equivalent to the desired claim that the expected value of the stock at a future date is just the forward value of the spot price. Again, this is an assumption. If, say, your stock pays dividends then it is untrue (without appropriate corrections).

Assuming all of that, then imagine the two scenarios sketched above:

Scenario A: use exercise at $t_1$ and just hold the asset until $t_2$.

Scenario B: you do not exercise early, instead you invest the $K_1$ at the riskless rate and wait until $t_2$.

Now, let's look at $t_2$. The asset is trading at X_2, say. Of course, your cash has grown to something more than $K_2$ (by your inequality on the strikes). If $X_2 ≥ K_2$ then of course you exercise the call, if you are in B. Thus, in both cases you own the asset, but in B you are slightly better off (because you made a little cash on your investment).

If, on the other hand, $X_2 < K_2$ then in scenario B you are even better off. YOu have a loss from scenario A which you are spared since you needn't exercise.

thus in call cases Scenario B dominates scenario A.

One might worry a bit about what happens between times $t_1$ and $t_2$. For example, if the asset shoots up the day after $t_1$ and crashes to $0$ some while after that, you can imagine that scenario A looks better (since you could sell the asset to exploit the bubble). But in that case the technical assumptions I stressed at the beginning of my answer kick in. You can sell the option just as easily as you could sell the asset and, as the call price is always at least the value of the forward purchase you can still exploit the bubble. Or you could short the stock against the forward call exercise.

As an exercise: find where, in the above argument, the assumption on the "riskless carry" was used.

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  • $\begingroup$ I don't understand the assumption that the asset carry at the riskless rate. Would you mind elaborating, please? $\endgroup$
    – Evan Aad
    Commented Aug 7, 2015 at 20:16
  • $\begingroup$ No problem. Look at the argument which shows that the forward price is just the forward value of the spot price (I'm sure your text has that worked out). That statement is true for any non-dividend paying asset that I can short frictionlessly (by a simple arbitrage argument). But it is false for assets which pay dividends (clearly) and it is false for assets which we can't short. "Difficult to Borrow" stocks trade at a premium to LIBOR because you can't short them without handing someone extra money. Accordingly the forward must be lower that the forward value of the cash. $\endgroup$
    – lulu
    Commented Aug 7, 2015 at 20:22
  • $\begingroup$ Oil is the opposite of a "Difficult to Borrow" stock. it costs a lot to store and insure oil, so (all else equal) the forward has to be higher then the sport adjusted by LIBOR (else everyone would buy it forward and leave the extra costs to someone else). $\endgroup$
    – lulu
    Commented Aug 7, 2015 at 20:24
  • $\begingroup$ Do you mean that the asset carries at the riskless rate iff whenever the asset is worth $x$ dollars at time $s$, it's worth $xe^{r(t-s)}$ dollars at time $t$, for $t > s$? $\endgroup$
    – Evan Aad
    Commented Aug 7, 2015 at 20:25
  • $\begingroup$ The prepayment option, which I alluded to in a comment, is a call on a cash flow. But the price of that cash flow may well down over time, not up, because if you buy it later you miss the cash flow over the intervening time. $\endgroup$
    – lulu
    Commented Aug 7, 2015 at 20:26
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Two conditions can obtain at $t_1$: either $S_1 \leq K_1$, and thus it is not optimal to exercise, or $S_1 \gt K_1$. In this case, one could exercise, but one can build a zero cost portfolio that yields a positive payoff with probability greater than zero (using the usual assumptions of no arbitrage: short selling, borrowing and lending at rate $r$, etc.).

In this second situation, build a zero-cost portfolio by keeping the call option and selling one unit of $S$ for $S_1$, and investing it at the risk-free rate. At $t_2$, the portfolio will be worth $(S_2 - K_2)^+ + S_1 e^{r (t_2 - t_1)} - S_2$.

Again, two conditions obtain:

If $S_2 \leq K_2$, the option is not exercised and the portfolio is worth:

$S_1 e^{r (t_2 - t_1)} - S_2 \geq S_1 e^{r (t_2 - t_1)} - K_2 \gt K_1 e^{r (t_2 - t_1)} - K_2 > 0$

If $S_2 > K_2$, the option is exercised and the portfolio is worth:

$S_2 - K_2 + S_1 e^{r (t_2 - t_1)} - S_2 = S_1 e^{r (t_2 - t_1)} - K_2 \gt K_1 e^{r (t_2 - t_1)} - K_2 > 0$

In both cases, we're using the fact that $S_1 \gt K_1$ and $K_1 e^{r (t_2 - t_1)} \gt K_2$. Since we get a positive payoff with probability 1, it is clear that it is not optimal to exercise the call early.

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This is an interest question. The payoff at time $t_1$ of the option is of the form \begin{align*} & \ \max\left( (S_{t_1}-K_1)^+, \ e^{-r(t_2-t_1)}E\Big((S_{t_2}-K_2)^+ \mid \mathcal{F}_{t_1}\Big) \right) \\ = & \ e^{-r(t_2-t_1)}E\Big((S_{t_2}-K_2)^+ \mid \mathcal{F}_{t_1}\Big) \\ & \ + \max\left( (S_{t_1}-K_1)^+ - e^{-r(t_2-t_1)}E\Big((S_{t_2}-K_2)^+ \mid \mathcal{F}_{t_1}\Big), \, 0\right), \end{align*} where $E$ is expectation operator under the risk-neutral measure, and $\mathcal{F}_{t_1}$ is the information set at time $t_1$. The option will be exercised at time $t_1$ if $$(S_{t_1}-K_1)^+ - e^{-r(t_2-t_1)}E\Big((S_{t_2}-K_2)^+ \mid \mathcal{F}_{t_1}\Big)>0.$$

However, if $K_1 > e^{-r(t_2-t_1)}K_2$, then by Jensen's inequality, \begin{align*} &\ (S_{t_1}-K_1)^+ - e^{-r(t_2-t_1)}E\Big((S_{t_2}-K_2)^+ \mid \mathcal{F}_{t_1}\Big) \\ \leq& \ (S_{t_1}-K_1)^+ - e^{-r(t_2-t_1)}\left(E\Big((S_{t_2}-K_2) \mid \mathcal{F}_{t_1}\Big) \right)^+\\ =&\ (S_{t_1}-K_1)^+ - e^{-r(t_2-t_1)}\left(e^{r(t_2-t_1)}S_{t_1}-K_2\right)^+\\ =&\ (S_{t_1}-K_1)^+ - \left( S_{t_1}-e^{-r(t_2-t_1)} K_2 \right)^+\\ \leq& \ 0. \end{align*} That is, it is never optimal to exercise at time $t_1$, if $K_1 > e^{-r(t_2-t_1)}K_2$.

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