Assume I need to price a Bermudan option which can be exercised at following dates: $t_1$, $t_2$, ..., $t_n$. I think that the price of such an option will be maximum of the prices of European options with maturities $t_1$, $t_2$, ..., $t_n$. Am I right or wrong?
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2 Answers
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You are wrong. Using the maximum of the prices of the European options is equivalent to choosing (and making that choice final) on $t=0$ the date $t_i$ on which you will exercise. As such a choice would be sub-optimal, you would be giving up value. Therefore the Bermuda option is worth more than the maximum of the prices of the European options.
answered Dec 20, 2017 at 12:30
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1$\begingroup$ This can be made a little more precise. For example, let $\mathscr{T}$ be the set of stopping times with values in $\{t_1, \ldots, t_n\}$. Then, the Bermudan option value is given by \begin{align*} \sup_{\tau \in \mathscr{T}}E\big(Discount(\tau)\,Option(\tau)\big) \ge \max_{1\le i \le n} E\big(Discount(t_i)\,Option(t_i)\big), \end{align*} as $t_i$, for $i=1, \ldots, n$, are particular stopping times. $\endgroup$– GordonDec 20, 2017 at 17:54
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Without the math, look at it this way: I give you a die to toss. You can toss it thrice and take the payoff as the number on the die. On each turn you can either accept the payoff or move on. At the last throw, you must accept the payoff.
Your logic would state that this product has value equivalent to tossing a die once and only once (here, all 'Europeans' are equally valuable, so your 'bermudan (3 tosses)' should equal the most expensive european (any one of the 3 tosses). You can see that this is wrong. To convince yourself more, think what would happen if I was allowed to toss the coin an infinite number of times.