Answer-first summary for fast verification
Answer: If the barrier (H) is $17.00, then the value of this down-and-in put option, p(di), is equal to the value of the vanilla put option; p(di) = p
**Correct answer: A** A down-and-in put only becomes active if the underlying price falls to or below the barrier. With \(S_0 = 18.40\) and \(H = 17.00\), the barrier has **not** been breached at inception, so the option is **not** immediately equivalent to a vanilla put. Why the others are true: - **B**: For knock-in and knock-out barriers with no rebate, the standard decomposition holds: \[ \text{Down-and-in} + \text{Down-and-out} = \text{Vanilla} \] so \(p(di) = p - p(do)\). - **C**: A knock-in option typically has lower vega than the corresponding vanilla option because part of its value depends on barrier activation rather than pure volatility exposure. - **D**: More frequent monitoring increases the chance that the barrier is observed as breached, so the value of the down-and-in option increases.
Author: Manit Arora
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Question-413.1. Let (p) be the value of a standard (non-exotic or vanilla) put option with a strike price of $16.00, where the underlying current underlying stock price, S(0), is $18.40. Consider an exotic knock-in barrier put option (aka, down-and-in put) also with a strike price of (K) of $16.00. Each of the following it true about the down-and-in put, EXCEPT which is false?
A
If the barrier (H) is $17.00, then the value of this down-and-in put option, p(di), is equal to the value of the vanilla put option; p(di) = p
B
If the barrier (H) is $14.00, then the value of this down-and-in put option, p(di), is equal to the value of the vanilla put option minus the value of an equivalent down-and-out put; p(di) = p - p(do)
C
The vega of the down-and-in put, p(di), is generally lower than the vega of the corresponding vanilla put; vega(di) < vega(p)
D
As we increase the frequency with which we observe the asset price in determining whether the barrier is reached, the value of the down-and-in put increases
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