结论：该衍生品价格为 $0.75$（至少有明确无套利上界为 0.75；在允许长期滚动空头融资的解释下也可视为精确定价）。

无套利上界：若该衍生品价格 $>0.75$，则卖出 100 份该衍生品，用 75 买入 1 股 IBM。

若 IBM 触及 100，则卖出股票得到 100，支付衍生品合计 100；仍可留下正现金（来自初始溢价），形成套利，矛盾。

因此价格不能超过 0.75；在“可无限期滚动空头/对冲头寸”的假设下，也可得到 0.75 作为合理价格。

---

Original Explanation

Most people incorrectly deduce that the derivative security is worth $1$. If you got this answer, go back right now and think some more. I present three solution techniques: the first uses standard no-arbitrage arguments; the second uses partial differential equations (PDEs); the third uses stopping times.42

FIRST SOLUTION

You are an investment banker. Assume there exists a derivative security that promises one dollar when IBM hits $100$ for the first time. If this security is marketable at more than $0.75$, then you should issue $100$ of them and use $75$ of the proceeds to buy one share of IBM. If IBM ever hits $100$, sell the stock and pay $1$ to each security holder as contracted. You sell the securities, perfectly hedge them, and still have money in your pocket. By no-arbitrage, the security cannot sell for more than $0.75$.

The converse is that if this security costs less than $0.75$, you should buy $100$ of these securities financed by a short position in one IBM share. For this to establish $0.75$ as a lower bound on the security price (and, therefore, to pinpoint the price at $0.75$ — the solution given to the interviewee by the Wall Street firm), you need to assume that you can roll over a short position indefinitely. This assumption seems reasonable for moderate amounts of capital. However, it is not clear to me that this is a reasonable interpretation of “ignore any short sale restrictions” when larger quantities of capital are involved. As one colleague said to me: “If it were possible to short forever, I’d short stocks with face value of a billion dollars, consume the billion, and roll over my short position forever.” This seems to be an arbitrage opportunity.

We conclude that $0.75$ is a clear upper bound by no-arbitrage, and thus $1$ cannot be the correct answer. Whether or not $0.75$ is also a lower bound is arguable (but it seems to make sense for moderate amounts of capital). The second solution technique also establishes $0.75$ as the value of the security.

SECOND SOLUTION

This technique is more advanced and may be beyond the average candidate.43 The derivative value $V$ must satisfy the Black–Scholes PDE (Wilmott et al. [1993]):

$$\frac{\partial V}{\partial t} +\frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} +rS\frac{\partial V}{\partial S} -rV = 0.$$

The boundary conditions that make sense for $V(S,t)$ are

$$V(S=100,\ \text{any } s>t) = \$1, \quad V(S=0,\ \text{any } s>t) = \$0.$$

Let us simplify by searching initially for a solution that is affine in $S$: $V(S,t)=k\,S(t)+\ell$, for some constants $k$ and $\ell$. The two boundary conditions imply

$$k\cdot \$100 + \ell = \$1, \qquad k\cdot \$0 + \ell = \$0.$$

From these we deduce that $k=\frac{1}{100}$ and $\ell=0$. The functional form $V(S,t)=\frac{1}{100}S(t)$ satisfies the Black–Scholes PDE and the two boundary conditions and is thus the derivative value. In the special case where $S(t)=\$75$, we get $V=\$0.75$, as for the first technique.

THIRD SOLUTION

Following Shreve (2004, p. 297), we take a “first passage” or “stopping time” approach. This technique is even more advanced than the previous one.

Let now be time $t=0$ and consider a derivative with lifespan $T$ that pays $1 if stock price $S(t)$ hits a barrier at level $B>S(0)$ before time $T$. In our case $B=\$100$ and $S(0)=\$75$. In the Black–Scholes world we have the risk-neutral SDE

$$dS(t)=r\,S(t)\,dt+\sigma\,S(t)\,d\widetilde{W}(t),$$

with solution

$$S(t)=S(0)\exp\!\Big(\big(r-\tfrac{1}{2}\sigma^2\big)t+\sigma\,\widetilde{W}(t)\Big),$$

where $\widetilde{W}(t)$ is a standard Brownian motion, and define the drift parameter $$\alpha \equiv \frac{1}{\sigma}\left(r-\frac{1}{2}\sigma^2\right).$$

Let $\widehat{M}(T)\equiv \max_{0\le t\le T}\widehat{W}(t)$, then $$\max_{0\le t\le T}S(t)=S(0)\exp\!\big(\sigma\,\widehat{M}(T)\big).$$

The stock price hits the barrier $B$ before time $T$ iff this maximum stock price exceeds $B$, i.e., $S(0)e^{\sigma \widehat{M}(T)}>B$. Simple algebra shows

$$\tilde{P}\Big(\max_{0\le t\le T}S(t)>B\Big) = 1-\tilde{P}\!\left(\widehat{M}(T)\le \frac{1}{\sigma}\ln\!\frac{B}{S(0)}\right) = 1-\tilde{P}\big(\widehat{M}(T)\le b\big),$$

where $b=\frac{1}{\sigma}\ln\!\left(\frac{B}{S(0)}\right)$. Using Shreve (2004, Cor. 7.2.2, p. 297),

$$\widehat{P}\big(\widehat{M}(T)\le b\big) = N\!\left(\frac{b-\alpha T}{\sqrt{T}}\right) - e^{2\alpha b}\,N\!\left(\frac{-b-\alpha T}{\sqrt{T}}\right), \quad b\ge 0.$$

It follows that the risk-neutral probability that we get our $1 is

$$\begin{aligned} \tilde{P}\Big(\max_{0\le t\le T} S(t) > B\Big) &= 1-\left[ N\!\left(\frac{b-\alpha T}{\sqrt{T}}\right) - e^{2\alpha b}\,N\!\left(\frac{-b-\alpha T}{\sqrt{T}}\right) \right] \\ &= N\!\left(\frac{-b+\alpha T}{\sqrt{T}}\right) + \left(\frac{B}{S(0)}\right)^{\left(\frac{2r}{\sigma^2}-1\right)} N\!\left(\frac{-b-\alpha T}{\sqrt{T}}\right) \\ &= N\!\left(\frac{\ln\!\left(\frac{S(0)}{B}\right)+\left(r-\frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}\right) + \left(\frac{S(0)}{B}\right)^{\left(1-\frac{2r}{\sigma^2}\right)} N\!\left(\frac{\ln\!\left(\frac{S(0)}{B}\right)-\left(r-\frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}\right). \end{aligned}$$

Equation $(\ast)$ gives the risk-neutral probability of the stock price touching the barrier and generating the payoff. If we multiply this probability by the $1 payoff, we get the risk-neutral expected future payoff. If the interest rate is zero, then we do not need to discount, and plugging $r=0$ into $(\ast)$ gives the value of the finitely-lived instrument that pays $1 when the stock price hits $B$. If we take the limit as $T\to\infty$, the first $N(\cdot)$ term $\to 0$ (argument $\to -\infty$), and the second $N(\cdot)$ term $\to 1$ (argument $\to +\infty$). We are left with $1\cdot \left(\frac{S(0)}{B}\right)$, which is just $0.75 in our case.

From equation $(\ast)$, you can deduce the risk-neutral probability that the stock price eventually hits the barrier as

$$\lim_{T\to\infty}\tilde{P}(\text{hit }B) = \begin{cases} 1, & r\ge \tfrac{1}{2}\sigma^{2},\\[6pt] \left(\dfrac{S(0)}{B}\right)^{\left(1-\frac{2r}{\sigma^{2}}\right)}, & 0\le r<\tfrac{1}{2}\sigma^{2}. \end{cases}$$

Extension. Some candidates have very recently been asked this question in the case $r>0$. My first two solutions do not rely upon $r=0$, so the answer must be the same. Can we demonstrate this using this third solution technique? Well, to discount the expected future value at rate $r$ we need to know how long to discount for. Let $\tau$ be the time at which $S(t)$ first hits $B$. This “first passage time” is distributed Inverse Gaussian with the following pdf:47

$$f(\tau)=\frac{\ln\!\left(\tfrac{B}{S(0)}\right)}{\sigma\sqrt{2\pi\,\tau^{3}}} \exp\!\left\{ -\frac{1}{2}\left( \frac{\ln\!\left(\tfrac{B}{S(0)}\right)-(r-\tfrac{1}{2}\sigma^{2})\,\tau}{\sigma\sqrt{\tau}} \right)^{2} \right\}, \quad \tau\ge 0.$$

The general functional form of the Inverse Gaussian is

$$f(x)=\sqrt{\frac{\lambda}{2\pi x^{3}}}\, \exp\!\left\{-\frac{\lambda}{2}\left(\frac{x-\mu}{\mu\sqrt{x}}\right)^{2}\right\}, \quad x\ge 0,\ \lambda>0,\ \mu>0.$$

In our case $x=\tau$, $\lambda=\dfrac{\big[\ln\!\left(\tfrac{B}{S(0)}\right)\big]^{2}}{\sigma^{2}}$ and $\mu=\dfrac{\ln\!\left(\tfrac{B}{S(0)}\right)}{\,r-\tfrac{1}{2}\sigma^{2}\,}$. With $B>S(0)$, the condition $\mu>0$ implies $r>\tfrac{1}{2}\sigma^{2}$, else we do not have a valid pdf for $f(\tau)$. If we proceed assuming $r>\tfrac{1}{2}\sigma^{2}$, our discount factor is the expected value of $e^{-r\tau}$ with respect to $f(\tau)$ such that $\tau\le T$:

$$ E!\