Chapter 6
Finance

It used to be common for candidates with no finance knowledge to get hired into quantitative finance positions. Although this still happens for candidates with specialized knowledge that is in high demand, it’s more likely that you are required, or at least expected, to have a basic grasp of topics in finance. So you should expect to answer some finance questions and be judged on your answers.

Besides classic textbooks,¹ there are a few interview books in the market to help you prepare for finance interviews.² If you want to get prepared for general finance problems, you may want to read a finance interview book to get a feel for what types of questions are asked. The focus of this chapter is more on the intuitions and mathematics behind derivative pricing instead of basic finance knowledge. Derivative problems are popular choices in quantitative interviews—even for divisions that are not directly related to derivative markets—because these problems are complex enough to test your understanding of quantitative finance.

6.1. Option Pricing

Let’s begin with some notations that we will use in the following sections.

$T$: maturity date; $t$: the current time; $\tau=T-t$: time to maturity; $S$: stock price at time $t$; $r$: continuous risk-free interest rate; $y$: continuous dividend yield; $\sigma$: annualized asset volatility; $c$: price of a European call; $p$: price of a European put; $C$: price of an American call; $P$: price of an American put; $D$: present value, at $t$, of future dividends; $K$: strike price; $PV$: present value at $t$.

Price direction of options

How do vanilla European/American option prices change when $S,K,\tau,\sigma,r,$ or $D$ changes?

Solution: The payoff of a call is $\max(S-K,0)$ and the payoff of a put is $\max(K-S,0)$. A European option can only be exercised at the expiration time, while an American option can be exercised at any time before maturity. Intuitively we can figure out that the price of a European/American call should decrease when the strike price increases 138 since a call with a higher strike has no higher—and sometimes lower—payoff than a call with a lower strike. Using similar analyses, we summarize the effect of changing market conditions on an option’s value in Table 6.1.

The impact of time to maturity on the price of a European call/put is uncertain. If there is a large dividend payoff between two different maturity dates, a European call with shorter maturity that expires before the ex-dividend date may be worth more than a call with longer maturity. For deep in-the-money European puts, the one with shorter maturity is worth more since it can be exercised earlier (time value of the money).

↑: increase; ↓: decrease; ?: increase or decrease

It is also worth noting that Table 6.1 assumes that only one factor changes value while all others stay the same, which in practice may not be realistic since some of the factors are related. For example, a large decrease in interest rate often triggers a stock market rally and increases the stock price, which has an opposite effect on option value.

Put-call parity

Put-call parity: $c+K^{-r\tau}=p+S-D$, where the European call option and the European put option have the same underlying security, the same maturity $T$ and the same strike price $K$. Since $p\ge0$, we can also derive boundaries for $c$, $S-D-Ke^{-r\tau}\le c\le S$, from the put-call parity.

For American options, the equality no longer holds and it becomes two inequalities: $S-D-K\le C-P\le S-K^{-r\tau}$.

Can you write down the put-call parity for European options on non-dividend paying stocks and prove it?

Solution: The put-call parity for European options on non-dividend paying stocks is $c+K^{-r\tau}=p+S$. We can treat the left side of the equation as portfolio $A$—a call and a zero-coupon bond with face value $K$—and the right side as portfolio $B$—a put and the underlying stock, which is a protective put. Portfolio $A$ has payoff $\max(S_T-K,0)+K=\max(S_T,K)$ at maturity $T$; portfolio $B$ has payoff $\max(K-S_T,0)+S_T=\max(S_T,K)$ at $T$. Since both portfolios have the same payoff at $T$ and no payoff between $t$ and $T$, the no-arbitrage argument³ dictates that they must have the same value at $t$. Hence, $c+K^{-r\tau}=p+S$.

If we rearrange the put-call parity equation into $c-p=S-K^{-r\tau}$, it will give us different insight. The portfolio on the left side of the equation—long a call and short a put—has the payoff $\max(S_T-K,0)-\max(K-S_T,0)=S_T-K$, which is the payoff of a forward with delivery price $K$. A forward with delivery price $K$ has present value $S-K^{-r\tau}$. So we again have the put-call parity $c-p=S-K^{-r\tau}$. This expression shows that when the strike price $K=S^{r\tau}$ (forward price), a call has the same value as put; when $K<S^{r\tau}$, a call has higher value; and when $K>S^{r\tau}$, a put has higher value.

American v.s. European options

A. Since American options can be exercised at any time before maturity, they are often more valuable than European options with the same characteristics. But when the stock pays no dividend, the theoretical price for an American call and European call should be the same since it is never optimal to exercise the American call. Why should you never exercise an American call on a non-dividend paying stock before maturity?

Solution: There are a number of solutions to this popular problem. We present three arguments for the conclusion.

Argument 1. If you exercises the call option, you will only get the intrinsic value of the call $S-K$. The price of the American/European call also includes time value, which is positive for a call on a non-dividend paying stock. So the investor is better off selling the option than exercising it before maturity.

In fact, if we rearrange the put-call parity for European options, we have $c=S-K^{-r\tau}+p=(S-K)+(K-K^{-r\tau})+p$. The value of a European call on a non-dividend paying stock includes three components: the first component is the intrinsic value $S-K$; the second component is the time value of the strike (if you exercise now, 140 you pay $K$ now instead of $K$ at the maturity date, which is lower in present value); and the third component is the value of the put, which is often considered to be a protection against falling stock price. Clearly the second and the third components are both positive. So the European call should be worth more than its intrinsic value. Considering that the corresponding American call is worth at least as much as the European call, it is worth more than its intrinsic value as well. As a result, it is not optimal to exercise the American call before maturity.

Argument 2. Let’s compare two different strategies. In strategy 1, we exercise the call option⁴ at time $t$ ($t<T$) and receive cash $S-K$. Alternatively, we can keep the call, short the underlying stock and lend $K$ dollars with interest rate $r$ (the cash proceedings from the short sale, $S$, is larger than $K$). At the maturity date $T$, we exercise the call if it’s in the money, close the short position and close the lending. Table 6.2 shows the cash flow of such a strategy:

It clearly shows that at time $t$, we have the same cash flow as exercising the call, $S-K$. But at time $T$, we always have positive cash flow as well. So this strategy is clearly better than exercising the call at time $t$. By keeping the call alive, the extra benefit can be realized at maturity.

Argument 3. Let’s use a mathematical argument relying on risk-neutral pricing and Jensen’s inequality—if $f(X)$ is a convex function,⁵ then $E[f(X)]\ge f(E[X])$. From Figure 6.1, it’s obvious that the payoff (if exercised when $S>K$) of a call option $C(S)=(S-K)^+$ is a convex function of stock price with property

Let $S_1=S$ and $S_2=0$, then $C(\lambda S)\le\lambda C(S)+(1-\lambda)C(0)=\lambda C(S)$ since $C(0)=0$.

If the option is exercised at time $t$, the payoff at $t$ is $C(S_t-K)$. If it is not exercised until maturity, the discounted expected payoff (to $t$) is $\widetilde E[e^{-r\tau}C(S_T)]$ under risk-neutral measure. Under risk-neutral probabilities, we also have $\widetilde E[S_T]=S_te^{r\tau}$.

So $\widetilde E[e^{-r\tau}C(S_T)]=e^{-r\tau}\widetilde E[C(S_T)]\ge e^{-r\tau}C(\widetilde E[S_T])=e^{-r\tau}C(e^{r\tau}S_t)$,

where the inequality is from Jensen’s inequality.

Let $S=e^{r\tau}S_t$ and $\lambda=e^{-r\tau}$, we have $C(\lambda S)=C(S_t)\le e^{-r\tau}C(e^{r\tau}S_t)\le e^{-r\tau}\widetilde E[C(S_T)]$.

Since the discounted payoff $e^{-r\tau}\widetilde E[C(S_T)]$ is no less than $C(S_t)$ for any $t\le T$ under the risk neutral measure, it is never optimal to exercise the option before expiration.

I should point out that the payoff of a put is also a convex function of the stock price. But it is often optimal to exercise an American put on a non-dividend paying stock. The difference is that $P(0)=K$, so it does not have the property that $P(\lambda S)\le\lambda P(S)$. In fact, $P(\lambda S)\ge\lambda P(S)$. So the argument for American calls does not apply to American puts.

Similar analysis can also show that early exercise of an American call option for dividend-paying stocks is never optimal except possibly for the time right before an ex-dividend date.

B. A European put option on a non-dividend paying stock with strike price $80 is currently priced at $8 and a put option on the same stock with strike price $90 is priced at $9. Is there an arbitrage opportunity existing in these two options?

Solution: In the last problem, we mentioned that the payoff of a put is a convex function in stock price. The price of a put option as a function of the strike price is a convex function as well. Since a put option with strike 0 is worthless, we always have $P(0)+\lambda P(K)=\lambda P(K)>P(\lambda K)$.

For this specific problem, we should have $8/9\times P(90)=8/9\times9=8>P(80)$. Since the put option with strike price $80 is currently price at 8, it is overpriced and we should short it. The overall arbitrage portfolio is to short 9 units of put with $K=\$80$ and long 8 units of put with $K=90$. At time 0, the initial cash flow is 0. At the maturity date, we have three possible scenarios:

$S_T\ge90$, payoff = 0 (No put is exercised.)

$90>S_T\ge80$, payoff $=8\times(90-S_T)>0$ (Puts with $K=90$ are exercised.)

$S_T<80$, payoff $=8\times(90-S_T)-9\times(80-S_T)=S_T>0$ (All puts are exercised.)

The final payoff $\ge0$ with positive probability that payoff $>0$. So it is clearly an arbitrage opportunity.

Black-Scholes-Merton differential equation

Can you write down the Black-Scholes-Merton differential equation and briefly explain how to derive it?

Solution: If the evolution of the stock price is a geometric Brownian motion, $dS=\mu S\,dt+\sigma S\,dW(t)$, and the derivative $V=V(S,t)$ is a function of $S$ and $t$, then applying Ito’s lemma yields:

where $W(t)$ is a Brownian motion.

The Black-Scholes-Merton differential equation is a partial differential equation that should be satisfied by $V$:

To derive the Black-Scholes-Merton differential equation, we build a portfolio with two components: long one unit of the derivative and short $\frac{\partial V}{\partial S}$ unit of the underlying stock. Then the portfolio has value $\Pi=V-\frac{\partial V}{\partial S}S$ and the change of $\Pi$ follows equation

It is apparent that this portfolio is risk-free since it has no diffusion term. It should have risk-free rate of return as well: $d\Pi=r(V-\frac{\partial V}{\partial S}S)dt$. Combining these results we have

which is the Black-Scholes-Merton differential equation.

The Black-Scholes-Merton differential equation is a special case of the discounted Feynman-Kac theorem. The discounted Feynman-Kac theorem builds the bridge between stochastic differential equations and partial differential equations and applies to all Ito processes in general:

Let $X$ be an Ito process given by equation $dX(t)=\beta(t,X)dt+\gamma(t,X)dW(t)$ and $f(x)$ be a function of $X$. Define function $V(t,x)=E[e^{-r(T-t)}f(X_T)\mid X_t=x]$, then $V(t,x)$ is a martingale process that satisfies the partial differential equation

and boundary condition $V(T,x)=f(x)$ for all $x$.

Under risk-neutral measure, $dS=rSdt+\sigma SdW(t)$. Let $S=X$, $\beta(t,X)=rS$ and $\gamma(t,X)=\sigma S$, then the discounted Feynman-Kac equation becomes the Black-Scholes-Merton differential equation $\frac{\partial V}{\partial t}+rS\frac{\partial V}{\partial S}+\frac12\sigma^2S^2\frac{\partial^2V}{\partial S^2}=rV$.

Black-Scholes formula

The Black-Scholes formula for European calls and puts with continuous dividend yield $y$ is:

where

$N(x)$ is the cdf of the standard normal distribution and $N'(x)$ is the pdf of the standard normal distribution: $\displaystyle N(x)=\int_{-\infty}^{x}\frac1{\sqrt{2\pi}}e^{-y^2/2}dy$ and $\displaystyle N'(x)=\frac1{\sqrt{2\pi}}e^{-x^2/2}$.

If the underlying asset is a futures contract, then yield $y=r$. If the underlying asset is a foreign currency, then yield $y=r_f$, where $r_f$ is the foreign risk-free interest rate.

A. What are the assumptions behind the Black-Scholes formula?

Solution: The original Black-Scholes formula for European calls and puts consists of the equations $c=SN(d_1)-Ke^{-r\tau}N(d_2)$ and $p=Ke^{-r\tau}N(-d_2)-SN(-d_1)$, which require the following assumptions:

1. The stock pays no dividends.
2. The risk-free interest rate is constant and known.
3. The stock price follows a geometric Brownian motion with constant drift $\mu$ and volatility $\sigma$: $dS=\mu Sdt+\sigma SdW(t)$.
4. There are no transaction costs or taxes; the proceeds of short selling can be fully invested.
5. All securities are perfectly divisible.
6. There are no risk-free arbitrage opportunities.

B. How can you derive the Black-Scholes formula for a European call on a non-dividend paying stock using risk-neutral probability measure?

Solution: The Black-Scholes formula for a European call on a non-dividend paying stock is

Under the risk-neutral probability measure, the drift of stock price becomes the risk-free interest rate $r(t)$: $dS=r(t)Sdt+\sigma SdW(t)$. Risk-neutral measure allows the option to be priced as the discounted value of its expected payoff with the risk-free interest rate:

where $V(T)$ is the payoff at maturity $T$.

When $r$ is constant, the formula can be further simplified as $V(t)=e^{-r\tau}E[V(T)\mid S(t)]$. Under risk-neutral probabilities, $dS=rSdt+\sigma SdW(t)$. Applying Ito’s lemma, we get

So $S_T=Se^{(r-\sigma^2/2)\tau+\sigma\sqrt\tau\varepsilon}$, where $\varepsilon\sim N(0,1)$. For a European option, we have

Let $\tilde\varepsilon=\varepsilon-\sigma\sqrt\tau$, then $d\varepsilon=d\tilde\varepsilon$, $\varepsilon=-d_2\Rightarrow\tilde\varepsilon=-d_2-\sigma\sqrt\tau=-d_1$ and we have

From the derivation process, it is also obvious that $1-N(-d_2)=N(d_2)$ is the risk-neutral probability that the call option finishes in the money.

C. How do you derive the Black-Scholes formula for a European call option on a non-dividend paying stock by solving the Black-Scholes-Merton differential equation?

Solution: You can skip this problem if you don’t have background in partial differential equations (PDE). One approach to solving the problem is to convert the Black-Scholes-Merton differential equation to a heat equation and then apply the boundary conditions to the heat equation to derive the Black-Scholes formula.

Let $y=\ln S$ ($S=e^y$) and $\tilde\tau=T-t$, then $\frac{\partial V}{\partial t}=-\frac{\partial V}{\partial\tilde\tau}$, $\frac{\partial V}{\partial S}=\frac{\partial V}{\partial y}\frac{dy}{dS}=\frac1S\frac{\partial V}{\partial y}$ and

⁶ The Black-Scholes-Merton differential equation $\frac{\partial V}{\partial t}+rS\frac{\partial V}{\partial S}+\frac12\sigma^2S^2\frac{\partial^2V}{\partial S^2}-rV=0$ can be converted to $-\frac{\partial V}{\partial\tilde\tau}+(r-\frac12\sigma^2)\frac{\partial V}{\partial y}+\frac12\sigma^2\frac{\partial^2V}{\partial y^2}-rV=0$.

Let $u=e^{r\tilde\tau}V$, the equation becomes $-\frac{\partial u}{\partial\tilde\tau}+(r-\frac12\sigma^2)\frac{\partial u}{\partial y}+\frac12\sigma^2\frac{\partial^2u}{\partial y^2}=0$.

Finally, let $x=y+(r-\frac12\sigma^2)\tilde\tau=\ln S+(r-\frac12\sigma^2)\tilde\tau$ and $\tau=\tilde\tau$, then $\frac{\partial u}{\partial y}=\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial\tilde\tau}=\frac{\partial u}{\partial\tau}+(r-\frac12\sigma^2)\frac{\partial u}{\partial x}$, which transforms the equation to

So the original equation becomes a heat/diffusion equation $\frac{\partial u}{\partial\tau}=\frac12\sigma^2\frac{\partial^2u}{\partial x^2}$. For heat equation $\frac{\partial u}{\partial\tau}=\frac12\sigma^2\frac{\partial^2u}{\partial x^2}$, where $u=u(x,\tau)$ is a function of time $\tau$ and space variable $x$, with boundary condition $u(x,0)=u_0(x)$, the solution is

⁷

For European calls, the boundary condition is $u_0(S_T)=\max(S_T-K,0)$.

$S=\exp(x-(r-0.5\sigma^2)\tau)$. When $x=\psi$ and $\tau=0$, $S_T=e^\psi$.

Let $\varepsilon=\frac{\psi-x}{\sigma\sqrt\tau}$, then $d\varepsilon=\frac{d\psi}{\sigma\sqrt\tau}$, $e^\psi=e^{x+\varepsilon\sigma\sqrt\tau}$, $\exp(-\frac{(x-\psi)^2}{2\sigma^2\tau})=e^{-\varepsilon^2/2}$ and when $\psi=\ln K$, $\varepsilon=\frac{\ln(K/S)-(r-\sigma^2/2)\tau}{\sigma\sqrt\tau}=-d_2$

Now, it’s clear that the equation for $u(S,\tau)$ is exactly the same as the equation for $E[V(T)\mid S]$ in question $B$. Hence, we have $V(S,t)=e^{-r\tau}u(S,\tau)=SN(d_1)-Ke^{-r\tau}N(d_2)$ as well.

D. Assume zero interest rate and a stock with current price at $1 that pays no dividend. When the price hits level $$H$ ($H>1$) for the first time you can exercise the option and receive $1. What is this option worth to you today?

Solution: First let’s use a brute-force approach to solve the problem by assuming that the stock price follows a geometric Brownian motion under risk-neutral measure: $dS=rSdt+\sigma SdW(t)$. Since $r=0$, $dS=\sigma SdW(t)\Rightarrow d(\ln S)=-\frac12\sigma^2dt+\sigma dW(t)$. When $t=0$, we have $S_0=1\Rightarrow\ln(S_0)=0$.

Hence, $\ln S=-\frac12\sigma^2t+\sigma W(t)\Rightarrow\frac{\ln S+\frac12\sigma^2t}{\sigma}=W(t)$ is a Brownian motion.

Whenever $S$ reaches $$H$, the payoff is $1. Because the interest rate is 0, the discounted payoff is also $1 under risk-neutral measure. So the value of the option is the probability that $S$ ever reaches $$H$, which is equivalent to the probability that $\ln S$ ever reaches $\ln H$. Again we can apply the exponential martingale $Z(t)=\exp\{\lambda W(t)-\frac12\lambda^2t\}$ as we did in Chapter 5:

To remove the terms including time $t$, we can set $\lambda=\sigma$ and the equation becomes $E[\exp(\ln S)]=1$. The Let $P$ be the probability that $\ln S$ ever reaches $\ln H$ (using $-\infty$ as the negative boundary for stopping time), we have

So the probability that $S$ ever reaches $$H$ is $1/H$ and the price of the option should be $$1/H$. Notice that $S$ is a martingale under the risk-neutral measure;⁸ but $\ln S$ has a negative drift. The reason is that $\ln S$ follows a (symmetrical) normal distribution, but $S$ itself follows a lognormal distribution, which is positively skewed. As $T\to\infty$, although the expected value of $S_T$ is 1, the probability that $S_T\ge1$ actually approaches 0.

It is simpler to use a no-arbitrage argument to derive the price. In order to pay $1 when the stock price hits $$H$, we need to buy $1/H$ shares of the stock (at $$1/H$). So the option should be worth no more than $$1/H$. Yet if the option price $C$ is less than $$1/H$ ($C<1/H\Rightarrow CH<1$), we can buy an option by borrowing $C$ shares of the stock. The initial investment is 0. Once the stock price hits $$H$, we will excise the option and return the stock by buying $C$ shares at price $$H$, which gives payoff $1-CH>0$. That means we have no initial investment, yet we have possible positive future payoff, which is contradictory to the no arbitrage argument. So the price cannot be less than $$1/H$. Hence, the price is exactly $$1/H$.

E. Assume a non-dividend paying stock follows a geometric Brownian motion. What is the value of a contract that at maturity $T$ pays the inverse of the stock price observed at the maturity?

Solution: Under risk-neutral measure $dS=rSdt+\sigma SdW(t)$. Apply Ito’s lemma to $V=1/S$:

So $V$ follows a geometric Brownian motion as well and we can apply Ito’s lemma to $\ln V$:

Hence, $\ln(V_T)\sim\ln(V_t)+N((-r+\frac12\sigma^2)\tau,\sigma^2\tau)$ and $E[V_T]=E[e^{\ln V_T}]=\frac1{S_t}e^{-r\tau+\sigma^2\tau}$.

Discounting the payoff by $e^{-r\tau}$, we have $V=e^{-r\tau}E[V_T]=\frac1{S_t}e^{-2r\tau+\sigma^2\tau}$.