4
$\begingroup$

My question concerns the Black-Scholes formula for the value of a European option, namely

\begin{align} C(S_t, t) &= N(d_1)S_t - N(d_2) Ke^{-r(T - t)} \\ d_1 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S_t}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)\right] \\ d_2 &= d_1 - \sigma\sqrt{T - t} \\ \end{align}

First, I want to ignore for purposes of this question the derivation of this formula using the Black-Scholes PDE dynamic hedging argument, which I haven't been through in detail, but I understand in principal. Rather, I'd like someone to explain to me why the following argument, which leads me to a (slightly) different answer than the one above, is wrong.

The Black-Scholes model of stock movements posits that the change $\Delta S$ in a stock price over a small time interval $\Delta t$ behaves as

$\Delta S = \mu S \Delta t + \sigma \sqrt{\Delta t} \varepsilon S$

where $\mu = \text{drift rate}$, $\sigma = \text{volatility}$ (constant), and $\varepsilon$ is a fair coin flip resulting in $1$ and $-1$ (I prefer this incremental equation to a stochastic one, I'm not up on Ito's lemma and all that). $S_T$, the stock price at time $T$, is then (for fixed $\Delta t$) the random variable

$S_T = S_0 \left(1+\mu \Delta t + \sigma \sqrt{\Delta t} \right)^X \left(1+\mu \Delta t - \sigma \sqrt{\Delta t}\right)^{N-X}$

where $X$ is a binomial R.V. counting the number of $1$'s from the coin flips and $N = T/\Delta t$. Using the normal approximation for $X$ and letting $\Delta t \rightarrow 0$ gets us

$S_T = S_0 e^{(\mu-\sigma^2/2)T}e^{\sigma \sqrt{T} Z}$

where $Z$ is standard normal (or we could replace $\sqrt{T} Z$ with brownian motion $W$ for a dynamic model).

Now, what is the fair value of a European option on this stock at strike price $K$ and time to expiry $T$? Well, the seller of such an option would expect to have to pay at expiry, on average

$E[\text{max}(S_T - K,0)]$

That is, the expected value of the payout of the option. And so to price it today, one would discount this expected payout by the risk-free rate:

$e^{-rT}E[\text{max}(S_T - K,0)]$

But, this is wrong, according to the Black-Scholes formula given a the beginning. I'll spare you the computations, but in fact, this computation returns the correct Black-Scholes formula only if we change our model $S_T$ to

$S_T = S_0 e^{(r-\sigma^2/2)T}e^{\sigma \sqrt{T} Z}$

That is, by replacing $\mu$, the drift rate of the stock, to $r$, the risk-free rate. And so my question is, why are we justified in doing this? Why is it okay, for purposes of this computation, to "pretend" that the drift rate of the stock is $r$, when it very well may not be in real life?

Two objections come to my mind regarding my computation. First, it assumes that the person valuing the option only cares about its expected payout, and not the volatility of the payout, due to the volatility of the stock itself (i.e. it assumes risk-neutrality, which may not be the case). Still, if someone were to sell enough of these options (on identical but independent stocks) for the law of averages to win out, then this would be a correct valuation. And secondly, this doesn't seem to be actually computing the market value of the option, but only the expected cost to the seller. That is, expected cost to the seller might not be exactly the same thing as value to the buyer. But still, if someone were to offer options at a price other than this one, obvious (statistical) arbitrage opportunities would exist (for a market participant with enough reserves to weather the volatility in the payouts).

I'm interested in your thoughts on these objections. But my big question is still: why does replacing $\mu$ with $r$ lead us to the correct formula, when $\mu$ and $r$ are almost never equal? Thanks you for any help.

$\endgroup$
3
$\begingroup$

There are a lot of different questions to address in your post.

I'd like someone to explain to me why the following argument, which leads me to a (slightly) different answer than the one above, is wrong.

Letting $\mathbb{P}$ designate real-world probabilities, the reason your reasoning is not correct is because you posit that:

the seller of such an option would expect to have to pay at expiry, on average $$ \mathbb{E}^{\mathbb{P}}[\max(S_T−K,0)] $$ That is the expected value of the payout of the option. And so to price it today, one would discount this expected payout by the risk-free rate.

Your reasoning would be right if the call option and the underlying asset on which it is written $-$ and which we use to hedge our exposure $-$ were not traded. To understand this, consider a way simpler product: a forward contract written on the stock, in which you promise to buy or sell one share of stock at maturity $T$ at a price $K$. The expected value of $S_T$ under your model is:

$$ \mathbb{E}^{\mathbb{P}}[S_T]=S_0e^{\mu T}$$

Is this a fair price for the forward contract:

$$ K = S_0e^{\mu T} \text{ ?}$$

Well, no: assume you can enter this contract with a client that, like yourself and other participants in the market, can borrow at a rate $r<\mu$. The client can enter into a forward sell agreement with you, so that you promise to buy the stock share at price $K$ above at time $T$. The client can then borrow at rate $r$ an amount $S_0$ $-$ i.e. equal to the stock's current price $-$ with maturity $T$ and buy the stock now. At $T$, your client will:

  • use the stock he purchased at $t=0$ to deliver the stock to you;
  • own an amount $S_0e^{rT}$ to his lender;
  • receive an amount $S_0e^{\mu T}$ from selling the stock to you.

So his final payoff will be:

$$ S_0(e^{\mu T}-e^{rT}) > 0 $$

which constitutes an arbitrage opportunity. The market will make those disappear because they constitute "free money". What is then the right price to charge for the forward contract? Well, the client has always the choice to borrow $S_0$ at rate $r$ and deliver you the share at maturity, hence the only logical price to avoid arbitrage strategies is:

$$ K_{\text{For}} = S_0e^{rT} $$

You see a few things here:

  • Like with the more complicated example of a call option, pricing a forward contract is equivalent to requiring the stock price to have a return rate of $r$ under some probability measure: it does not matter that the stock's true average return is $\mu$, the fact that you can hedge your exposure by borrowing at a rate $r$ overrides this property.
  • When the derivative $-$ call option, forward contract $-$ or its underlying can be traded and hence the risk can be hedged, then the law of large numbers that you call in is overridden by the requirement that there must not be arbitrage opportunities (in real life, these fade away very quickly). For more details on this, you can check my answer to question "Reference for why a derivative is a derivative and not say an insurance contract".

Still, if someone were to sell enough of these options (on identical but independent stocks) for the law of averages to win out, then this would be a correct valuation.

The fact is that, by valuing the derivative that way, he will be bleeding money away by leaving market participants arbitrage him at each contract.

But my big question is still: why does replacing $\mu$ with $r$ lead us to the correct formula, when $\mu$ and $r$ are almost never equal?

Intuitively, you can think of it this way: it is not the rate of return of the underlying asset that matters, but rather the rate at which you can borrow the underlying asset to hedge yourself.

Mathematically, the shortest way to understand this is through a PDE approach. You might know that through dynamic hedging arguments, you can derive a PDE that determines the option's value; letting $C_t$ be the price of the call option and $\Phi(S_T)=\max(S_T-K,0)$ its payoff, this PDE is:

$$ \frac{\partial C_t}{\partial t} + \frac{\partial C_t}{\partial S_t}rS_t +\frac{1}{2}\frac{\partial^2C_t}{\partial S_t^2}\sigma^2S_t^2-rC_t = 0 $$

Anyway, the main point here is that we can apply to that PDE equation the Feynman-Kac Theorem which states its solution, i.e. the function $C_t$ that follows those dynamics with terminal condition $C_T=\Phi(S_T)$, can be computed as the following expectation $-$ under a probability measure $\mathbb{Q}$ where the drift of the stock price is $r$:

$$ C_0 = \mathbb{E}^{\mathbb{Q}}\left[e^{-\int_0^Trdt}\Phi(S_T)|S_0\right] = e^{-rT}\mathbb{E}^{\mathbb{Q}}\left[\max(S_T-K,0)|S_0\right] $$

The fact that pricing a derivative contract is equivalent to computing the expectation of its payoff under a probability measure $\mathbb{Q}$ where the stock's price drift is $r$ instead of $\mu$ is little more than a mathematical trick resulting from the Feynman-Kac Theorem.

$\endgroup$
  • $\begingroup$ Very nice answer. $\endgroup$ – Quantuple Jul 10 '17 at 7:23
3
$\begingroup$

You make a crucial mistake when you assume that the price of the option is $$C=e^{-rT}E^{\mathbb{P}}[(S_T-K)^+]$$ that is, it is not true that the price of an asset is the discounted (using the risk-free rate) payoff under the physical measure $\mathbb{P}$. The following two statements, on the other hand, are true: $$C=E^{\mathbb{P}}\left[\frac{\xi_T}{\xi_0}(S_T-K)^+\right]$$ $$C=e^{-rT}E^{\mathbb{Q}}[(S_T-K)^+]$$ where $\xi_t$ is the state price density, i.e. the stochastic process $\xi_t$ under which the product with any asset $\xi_t S_{it}$ is a $\mathbb{P}$-martingale, or $\mathbb{Q}$ is the risk-neutral measure under which any asset has drift $r$, i.e. the any asset follows the following SDE $$\frac{dS_{it}}{S_{it}}=rdt+\sigma dZ^\mathbb{Q}$$.

In my opinion, the easiest way to show that Black-Scholes follows, is to solve the same integral noticing that under $\mathbb{Q}$ the stock price is $S_T=S_0 e^{(r-\frac{1}{2}\sigma^2)T+\sigma \sqrt{T}Z}$. Alternatively, you can show that the SPD follows $\frac{d\xi}{\xi}=-rdt-\frac{\mu-r}{\sigma}dZ^{\mathbb{P}}$, use Ito's lemma on $f(t,S,\xi)=S\xi$ and obtain the same result again.

$\endgroup$
1
$\begingroup$

The answers above address the issue of which probability measure should be used when calculating the expected value of a call option under Black-Scholes assumptions. For perspective, you should also review a text like Musiela and Rutkowski.

If you would like to see the detailed calculation, I have worked through it using Mathematica. See ntgladd.com, tab = Finance, section = Black-Scholes Formalism notebook = 17-9 Derivation of Black-Scholes formula by calculating an expectation. The same formula is derived from the Black-Scholes PDE in 17-10 Solving BS PDE for call option.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.