I'm teaching an applied math class this summer and I want to take a short detour into finance (not my specialty at all); specifically the Black-Scholes model of stock movements. I want my students to be able to simulate stock movements and create some fun graphs. But I still have a couple basic questions about it myself.
If I have this right, the change $\Delta S$ in a stock price over a small time interval $\Delta t$ is posited to behave 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). From this we can compute the following expected values:
$\ln(E[S_T]) = \ln(S_0) + \mu T$
$E[\ln(S_T)] = \ln(S_0) + (\mu - \sigma^2/2) T$
First, a dumb question: if volatility isn't a concern, it's the second of these quantities that an investor is concerned with when deciding to purchase a stock, yes? i.e., the expected value of logarithmic returns is the correct measure of the performance of a stock, not log of the expected value?
My second question maybe isn't as dumb. Assuming what I just said is correct, does this imply that, in a world where this model held perfectly and that $\mu$ and $\sigma$ were known for all stocks and to all investors, that $\mu - \sigma^2/2 = r$, the risk-free rate? It would seem to, since a riskless bond with $\mu = r$ and $\sigma = 0$ is always available to investors. I'm asking because I'm curious and I'd like to say something intelligent to my students about the relationship between $\mu$ and $\sigma$ in this model, e.g higher $\sigma$ means higher $\mu$.
Another question. I'm told there's a magic wand called risk-neutral valuation which allows me instead to write
$\Delta S = r S \Delta t + \sigma \sqrt{\Delta t} \varepsilon^\star S$
where we've replaced $\mu$ with the risk-free rate and $\varepsilon^\star$ is a different random variable derived from the risk-neutral probability measure. I'll buy that for the moment. What confuses me is how, when deriving the Black-Scholes formula for European options, one arrives at the correct formula, even though we've replaced $\mu$ with $r$ but not replaced $\varepsilon$ with $\varepsilon^\star$.
What I mean is, suppose you write
$S_T = S_0 e^{(r-\sigma^2/2)T}e^{\sigma \sqrt{T} Z}$
That is, replacing $\mu$ with $r$ but not changing $\sigma$ to $0$ or changing anything in the second exponential (i.e. changing it to a different version of browning motion) If you use this expression to compute the discounted expected value of the payout of a European call option at strike price $K$
$e^{-rT}E[\text{max}(S_T - K,0)]$
one arrives at the correct formula for $C$, the Black-Scholes price of a European call option. But why should this be? Why should the fair value of such an option be arrived at by assuming $\mu = r$, but still assuming $\sigma \neq 0$? I realize that letting $\sigma = 0$ makes an option pointless to begin with, but I really don't get why we are justified in letting $\mu = r$, instead of what $\mu$ actually is (whatever it may be).
Finally, anything intelligent you can tell me about how actual investors react to (their estimation of) $\mu$ and $\sigma$, within the context of this model, would be helpful.
Sorry for my naivety, the closest I've been to a stock market is flipping past CNBC on my couch. Thanks for any help.
UPDATE:
What you've both said makes good sense to me. A quick aside: is the Riesz Representation theorem the essential ingredient one uses to prove the existence of risk-neutral measure?
I'm still fuzzy on one thing though. I've not been through the Black-Scholes PDE/dynamic hedging argument in detail but I get the gist; setting up a self-financing risk-less portfolio by trading back and forth the derivative and the stock. And I'm sure that's the most conceptually sound and insightful way to derive the Black-Scholes formula for European options. But I'm not going to have time to go into this in class, so let's suppose instead we didn't know any of this Black-Scholes PDE stuff, nor the Feynman-Kac formula. Again assuming the model
$S_T = S_0 e^{(\mu - \sigma^2/2)T} e^{\sigma \sqrt{T} Z}$
is there a way to argue from simpler principles that the computation
$e^{-rt} E[\text{max}(S_T-K,0)]$
is a valid pricing for a European option, after replacing $\mu$ with $r$? Because honestly, if I'm actually out there selling these options and need to price them, and I have no education about any of this other than this model, and I know I can sell enough of them for the law of averages to win out, I'm doing this exact computation but leaving $\mu$ right where it is (this was my original guess as to how to price an option by the way, before learning the actual formula or the hedging argument). In fact, it seems to me that, whatever probabilistic model $S_T$ you believe in for a particular stock, this computation should lead you to your best guess as to the value of the option, without any alterations such as letting $\mu = r$. Where am I going wrong here?
Finally, could you recommend some realistic values of $\mu$ and $\sigma$ for me to play around with my students? Do practical traders actually bother trying to estimate $\mu$ and $\sigma$?
Thanks for both of your help.
Answer
Let's take your questions in turn -
If volatility isn't a concern, an investor is concerned with $$ E[\log S_T] = \log S_0 + (\mu - \tfrac{1}{2}\sigma^2)T $$ when deciding to purchase a stock, yes? i.e. the expected value of logarithmic returns is the correct measure of the performance of a stock, not log of the expected value?
Yes, for long-term performance you should focus on the log-returns, as these incorporate the volatility drag that penalizes very volatile stocks.
Assuming what I just said is correct, does this imply that, in a world where this model held perfectly and that $\mu$ and $\sigma$ were known for all stocks and to all investors, that $\mu-\tfrac{1}{2}\sigma^2=r$, the risk-free rate?
In a world where the model held perfectly, and $\mu$ and $\sigma$ were known to all investors, and investors are risk-neutral, then this equation would hold. More generally you could have $$ \mu = r + \tfrac{1}{2}\lambda \sigma^2 $$ where $\lambda$ is the market price of risk. If $\lambda>1$ then investors are risk-averse i.e. they demand to be compensated with additional return for bearing risk. If $\lambda<1$ then investors are risk-seeking. A more sophisticated model is the capital asset pricing model (CAPM) which incorporates covariance with the market as an additional source of risk. Note that the CAPM is itself seen as a very simplistic and naive model of asset returns.
Why should the fair value of an option be arrived at by assuming $\mu=r$ but still assuming $\sigma\neq 0$?
The argument that leads to the Black-Scholes option pricing formula is a dynamic hedging argument. By following a particular trading strategy (i.e. buying and selling the stock in specified amounts) the investor can replicate the payout of a call or put option. Therefore, the price of the option must be equal to the cost of implementing this trading strategy (following the law of one price).
To determine the cost of implementing the trading strategy (in the limit of continuous pricing) you set up a PDE, the Black-Scholes equation,
$$ V_t + rSV_S + \tfrac{1}{2}\sigma^2S^2V_{SS} = rV $$
whose solution gives the price of the option. Note that $\mu$ does not appear in this equation (it drops out due to dynamic hedging) so the price of the option cannot depend on $\mu$.
Finally, the link to expectations (and the reason you can solve for the price of the option using Monte Carlo) is provided by the Feynman-Kac formula which translates the solutions of a certain class of parabolic PDEs (of which the Black-Scholes equation is one) to expectations of a particular stochastic process. Since the Black-Scholes equation does not depend on $\mu$, we know that the stochastic process cannot depend on $\mu$ either, and in fact the stochastic process turns out to be
$$ \frac{dS_t}{S_t} = r dt + \sigma dZ_t $$
with a terminal condition specified by the payout of the option (we use a terminal condition rather than an initial condition because the Black-Scholes equation has time reversed compared to a traditional, heat-like parabolic PDE).
That's the mathematical answer. The intuitive answer is "we can hedge out all of the price movements related to the drift of the stock, therefore the price of the option can't depend on the drift".
That is, even if you had two perfectly correlated stocks with a different drift term, and you bought a call option on each of them, the options would have a different terminal value (the one with higher drift would be worth more) but the costs of dynamically hedging the stock movements for the stock with greater drift would also be higher, by an amount which exactly offsets the difference in option terminal values, so the options have the same value now.
No comments:
Post a Comment