Expected Growth

The model assumption of the Black-Scholes formula has two parameters for the geometric Brownian motion, the volatility $\sigma$ and the expected growth $\mu$ (which disappears in the option formulae). How can this parameter $\mu$ be estimated?

-

4 Answers

As you said, $\mu$ is the expected return that is the expected value (mathematical expectation) of the random variable "stock return" under the objective probability measure. Assuming that returns are stationary*, the obvious way to estimate it is to compute a large number $N$ of returns $R_i$, then to average them. You also want to annualize this average (multyply by 252).

Now if you are using consecutive periods, and logarithmic returns, this simply amounts to computing the overall return $\log S_{T_N} / S_{T_0}$ and dividing it by the time lapse $T_N-T_0$ (in year fraction).

(*) : This is a dubious hypothesis, and the estimate will indeed be very unstable.

-

In the risk-free measure, $\mu$ is equal to the risk-free rate. In the real-world measure, $\mu$ is uknown and must be estimated using statistical methods: either directly from historical data, or from a model calibrated to historical data. Note that, in general, there is nothing preventing you from using today's (or past) prices as inputs to this model. For example, no-arbitrage relationships ought to hold in both "worlds".

-

The drift term, $\mu$ is assumed to follow from the 'no-arbitrage' assumption. That is, if $\mu$ were greater than the risk-free rate, one would borrow at the risk-free rate, invest in the stock, and collect the difference. If the stock may be freely borrowed, and $\mu$ is less than the risk free rate, one would short the stock and invest in the risk-free rate, and collect the difference.

-
I understand that it becomes the risk free rate after an appropriate change of measure. What I am asking for is whether I can estimate the non-risk free value? – John Smith Feb 8 '11 at 4:20
Plus, what shabbychef describes is a bona fide arbitrage opportunity only if the stock has a deterministic rate of return, equal to $\mu$. As soon as the future value of the stock is random, the probability of a negative terminal value is nonzero. What the reasoning does prove is that the forward price of the stock does not depend on $\mu$. – egoroff Feb 8 '11 at 17:01

I take it that μ is the drift of the long-term equilibrium price.

let's take a lognormal model as an example,

dS = μ x dt + σ x S x dz

where:

S= spot, t = time, T-t = length of time μ = drift rate, σ = volatility, dz= random variable,

In order to solve for μ, you might first want to look for the expected spot price:

given X = ln(S),

dX = ((μ - σ)dt)/2 + σdz

This allows us to solve for X,

ST = St e ^((μσ^2/2)(T-t) -σdz)

Taking the expected value of both Sides:

E[ST] = S e ^(μ(T-t))

This equation can be used to look for your expected growth, μ

-