The continuously compounded rate of return for a risky asset was first discussed in Section 5.5. We assumed there that over a period , follows a normal distribution with an expected value of and a standard deviation of , for some constants and . In fact, the assumption of a normal distribution is redundant.
The assumption
Footnote11 that over periods of equal length, the rates of return are i.i.d. random variables already implies that
follows the normal distribution. Dividing a given time period into
equal periods so that the rate of return over the original period becomes the sum of the returns over these small periods produces this conclusion. Letting
approach infinity and applying to the Central Limit Theorem provides the required result. We have already stipulated a way of choosing the parameters in the binomial model so that in the limit
follows the normal distribution. Hence, the price process,
follows the lognormal distribution.
Equations (14.6) and (14.7) present a sufficient condition under which the limiting distribution produces an expected value of
and a standard deviation of
. As we can see in the MAPLE commands below, the general solution,
,
, and
, to these equations does not uniquely determine the value of the parameters.
> u:='u':v:='v':N:='N':qu:='qu':h:='h':Ex:='Ex':Var:='Var':
> solve({mu*T=N*(qu*u+(1-qu)*v),sigma^2*T=\
> N*((qu-qu^2)*u^2-2*(1-qu)*v*qu*u+(1-qu)*v^2-(1-qu)^2*v^2)},{u,v,qu});
Moreover, these conditions are too strong. There is no need for the equations to be satisfied for each . Rather they should be satisfied only in the limit as approaches infinity. A solution, for which equation (14.7) is satisfied only in the limit is presented below. It makes it easier to motivate some properties of the Brownian motion.
Consider a solution Footnote12 where and , i.e.,
> u:=sigma*sqrt(T/N);
> v:=-u;
Note that a property of this solution is that the absolute value of the change in the continuously compounded rate of return is with certainty, and that it depends on the time interval via the square root of its length. These proprieties highlight some of the properties of the Brownian motion as we shall soon see. The expected value of is given by
> Ex:=simplify(N*(qu*u+(1-qu)*v));
and its variance is given by
> Var:=collect(N*((qu-qu^2)*u^2-2*(1-qu)*v*qu*u+(1-qu)*v^2-(1-qu)^2*v^2),qu);
Let us assume that is given by equation (14.47),
,
(14.47)
and assign its value so we can calculate the induced expected value (denoted ) and variance (denoted ) of . We can thereby check if equations (14.6) and (14.7) are satisfied.
> qu:=(1+mu*sqrt(T/N)/sigma)/2;
> simplify(Ex);
> expand(Var);
We therefore see that equation (14.6) is satisfied but equation (14.7) is not. However, as goes to infinity and the length of the time interval approaches zero, indeed the variance approaches as desired. This is demonstrated in the MAPLE command below.
> limit(Var,N=infinity);
In both solutions, the one presented here and the one presented in Section 14.1.1, the possible realization of the change in , , and , over a period of length , depends on . It ensures that the variance over the interval , as approaches infinity, depends on and not on . It therefore explains why Assumption II in Section 5.4 stipulates that depends, via its standard deviation on . Furthermore, with the specifications as in equation (14.47), the absolute value of the change in over a period, in the binomial model, is with certainty since . Hence the expected value of the absolute change over a period in the binomial model is . Consequently, the total expected change of over is , which approaches infinity as goes to infinity.
> assume(T>0,sigma>0);
> limit(N*sigma*sqrt(T/N),N=infinity);
This explains why the graph of a realization of a Brownian motion versus "travels" an infinite distance over a finite time Footnote13 interval and therefore looks very jagged. Furthermore, the absolute value of the rate of change in over a period in the binomial model is given by , which as goes to infinity, as shown below, explodes.
> limit(sigma*sqrt(T/N)/(T/N),N=infinity);
Hence it explains why the (calculus) derivative of the function with respect to does not exist.
Readers who opt to skip the next Chapter may wish to execute the procedure Simudif below now. It is a simulation of the realization of a Brownian motion as shown in Figure 14.7. The parameters for this procedure are given below:
, the current value of ; 0.20 in the example below
and the left and right end points of the time interval, respectively; 0 and 1 in our example
N, , and are as defined above, which are set to 1000, 0.20, and 0.18, respectively, in the example
The last parameter "plot", if omitted, causes the procedure to produce an animation instead of a static graph.
> Simudif(.20,0,1,1000,.20,.18,"plot");
Figure 14.7: A Realization of a Brownian Motion, Y(t) versus t
The linear graph in Figure 14.7 is of the expected value,
, while the jagged graph is a realization of the path taken by the Brownian motion
Footnote 14. by the Brownian motion. The reader may want to change the parameter values and rerun the procedure to gain a better appreciation of the process. Figure 14.7 makes visual the reasons why the tools used in deterministic calculus are not able to handle such circumstances. The next Chapter introduces the reader to methods by which the Black-Scholes formula can be derived in this stochastic environment.
Footnotes
See the assumption at the end of Section 5.5 and equation (5.30) and (5.31).
The solution presented here follows [20]. The reader is asked, in the exercises, to utilize the solution presented in Section 14.1.1 in order to motivate the properties of the Brownian motion that are discussed in the rest of this section.
See our discussion of total variation in the next Chapter.
This animation is produced by sampling a normal random variable, instead of a binomial variable, at the beginning of each period, the length of which is . A very similar graph would have been produced if the sampling were done with a binomial variable adopting one of the solutions presented in this Chapter.