6.3. PROPERTIES OF GEOMETRIC BROWNIAN MOTION 209
Figure 6.1: The p.d.f. for a lognormal random variableNow differentiating with respect tox, we obtain that
fX(x) =1
√
2 πσx√
texp((− 1 /2)[(ln(x)−ln(z 0 )−μt)/(σ√
t)]^2 ).Calculation of the Mean
We can calculate the mean of Geometric Brownian Motion by using the m.g.f.
for the normal distribution.
Theorem 4 .E[z 0 exp(μt+σW(t))] =z 0 exp(μt+ (1/2)σ^2 t)
Proof.
E[X(t)] =E[z 0 exp(μt+σW(t))]
=z 0 exp(μt)E[exp(σW(t))]
=z 0 exp(μt)E[exp(σW(t)u)]|u=1
=z 0 exp(μt) exp(σ^2 tu^2 /2)|u=1
=z 0 exp(μt+ (1/2)σ^2 t)sinceσW(t)∼N(0,σ^2 t) andE[exp(Y u)] = exp(σ^2 tu^2 /2) whenY∼N(0,σ^2 t).
See Moment Generating Functions, Theorem 4.