How Do You Calculate the Expected Value of Geometric Brownian Motion?

[motherh]

Hi, I am trying to answer the following question:

Consider a geometric Brownian motion S(t) with S(0) = S_0 and parameters μ and σ^2. Write down an approximation of S(t) in terms of a product of random variables. By taking the limit of the expectation of these compute the expectation of S(t).

Let Z_i = σ√Δ (with probability 1/2(1+ \frac{μ}{σ}√Δ)) or -σ√Δ (with probability 1/2(1- \frac{μ}{σ}√Δ)).

Then logS(t) ≈ Z_1 + ... + Z_(\frac{t}{Δ}) (as logS(t) is Brownian motion)

And this approximation gets better as \frac{t}{Δ} tends to infinity

S(t) ≈ exp(Z_1) * ... * exp(Z_(\frac{t}{Δ}))

So now exp(Z_i) = exp(σ√Δ) (with probability 1/2(1+ \frac{μ}{σ}√Δ)) or exp(-σ√Δ) (with probability 1/2(1- \frac{μ}{σ}√Δ)).

E[S(t)] ≈ (E[exp(Z_1)])^\frac{t}{Δ}

I know that the answer I am looking for is S_0*exp(tμ+\frac{tσ^2}{2}). Also

E[exp(Z_1)] =1/2exp(σ√Δ)*(1+ \frac{μ}{σ}√Δ) -1/2exp(-σ√Δ)*(1- \frac{μ}{σ}√Δ).

So I believe that

E[S(t)] = lim [ 1/2exp(σ√Δ)*(1+ \frac{μ}{σ}√Δ) -1/2exp(-σ√Δ)*(1- \frac{μ}{σ}√Δ) ]^\frac{t}{Δ} (as \frac{t}{Δ} tends to infinity)

Where do I go from here? I want to make use of lim (1+x/t)^t = exp(x) but how do I go from here? Help is MUCH appreciated!

 

[Ray Vickson]

Hi, I am trying to answer the following question:

Consider a geometric Brownian motion S(t) with S(0) = S_0 and parameters μ and σ^2. Write down an approximation of S(t) in terms of a product of random variables. By taking the limit of the expectation of these compute the expectation of S(t).

Let Z_i = σ√Δ (with probability 1/2(1+ \frac{μ}{σ}√Δ)) or -σ√Δ (with probability 1/2(1- \frac{μ}{σ}√Δ)).

Then logS(t) ≈ Z_1 + ... + Z_(\frac{t}{Δ}) (as logS(t) is Brownian motion)

And this approximation gets better as \frac{t}{Δ} tends to infinity

S(t) ≈ exp(Z_1) * ... * exp(Z_(\frac{t}{Δ}))

So now exp(Z_i) = exp(σ√Δ) (with probability 1/2(1+ \frac{μ}{σ}√Δ)) or exp(-σ√Δ) (with probability 1/2(1- \frac{μ}{σ}√Δ)).

E[S(t)] ≈ (E[exp(Z_1)])^\frac{t}{Δ}

I know that the answer I am looking for is S_0*exp(tμ+\frac{tσ^2}{2}). Also

E[exp(Z_1)] =1/2exp(σ√Δ)*(1+ \frac{μ}{σ}√Δ) -1/2exp(-σ√Δ)*(1- \frac{μ}{σ}√Δ).

So I believe that

E[S(t)] = lim [ 1/2exp(σ√Δ)*(1+ \frac{μ}{σ}√Δ) -1/2exp(-σ√Δ)*(1- \frac{μ}{σ}√Δ) ]^\frac{t}{Δ} (as \frac{t}{Δ} tends to infinity)

Where do I go from here? I want to make use of lim (1+x/t)^t = exp(x) but how do I go from here? Help is MUCH appreciated!

Use standard probability results to figure out what is the distribution of $\sum_i Z_i$ in the limit as $n \to \infty, \Delta \to 0$ with $t = n \Delta$ fixed.