Why Geometric Brownian Motion Gives the Same Terminal Price Across Time Steps

# Simulating stock prices with and without intermediate paths


# Simulating stock prices with and without intermediate paths












So I am simulating stock prices with what I believe to be geometric Brownian motion using parameters from the usual Black-Scholes framework (Please correct me if I am wrong) with the following formula:

$$S_{t} = S_{0}e^{(r-\delta -\frac{1}{2}\sigma^{2})t +z\sigma \sqrt{t}} ,$$

where St is the stock price at time t, r is the risk-free rate, delta is the dividend rate, sigma is the volatility, and z is a draw from the standard normal distribution.

However, when I simulate the stock prices one year from now by plugging in t=1 vs plugging in t=1/12 (and simulate 12 successive runs), I get drastically different ending prices.

The simulated stock prices from the single step (t=1) has much higher variation than stock prices simulated from the 12 time step versions.

I am wondering if I am missing something from this equation.

`A somewhat related question-------maybe too simple to start a new topic------- is the following:`

I remembered back in school that when simulating stock prices, one should use alpha---the real rate of return, as opposed to the risk free rate in the equation (using r in the simulation equation implying we're in the risk-neutral world?). (http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?t=216817)

However, when I use this equation to simulate stock prices I am able to get option prices very close to the B-S-M theoretical prices.

So my question is why can't we simulate stock prices with alpha and discount at some other rate to price the same option? (Is it because alpha is unknown, or the other discount rate is unknown?).

Thanks for reading through!

## Answer by ZRH (score 1, accepted)

https://quant.stackexchange.com/a/44631

Variance should be precisely the same, for the following reason: Imagine you partition your time interval $t$ into $n$ instalments of $t/n$ each.

So basically: $S_{t/n}=S_0e^{(r-\delta-0.5\sigma^2)t/n+z\sigma\sqrt{t/n}}$

In order to arrive at the result you care about ($S_t$), you go from $S_0$ to $S_{t/n}$ to $S_{2t/n}$ etc until you have arrived at $S_t$. Successive multiplication during simulation results in the following:

$S_t=S_0\Pi_{i=1}^{n} e^{(r-\delta-0.5\sigma^2)t/n} e^{z\sigma\sqrt{t/n}}=S_0e^{(r-\delta-0.5\sigma^2)t}\Pi_{i=1}^n e^{z\sigma\sqrt{t/n}}=S_0e^{(r-\delta-0.5\sigma^2)t+z\sigma\sqrt{t}}$

Note that the last step requires $\Pi_{i=1}^n e^{z\sigma\sqrt{t/n}}=e^{\sqrt{n}z\sigma\sqrt{t/n}}=e^{z\sigma\sqrt{t}}$, as the sum of $n$ $(0,1)$-distributed random numbers is distributed as $(0,\sqrt{n})$