Consider a non-dividend paying stock subject to risk-free force of interest

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Consider a non-dividend paying stock subject to risk-free force of interest, r = 0.045 as well as volatility, σ = 0.24 and time horizon three months, i.e., T = 1/4 for a stock with current price S0 = $45. Recall that dSt = St(rdt + σdBt) where {Bt}t≥0 is standard Brownian motion. Furthermore, recall that Brownian motion is Gaussian with mean 0 and variance t. That is, Bt ∼ N (0, t) for each t fixed, 0 ≤ t ≤ T. It follows that we can exploit this normality, thus leading to the discretisation Sti − Sti−1 = Sti−1(r∆t + σ√∆t) for ∆t = ti − ti−1 where we assume that is a random draw from standard Gaussian distribution, i.e. ∼ N (0, 1) (4)

Given our time horizon of 3 months, we would like to simulate prices for ∆t = 0.005, we simulate iteratively. In other words, after the first time period, we get that St1 − S0 = S0(r∆t + σ√ ∆t)

Where we repeat until the final time step, T = 0.25.

Your task is to simulate 100 such sample paths.

Attempt at Question:

So I understand that we must use the formula St1 − S0 = S0(r∆t + σ√ ∆t) for ∆t=0.005 so i created an excel spreadsheet and subbed in the values and went up by 0.005. I have attached an image of my excel spreedsheetenter image description heret.

but I only got 50 samples instead of 100. Im a little confused with the random draw from the standard gaussian distribution as that is multiplied by the volatility in the formula. Does that alternate between 0 and 1 and i have to do it for both 0 and 1 to get two seperate data sets that will add up to 100. Thanks for any help.

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You are asked to produce a time series of simulated prices at times from $T=0$ years to $T=0.25$ years at intervals of $\Delta T = 0.005$. So each time series will contain $51$ values (if we count the values at both ends).

You are then asked to repeat this $100$ times, so you get $100$ time series with $51$ values in each one.

A random variable that follows a standard Gaussian distribution $N(0,1)$ can take any real value, but the distribution of its values will have a mean of $0$ and a variance of $1$.