The process ${W(t)}{t\geq0}$ is said to be (standard) Brownian Motion if the following are satisfied:
$W(0)=0$
For $s,t\geq0$ the random variable $W(s+t)-W(s) \sim N(0,t)$
Whenever $0\leq t_0\leq t_1\leq....<t_n$ , the quantities $W(t_1)-W(t_0),W(t_2)-W(t_1),....,W(t_n)-W(t{n-1})$ are independent
$W(t)$ is a continuous function of $t$ with probability $1$
Deriving Ito's Lemma (Time Independent)
Let us suppose that the asset price $S$ satisifies the Stochastic Differential Equation (SDE)
$$ dS = \mu dt + \sigma dW $$
where $\mu (t)$ and $\sigma(t)$ depends on the time-interval we look at (say 10D period) and the $W(s)$ for $s\leq t$ is a Brownian motion i.e the random fluctuation in a stock-price $S$
Now consider a function $f(S(t),t)$ of asset price where $f$ has a continuoous second derivative (i.e $f\in C^2 (0,T)$). For simplicity, let us assume that $f$ is independent of time i.e $f=f(S)$. Then by Taylor's Theorem:
Since $W(t)$ is a Brownian Motion $\implies$$dW=W(t+dt) - W(t)$ is a Brownian Motion with $dW \sim N(0,dt) \ \implies E[(dW)^2]= Var(dW)+0 \ \implies =dt$
So in the limit and replacing $(dW)^2$ by $dt$ , we get:
Let $f(S)$ be a continuous twice differentiable and suppose that:
$$ dS = \mu dt + \sigma dW$$
Then:
$$df = \frac{df}{dS}(\mu dt+\sigma dW) + \frac{1}{2}\frac{d^2f}{dS^2}\sigma^2dt$$
When written out in the Integral form:
$$ f(S(T)) - f(S(0)) = \int_{0}^{T} \left(\mu dt+\sigma dW\right) dt + \int_{0}^{T} \sigma\frac{df}{dS} dW$$
Thus we get a relationship between a Stochastic Integral and a Standard Integral with respect to time.
A model for stock price
Consider an asset with price $S(t)$ that evolves according to the SDE
$$ dS = \mu Sdt + \sigma SdW$$
Over a period $dt$, the price changes by a deterministic quantity$\mu Sdt$ (representing some underlying deterministic growth) and a random quantity$\sigma SdW$ (where $\sigma$ measures the volatility of the asset).
It is useful to work in terms of $log(S(t))$ so we define:
$$ f(S) = log(S) \ f'(S) = \frac{1}{S} \ f''(S) = -\frac{1}{S^2} $$
So after plugging the following in the SDE we get:
$$ log\left(\frac{S(T)}{S(0)}\right) \sim N\left((\mu - \frac{1}{2}\sigma^2)T,\sigma^2T\right) $$
The above equation can be generalised to give the following:
So we say that $Y = \frac{S(t)}{S(0)}$ is log-normally distributed with:
$$ E[Y] = exp\left[{\eta+\frac{1}{2}\sigma^2}t\right] \ $$
where $\eta = (\mu - \frac{1}{2}\sigma^2)t$
where $\mu$ is 1+log-return's mean and $\sigma$ is 1+log-return's standard deviation
In my algorithm:
$$ \Delta t = 1 \quad \text{since I have daily data} \
S_{t+1} = S_t*exp\left[\left(\mu - \frac{1}{2}\sigma^2\right) + \sigma W_t\right] $$
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