In the previous blog I have told little bit about binomial asset pricing model and the general idea. We are now going to find the limiting distribution of the binomial asset pricing. We are going to prove this argument with some predefined assumptions: the risk free interest rate, r = 0.
We will show that binomial model is a discrete time version of geometric Brownian Motion model, which we will learn later is the ground work of Black-Scholes-Merton option pricing formula.
Let us build a model for a stock price on the time interval from 0 to t by choosing an integer n and constructing a binomial model for the stock price that takes n steps per unit time. We are choosing n and t such that nt is an integer.
\[\begin{aligned} &\text{Up factor}:= u_n = 1+\frac{\sigma}{\sqrt{n}} \\ &\text{Down factor}:= d_n = 1-\frac{\sigma}{\sqrt{n}} \\ \end{aligned}\]
Here, \(\sigma\) is a positive constant.
So, in this case the risk neutral probabilities are- \[\begin{aligned} \tilde {p} &= \frac{1+r-d_n}{u_n-d_n} \\ &= \frac{1-d_n}{u_n-d_n} \qquad \text{[as r = 0]} \\ &= \frac{\sigma/\sqrt{n}}{2\sigma/\sqrt{n}} \\ &= \frac{1}{2} \\ \end{aligned}\]
Similarly we have- \(\tilde{q}= 1-\tilde{p}= \frac{1}{2}\).
The price of the stock at time t is determined by the initial stock price, i.e. \(S_0\) or \(S(0)\) and nt, number of tosses. Let us first define few notations.
So we can write the following two equations.
\[\begin{aligned} H_{nt}+ T_{nt} &= nt \\ H_{nt}- T_{nt} &= M_{nt} \\ \end{aligned}\]
From these we have- \[H_{nt} = \frac{1}{2}(nt+M_{nt}) \qquad \text{and} \qquad T_{nt} = \frac{1}{2}(nt-M_{nt}) \]
So, the stock price at time t is- \[\begin{aligned} S_n(t) = S(0)u_n^{H_{nt}}d_n^{T_{nt}} &= S(0)(1+\frac{\sigma}{\sqrt{n}})^{H_{nt}}(1-\frac{\sigma}{\sqrt{n}})^{T_{nt}} \\ &= S(0)(1+\frac{\sigma}{\sqrt{n}})^{\frac{1}{2}(nt+M_{nt})}(1-\frac{\sigma}{\sqrt{n}})^{\frac{1}{2}(nt-M_{nt})} \end{aligned}\]
We will find the distribution of of \(S_n(t)\) as \(n \rightarrow \infty\).
Theorem: As \(\text{n} \rightarrow \infty\) the distribution of \(S_n(t)\) stated before converges to the distribution of- \[S(t) = S(0).\exp\{\sigma. W(t)-\frac{1}{2}\sigma^2t\}\] where, W(t) is a normal random variable with mean = 0 and variance = t.
The distribution of S(t) is called log-normal. Suppose random variable X follows normal distribution, then \(Y = c.e^X\) follows log-normal distribution. To know more go to- https://en.wikipedia.org/wiki/Log-normal_distribution.
We are going to prove as \(\text{n} \rightarrow \infty\) distribution of \(S_n(t)\) converges to the distribution of \(S(t)\). It is enough to show- \[\lim_{n \rightarrow \infty} \log(S_n(t)) = \log(S(t))\]
Previously, we have assumed that- \[\begin{aligned} &S_n(t) = S(0)(1+\frac{\sigma}{\sqrt{n}})^{\frac{1}{2}(nt+M_{nt})}(1-\frac{\sigma}{\sqrt{n}})^{\frac{1}{2}(nt-M_{nt})} \\ \implies &\log(S_n(t)) = \log(S(0))+ {\frac{1}{2}(nt+M_{nt})}\log(1+\frac{\sigma}{\sqrt{n}}) + {\frac{1}{2}(nt-M_{nt})}\log(1-\frac{\sigma}{\sqrt{n}}) \\ \end{aligned}\]
Now- \[\log(S(t)) = \log(S(0))+ \sigma W(t) -\frac{1}{2}\sigma^2t\] Where- \(W(t) \sim N(0,t)\).
To do this, we need the Taylor series expansion of \(f(x) = \log(1+x)\). And- \(f'(x) = \frac{1}{1+x} \quad \text{and} \quad f''(x) = -\frac{1}{(1+x)^2}\). From this we have- \(f'(0) = 1 \quad \text{and} \quad f''(0) = -1\). So by Taylor’s Theorem we have- \[\begin{aligned} \log(1+x) &= f(0)+f'(0)x+ \frac{1}{2}f''(0)x^2+O(x^3) \\ &= x-\frac{1}{2}x^2+O(x^3) \end{aligned}\] where \(O(x^3)\) is terms of order \(x^3\). Now we are going to apply this on \(\log(S_n(t))\) expression. Taking \(x=\frac{\sigma}{\sqrt{n}}\) and \(x=-\frac{\sigma}{\sqrt{n}}\), we get-
\[\begin{aligned} \log(S_n(t)) &= \log(S(0)) + \frac{1}{2}(nt+M_{nt})(\frac{\sigma}{\sqrt{n}}-\frac{\sigma^2}{2n}+O(n^{-3/2})) + \frac{1}{2}(nt-M_{nt})(-\frac{\sigma}{\sqrt{n}}-\frac{\sigma^2}{2n}+O(n^{-3/2})) \\ &= \log(S(0)) + nt(-\frac{\sigma^2}{2n}+O(n^{-3/2}))+ M_{nt}(\frac{\sigma}{\sqrt{n}}+O(n^{-3/2})) \\ &= \log(S(0)) + -\frac{1}{2}\sigma^2t+ O(n^{-1/2}))+ \frac{\sigma}{\sqrt{n}}M_{nt}+ \frac{1}{\sqrt{n}}M_{nt}.O(n^{-1}) \\ \end{aligned}\]
Previously in Scaled symmetric random walk, we have assumed \(W^{(n)}(t) = \frac{1}{\sqrt{n}}M_{nt}\); and limiting distribution of \(W^{(n)}(t)\) is \(N(0,t)\). So the above expression becomes-
\[\begin{aligned} \log(S_n(t)) &= \log(S(0)) + -\frac{1}{2}\sigma^2t+ O(n^{-1/2})+ \frac{\sigma}{\sqrt{n}}M_{nt}+ \frac{1}{\sqrt{n}}M_{nt}.O(n^{-1}) \\ &= \log(S(0)) + -\frac{1}{2}\sigma^2t+ O(n^{-1/2})+ \sigma.W^{(n)}(t)+ W^{(n)}(t).O(n^{-1}) \\ \end{aligned}\]
As \(n \rightarrow \infty\), \[\begin{aligned} \lim_{n \rightarrow \infty} \log(S_n(t)) \rightarrow \log(S(0)) -\frac{1}{2}\sigma^2t+ \sigma.W(t) \\ \end{aligned}\]
Now, \(\sigma.W(t)-\frac{1}{2}\sigma^2t \sim N(-\frac{1}{2}\sigma^2t, \sigma^2t)\). Implies,
\[\lim_{n \rightarrow \infty} S_n(t) = S(0). \exp\{\sigma.W(t)-\frac{1}{2}\sigma^2t \} \qquad \text{Q.E.D}\]
Next, we are going to mathematical definition of Brownian Motion, it’s distribution characteristic etc. Have fun reading!!