In the previous blog, we have seen how to construct Ito integral for general integrands and it’s properties. Now, we are going to use these properties to find some interesting results.
Suppose we want to compute-
\[\mathbb{I(T) = \int_{0}^{T}W(t).dW(t)}\] In this Ito integral, we have \(\mathbb{\Delta(t) = W(t)}\). So according to the definition of Ito integral for general integrands, we need to find- a sequence \(\mathbb{\{\Delta_{n}(t)\}}\) of simple processes, such that as- \(\mathbb{n} \rightarrow \infty\) these processes converge to continuously varying \(\mathbb{\Delta(t)}\).
\[\begin{equation} \mathbb{\Delta_n(t)} = \begin{cases} \mathbb{W(0)} & \text{if} \quad \mathbb{0 \leq t < \frac{T}{n}} \\ \mathbb{W(\frac{T}{n})} & \text{if} \quad \mathbb{\frac{T}{n} \leq t < 2\frac{T}{n}} \\ . \\ . \\ . \\ \mathbb{W(\frac{(n-1)T}{n})} & \text{if} \quad \mathbb{\frac{(n-1)T}{n} \leq t < T} \\ \end{cases} \end{equation}\] Since Brownian motion has a continuous path (or it’s continuous) we must have- \[\mathbb{\underset{n \rightarrow \infty}{\lim} E \int_{0}^{T}{|\Delta_n(t)-W(t)|^2}.dt = 0}\]
So the condition, the sequence of simple process \(\mathbb{\Delta_{n}(t)}\) converges to \(\mathbb{W(t)}\) is satisfied. Hence we can write-
\[\begin{aligned} \mathbb{I(T)} &= \mathbb{\int_{0}^{T} W(t).dW(t)} = \mathbb{\int_{0}^{T} \Delta_n(t).dW(t)} \\ &= \mathbb{\sum_{j=0}^{n-1}W(\frac{jT}{n}).[W(\frac{(j+1)T}{n})-W(\frac{jT}{n})]} \end{aligned}\]
To simplify notation, we denote-
\[\begin{aligned} &\mathbb{W_j} = \mathbb{W(\frac{jT}{n})} \qquad \text{for all} \ \mathbb{j = 0,1, \cdots, n} \\ &\mathbb{W_0} = \mathbb{W(0)} = 0 \\ \end{aligned}\]
Then we can write-
\[\begin{aligned} \mathbb{\frac{1}{2}\sum_{j=0}^{n-1}(W_{j+1}-W_{j})^2} &= \mathbb{\frac{1}{2}\sum_{j=0}^{n-1}W_{j+1}^2 - \sum_{j=0}^{n-1}W_{j+1}W_{j} + \frac{1}{2}\sum_{j=0}^{n-1}W_{j}^2} \\ &= \mathbb{\frac{1}{2}\sum_{j=1}^{n}W_{k}^2 - \sum_{j=0}^{n-1}W_{j+1}W_{j} + \frac{1}{2}\sum_{j=0}^{n-1}W_{j}^2} \\ &= \mathbb{\frac{1}{2}W^2_n} + \mathbb{\frac{1}{2}\sum_{j=0}^{n-1}W_{k}^2 - \sum_{j=0}^{n-1}W_{j+1}W_{j} + \frac{1}{2}\sum_{k=0}^{n-1}W_{j}^2} \qquad \text{[since,} \ \mathbb{W_0 = 0}] \\ &= \mathbb{\frac{1}{2}W^2_n} + \mathbb{\sum_{j=0}^{n-1}W_{j}^2 - \sum_{j=0}^{n-1}W_{j+1}W_{j}} \\ &= \mathbb{\frac{1}{2}W^2_n} + \mathbb{\sum_{j=0}^{n-1}W_{j}(W_{j}- W_{j+1})} \\ \end{aligned}\]
From that we can write- \[\mathbb{\sum_{j=0}^{n-1}W_{j}(W_{j+1}- W_{j})} = \mathbb{\frac{1}{2}W^2_n} - \mathbb{\frac{1}{2}\sum_{j=0}^{n-1}(W_{j+1}-W_{j})^2}\] In original notation, it will be-
\[\mathbb{\sum_{j=0}^{n-1}W(\frac{jT}{n})[W(\frac{(j+1)T}{n})- W(\frac{jT}{n})]} = \mathbb{\frac{1}{2}W^2(T)} - \mathbb{\frac{1}{2}\sum_{j=0}^{n-1}[W(\frac{(j+1)T}{n})-W(\frac{jT}{n})]^2}\]
If \(\mathbb{n \rightarrow \infty}\), the above equation goes to-
So the whole thing becomes- \[\begin{aligned} &\mathbb{\underset{n \rightarrow \infty}{\lim}} \ \mathbb{\sum_{j=0}^{n-1}W(\frac{jT}{n})[W(\frac{(j+1)T}{n})- W(\frac{jT}{n})]} = \mathbb{\underset{n \rightarrow \infty}{\lim}} \ \mathbb{\frac{1}{2}W^2(T)} - \mathbb{\frac{1}{2}\sum_{j=0}^{n-1}[W(\frac{(j+1)T}{n})-W(\frac{jT}{n})]^2} \\ \implies & \mathbb{\int_{0}^{T}W(t).dW(t)} = \mathbb{\frac{1}{2}W^2(T)} - \mathbb{\frac{1}{2}[W,W](T)} \\ \implies & \mathbb{\int_{0}^{T}W(t).dW(t)} = \mathbb{\frac{1}{2}W^2(T)} - \mathbb{\frac{1}{2}T} \end{aligned}\]
Now you see the difference between the ORDINARY CALCULUS and the STOCHASTIC CALCULUS.
If \(\mathbb{g(t)}\) is a differentiable function with \(\mathbb{g(0) = 0}\),, then- \[\mathbb{\int_{0}^{T}g(t).dg(t) = \frac{1}{2}g^2(T)}\]
The extra term \(-\frac{1}{2}\mathbb{T}\) in the above Ito integral comes from the non-zero quadratic variation of Brownian motion. The way we constructed this Ito integral is the Left Riemann sum; always evaluating the integrand at the left-hand endpoint of the sub-interval. But if we were instead to evaluate at the midpoint, i.e. Mid-point rule of Riemann sum, mathematically- \[\mathbb{I_{M}(t)} = \mathbb{\underset{n \rightarrow \infty}{\lim}} \quad \mathbb{\sum_{j=0}^{n-1}W(\frac{(j+\frac{1}{2})T}{n})[W(\frac{(j+1)T}{n})-W(\frac{jT}{n})]}\]
Here \(\mathbb{I(t)}\) and \(\mathbb{I_{M}(t)}\) are not equal. In case of ordinary calculus- the limiting version of Left-hand Riemann sum, Right-hand Riemann sum and Mid-point rule of Riemann sum are equal. But its not true for Stochastic calculus.
\(\mathbb{I_{M}(t)}\), the mid-point rule version of Ito integral is called Stratonovich integral.
Let \(\mathbb{W(t)}, \ \mathbb{t \geq 0}\) be a Brownian motion. Let \(\mathbb{T > 0}\) is fixed and let \(\mathbb{\Pi = \{t_0, t_1, \cdots, t_n\}}\) be a partition of \(\mathbb{[0,T]}\); i.e. \[\mathbb{0 = t_0 < t_1 < \cdots < t_n = T}\]
For each \(\mathbb{j}\), let us define- \(\mathbb{t_j^* = \frac{t_j+t_{j+1}}{2}}\), i.e. the mid-point of interval \(\mathbb{[t_j,t_{j+1}]}\).
Define the Half-sample quadratic variation corresponding to partition \(\mathbb{\Pi}\) as- \[\mathbb{Q_{\Pi/2}} = \mathbb{\sum_{j=0}^{n-1}(W(t_j^*)-W(t_j))^2}\] First we want to that- \[\mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \mathbb{Q_{\Pi/2} = \frac{T}{2}}\] For that, it is enough to show that-
Let us start with the easier one, the expectation.
\[\begin{aligned} \mathbb{E(Q_{\Pi/2})} &= \mathbb{E(\mathbb{\sum_{j=0}^{n-1}(W(t_j^*)-W(t_j))^2})} \\ &= \mathbb{\sum_{j=0}^{n-1}E[(W(t_j^*)-W(t_j))^2]} \\ \end{aligned}\]
Now, since for all values of \(\mathbb{j}\) we have- \(\mathbb{t_j^* > t_j}\), we can write- \[\mathbb{W(t_j^*)-W(t_j) \sim N(0,t_j^*-t_j)}\] and \(\mathbb{t_j^* - t_j = \frac{t_j+t_{j+1}}{2}-t_j = \frac{t_{j+1}-t_{j}}{2}}\).
So we can write- \[\begin{aligned} \mathbb{E(Q_{\Pi/2})} &= \mathbb{\sum_{j=0}^{n-1}E[(W(t_j^*)-W(t_j))^2]} \\ &= \mathbb{\sum_{j=0}^{n-1}\frac{t_{j+1}-t_{j}}{2}} \\ &= \mathbb{\frac{T}{2}} \qquad \text{[so the first part is proved.]} \end{aligned}\] Now lets move towards the variance art of the proof. For that we will use few very simple tricks.
A Brownian motion, with \(\mathbb{t > 0, \ W(t) \sim N(0,t)}\). The kurtosis of Normal distribution is = 3. i.e. \[\mathbb{\frac{\mu_4}{\sigma^4} = 3}\] and \(W(t_j^*)-W(t_j) \quad \underset{\equiv}{\mathbb{d}} \quad W(t_j^*-t_j) \sim \mathbb{N(0,t_j^*-t_j)}\)
This will help us to evaluate- \[\begin{aligned} \mathbb{E[(W^2(t)-t)^2]} &= \mathbb{E[(W^4(t)-2W^2(t).t+ t^2)]} \\ &= \mathbb{3t^2-2t^2+t^2} = \mathbb{2t^2} \\ \end{aligned}\]
So, \[\begin{aligned} & \quad \mathbb{Var(Q_{\Pi/2})} \\ &= \mathbb{E[(\mathbb{\sum_{j=0}^{n-1}(W(t_j^*)-W(t_j))^2}- \mathbb{\frac{T}{2}})^2]} \\ &= \mathbb{E[(\mathbb{\sum_{j=0}^{n-1}(W(t_j^*)-W(t_j))^2}- \mathbb{\sum_{j=0}^{n-1}\frac{t_{j+1}-t_{j}}{2}})^2]} \\ &= \mathbb{\sum_{j,k = 0}^{n-1}} \mathbb{E[((W(t_j^*)-W(t_j))^2 - \frac{t_{j+1}-t_j}{2}).((W(t_k^*)-W(t_k))^2 - \frac{t_{k+1}-t_k}{2})]} \\ &= \mathbb{\sum_{j,k = 0; j\neq k}^{n-1}} \mathbb{E[((W(t_j^*)-W(t_j))^2 - \frac{t_{j+1}-t_j}{2}).((W(t_k^*)-W(t_k))^2 - \frac{t_{k+1}-t_k}{2})]} \\ & \quad + \mathbb{\sum_{j=0}^{n-1}} \mathbb{E[((W(t_j^*)-W(t_j))^2 - \frac{t_{j+1}-t_j}{2})^2]} \\ &= \mathbb{\sum_{j,k = 0; j\neq k}^{n-1}} \mathbb{E[((W(t_j^*)-W(t_j))^2 - \frac{t_{j+1}-t_j}{2})].E[((W(t_k^*)-W(t_k))^2 - \frac{t_{k+1}-t_k}{2})]} \\ & \quad + \mathbb{\sum_{j=0}^{n-1}} \mathbb{E[((W(t_j^*)-W(t_j))^2 - \frac{t_{j+1}-t_j}{2})^2]} \\ &= 0 + \mathbb{\sum_{j=0}^{n-1}} \mathbb{E[((W(t_j^*)-W(t_j))^2 - \frac{t_{j+1}-t_j}{2})^2]} \qquad \text{[due to independence]} \\ &= \ \mathbb{\sum_{j=0}^{n-1}} \mathbb{E[(W^2(t_j^*-t_j) - \frac{t_{j+1}-t_j}{2})^2]} \\ &= \mathbb{\sum_{j=0}^{n-1} 2.(\frac{t_{j+1}-t_j}{2})^2} \\ &\leq \mathbb{\underset{0 \leq j \leq n-1}{\max}(t_{j+1}-t_j)}.\mathbb{\sum_{j=0}^{n-1} (\frac{t_{j+1}-t_j}{2})} = \mathbb{\underset{0 \leq j \leq n-1}{\max}(t_{j+1}-t_j)}.\mathbb{\frac{T}{2}} \rightarrow 0 \\ \end{aligned}\]
Combining these two, we can conclude- \[\mathbb{\underset{||\Pi|| \rightarrow 0}{\lim} Q_{\Pi/2} \rightarrow \frac{T}{2}} \quad \text{in} \ \mathbb{L^2(P)}\]
Now, notation-wise let’s define the Stratonovich integral of \(\mathbb{W(t)}\) with respect to \(\mathbb{W(t)}\) as- \[\mathbb{\int_{0}^{T}W(t) \ o \ dW(t)} = \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim} \ \sum_{j=0}^{n-1} \mathbb{W(t_j^*)[W(t_{j+1})-W(t_j)]}}\]
We can write that as- \[\begin{aligned} &\mathbb{\sum_{j=0}^{n-1} \mathbb{W(t_j^*)[W(t_{j+1})-W(t_j)]}} \\ &\mathbb{\sum_{j=0}^{n-1} \mathbb{W(t_j^*)\{[W(t_{j+1})-W(t_j^*)]+[W(t_{j}^*)-W(t_j)]\}}} \\ &\mathbb{\sum_{j=0}^{n-1} \mathbb{\{W(t_j^*)[W(t_{j+1})-W(t_j^*)]+W(t_j^*)[W(t_{j}^*)-W(t_j)]\}}} + \mathbb{\sum_{j=0}^{n-1}} \mathbb{(W(t_j^*)-W(t_j))^2} \\ \end{aligned}\]
By the construction of Ito integral, \[\mathbb{\underset{||\Pi^*|| \rightarrow 0}{\lim}} \ \mathbb{\sum_{j=0}^{n-1} \mathbb{\{W(t_j^*)[W(t_{j+1})-W(t_j^*)]+W(t_j^*)[W(t_{j}^*)-W(t_j)]\}}} = \mathbb{\int_{0}^{T}W(t).dW(t)}\] where \(\mathbb{\Pi^*}\) is the partition- \[\mathbb{0= t_0 < t_{1}^* < t_1 < t_2^* < \cdots < t_{n-1}^* < t_n = T}\] Hence, we can write- \[\begin{aligned} \mathbb{\int_{0}^{T}W(t) \ o \ dW(t)} &= \mathbb{\int_{0}^{T}W(t).dW(t) \ + \frac{T}{2}} \\ &= \mathbb{\frac{1}{2}W^2(T)} \end{aligned}\]
So you see the Left-hand sum and Mid-point sum are not equal.
However since we are learning all these with respect to financial world. The mid-point definition is not very useful in financial world. Consider the following situation.
In finance, the integrand represents a position in an asset and the integrator represents the price of that asset. i.e. \[\begin{aligned} &\mathbb{\Delta(t)} = \ \text{Position in a asset i.e. amount of asset at the time} \ \mathbb{t} \\ &\mathbb{dW(t)} = \ \text{Instantaneous difference of price at time} \ \mathbb{t} \\ \end{aligned}\] For example- let at 10:00 a.m. you buy 100 units of an asset at price \(\$10/ \text{unit}\). At 11:00 a.m. the price of that asset becomes \(\$12/ \text{unit}\). Then the amount of profit (loss) is \(\$\) 100*(12-10) = \(\$\) 200. This is nothing but- \(\mathbb{\Delta(t).dW(t)}\).
Now, instead of one hour- like 10:00 a.m. to 11:00 a.m. decrease the time interval and sum it over 0 to time \(\mathbb{T}\), i.e. the time up to which we are keeping the asset. This is the Left-hand sum of Ito integral.
We cannot decide at 10:00 a.m. which position we took at 9:00 a.m. We must decide the position at the beginning of each time interval, and the Ito integral is the limit of the gain achieved by that kind of trading as the time between trades approaches zero.
In the next blog, we are going to see the famous Ito-Doeblin formula. Happy reading.