I am trying to understand why the Quadratic Variation property of Brownian Motion is true.
That is, if $W$ is a Wiener Process (https://en.wikipedia.org/wiki/Wiener_process), then a very small change in the time results in an equally small change in position squared.
Mathematically, we can express the Quadratic Variation property as:
$$(dw^2) = dt$$
Here is my attempt to prove this:
We start with the following:
- A standard Brownian motion $W(t)$
- From the theoretical properties of a Wiener Process , we know that each increment $W(t_{j+1}) - W(t_j)$ is a random variable with a normal distribution with mean 0 and variance $t_{j+1} - t_j$.
- An $n$ number of partitions $\Pi = \{0 = t_0 < t_1 < \cdots < t_n = t\}$ of the interval $[0, t]$
Now, we need to prove:
$$[W, W](t) = \lim_{|\Pi| \rightarrow 0} \sum_{j=0}^{n-1} (W(t_{j+1}) - W(t_j))^2$$
Let's first define the change in position squared as:
$$Q_n = \sum_{j=0}^{n-1} (W(t_{j+1}) - W(t_j))^2$$
As a quick refresher, note the following identity for random variables in probability:
- $Var(X) = E(X^2) - [E(X)]^2$
- If $E(X)=0$, then the above identity directly relates variance to the expected value squared: $Var(X) = E(X^2)$
If we define a Wiener Process $X_t$ as:
$$X_t : W_{t+u} - W_t \sim N(0,u)$$
This basically implies that:
- $E(X_t) = 0$
- $Var(X_t) = u = E(X^2)$.
Going back, if we take the Expected Value of $Q_n$, we get (note: the Expected Value operator can be shifted inside the summation):
$$E[Q_n] = E\left[\sum_{j=0}^{n-1} [W(t_{j+1}) - W(t_j)]^2\right] = \sum_{j=0}^{n-1} E[(W(t_{j+1}) - W(t_j))^2] = \sum_{j=0}^{n-1} (t_{j+1} - t_j) = t$$
This is the equivalent of saying: $$E[Q_n] = t$$ $$E[(dw^2)] = dt$$
However, the original proof requires a conclusion on $Q_n$, not on $E(Q_n)$.
As a formality, we take the limit:
$$\lim_{n \to \infty} E[Q_n] = Q_n = t$$
Thus, we have shown that for an infinite number of infinitesimally small partitions $n$ (i.e. limit as n approaches infinity), the quadratic difference is equal to $t$.
Is my proof correct?
Thanks!
- Note: Another way of seeing this is - for a single increment:
$$E[(W(t_{j+1}) - W(t_j))] = 0$$ $$[E[(W(t_{j+1}) - W(t_j))]]^2 = 0^2 = 0$$
$$Var[(W(t_{j+1}) - W(t_j))] = E[(W(t_{j+1}) - W(t_j))^2] - [E[(W(t_{j+1}) - W(t_j))]]^2 = E[(W(t_{j+1}) - W(t_j))^2] - 0 = E[(W(t_{j+1}) - W(t_j))^2]$$
In summary:
$$E[(W(t_{j+1}) - W(t_j))^2] = Var[(W(t_{j+1}) - W(t_j))]$$
We always know that the variance of a a Wiener Process is directly proportional to its time difference: $Var[W_a - W_b] = b-a$
So in our case, the variance between $t_{j+1}$ and $t_{j}$ becomes:
$$Var[(W(t_{j+1}) - W(t_j))] = (j+1) - j = 1$$ $$E[(W(t_{j+1}) - W(t_j))^2] = Var[(W(t_{j+1}) - W(t_j))] = 1$$
Now if we sum this over $t$ increments, we get:
$$ Var[(W(t_{j+1}) - W(t_j))] = t \cdot 1 = t$$
This is another way to confirm the same result
