Why is $dW^2=dt$ in stochastic calculus?(do not use Ito’ Lemma),

Question

I'm trying to calculate the following integral: $$\int_0^t W_tdW_t$$ Without using Ito’ Lemma. I am confused about how $dW^2=dt$ when Ito's Lemma is not used.

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Hint

Let W be a standard Wiener process and t an arbitrary positive real number. For each $n$, and $t_i=it2^{-n}$, then:

1. show that $\sum_i(\Delta W(t_i))^2$ converges to $t$ as $n$ grow.
2. show that the terms in the sum are IID and that their variance shrinks sufficiently fast as $n$ grows and use the fourth moment of a Gaussian distribution.

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So, I am left with the following:

$$E(\sum_i(\Delta W(t_i))^2)=E(\sum_i(W(t_{i+1})-W(t_{i}))^2)=?$$

What I have done

So, can I say since $(W(t_{i+1})-W(t_{i}))$ follow $\sqrt {t_{i+1}-t_{i}}N(0,1)$ and by strong law of Large number $\sum_i(W(t_{i+1})-W(t_{i}))^2$ converges to $\sum_iE(W(t_{i+1})-W(t_{i}))^2$=1/2 $\sum_i(t_{i+1}-t_{i})$=t?

Is this correct?

Answer 1

Recall Riemann sum approximation is:

$$X_t^m = \sum_{t_j<t}f_{t_j}\Delta W_j$$

where $\Delta t = 2^{-m}$ and $W_t$ is a standard Brownian motion, and

$$\Delta W_j = W_{t_{j+1}}-W_{t_j}$$

The pathwise convergence is for every Brownian motion and the Riemann approx. converges to the limit such that the limit is measurable for the filtration up till $t$ because $X_t$ is a continuous function of $W_{[0,t]}$. Here I will skip a lot of explanation on Reimann sum approx. (and will take $\Delta W_j$ as Gaussian with mean $\Delta t$ and variance $2\Delta t^2$ (essentially above hint) see this) and go to the problem at hand which is:

$$X_t=\int_{0}^t W_s dW_s$$

By Riemann sum approx. we write

$$X_t^m=\sum_{t_j<t}W_{t_j}(W_{t_{j+1}}-W_{t_{j}})$$

Noticing that:

$$W_{t_j}=\frac{1}{2}(W_{t_{j+1}}+W_{t_j})-\frac{1}{2}(W_{t_{j+1}}-W_{t_j})$$

So we can write:

$$X_t^m=\frac{1}{2}\underbrace{\sum_{t_j<t}W_{t_j}(W_{t_{j+1}}+W_{t_j})(W_{t_{j+1}}-W_{t_j})}_{\text{first sum}}-\frac{1}{2}\underbrace{\sum_{t_j<t}W_{t_j}(W_{t_{j+1}}-W_{t_j})(W_{t_{j+1}}-W_{t_j})}_{\text{second sum}}$$

Notice, $$(W_{t_{j+1}}+W_{t_j})(W_{t_{j+1}}-W_{t_j})=W^2_{t_{j+1}}-W_{t_j}^2$$

By telescoping sum and by letting $t_n=\max\{t_j|t_j<t\}$ the first sum term becomes $\frac{1}{2}(W^2_{t_{n+1}}-W_0^2)$. Also, because $W_0=0$, we get $\frac{1}{2}W^2_{t_{n+1}}$ and $W_{t_{n+1}}\rightarrow W_t$ as $\Delta t \rightarrow 0$. Now the second sum term is $\sum_{t_j<t}\Delta W_j^2=S$ and since $E(\Delta W_j^2)=\Delta t$, so

$$E(\sum_{t_j<t}\Delta W_j^2)=E(S)=\sum_{t_j<t}\Delta t=t_n$$

This answer's the ? in the above question

$t_n\rightarrow t$ as $\Delta t\rightarrow 0$. Similarly for variance

$$var(S)=2\Delta t \sum_{t_j<t}\Delta t=2\Delta t t_n \leq 2 t 2^{-m}$$

Thus,

$$\int_0^t W_s dW_s = \frac{1}{2}(W_t^2 - t)$$

This is the problem statement

Notice, if $W_t$ was differentiable we would obtain $\frac{1}{2}W_t^2$ instead.