Cfrgtkky

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.

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.

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?

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 $2Delta t^2$ (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

$t_nrightarrow t$ as $Delta trightarrow 0$. Similarly for variance

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

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.