2 the formula corrected, a reference added

I find the intuitive explanation by Paul Wilmott particularly appealing.

Stochastic integrals

Fix a small $$\int_0^t f(W_{\tau})h>0. The stochastic integral$$\int_0^{h} f(W(t))\ dW_{\tau}$$involve dW(t)=\lim\limits_{N\to\infty}\sum\limits_{j=1}^{N} f\left(W(t_{j-1})\right)\left(W(t_{j})-W({t_{j-1}})\right),\quad t_j= h\frac{j}{N},$$ involves adding up an infinite number of random variables. There Let's substitute every term $f\left(W(t_{j-1})\right)$ with its formal Taylor expansion. Then there are several contributions to the sum: those that are a sum of random variables and those that are a sum of the squares of random variables, and then there are higher-order terms.

Add up a large number of independent random variables and the Central Limit Theorem kicks in, the end result being a normally distributed random variable. But what is Let's calculate its mean and standard deviation?.

When we add up $N$ terms that are normal, each with a mean of $0$ and a standard deviation of $\sqrt{\delta t/N}$, \sqrt{h/N}$, we end up with another normal, with a mean of$0$and a standard deviation of$\sqrt{\delta t}$. \sqrt{h}$. This is our $dW$. Notice how the $N$ disappears in the limit.

Then

Now, if we add up the $N$ squares of the same normal terms then we get something which is normally distributed with a mean of $$N\left(\sqrt{\frac{\delta t}{N}}\right)^2=\delta t$$$N\left(\sqrt{\frac{h}{N}}\right)^2=h$$and a standard deviation which is \delta t\sqrt{2/N}. h\sqrt{2/N}. This tends to zero as N gets larger. In this limit we end up with, in a sense, our dW^2_t=dt, dW^2(t)=dt, because the randomness as measured by the standard deviation disappears leaving us just with the mean dt. The higher-order terms have means and standard deviations that are too small, disappearing rapidly in the limit as N\to\infty. 1 I find the intuitive explanation by Paul Wilmott particularly appealing. Stochastic integrals$$\int_0^t f(W_{\tau})\ dW_{\tau}$$involve adding up an infinite number of random variables. There are several contributions to the sum: those that are a sum of random variables and those that are a sum of the squares of random variables, and then there are higher-order terms. Add up a large number of independent random variables and the Central Limit Theorem kicks in, the end result being a normally distributed random variable. But what is its mean and standard deviation? When we add up N terms that are normal, each with a mean of 0 and a standard deviation of \sqrt{\delta t/N}, we end up with another normal, with a mean of 0 and a standard deviation of \sqrt{\delta t}. This is our dW. Notice how the N disappears in the limit. Then if we add up the N squares of the same normal terms then we get something which is normally distributed with a mean of$$N\left(\sqrt{\frac{\delta t}{N}}\right)^2=\delta t$$and a standard deviation which is$\delta t\sqrt{2/N}.$This tends to zero as$N$gets larger. In this limit we end up with, in a sense, our$dW^2_t=dt$, because the randomness as measured by the standard deviation disappears leaving us just with the mean$dt$. The higher-order terms have means and standard deviations that are too small, disappearing rapidly in the limit as$N\to\infty\$.