Consider a mesh $\{t_j^{(n)}\}$ for $1\le j\le k(n)$.

Mörters and Peres (Brownian motion, Remark 7.7) give a condition that turns out to be equivalent to
$$\tag{1}\label{1}
\sum_{n=1}^\infty \sum_{j=1}^{k(n)} (t^{(n)}_{j+1} – t^{(n)}_j)^2 <\infty
$$
(which is used explicitly in Exercise 1.16) and show that it guarantees that almost surely
$$
\sum g(B_{t_j})\Delta B_{t_j} \rightarrow \int g(B_s)dB_s
$$
where $\Delta B_{t_j} = B_{t_{j+1}} – B_{t_j}$ and $\Delta t_j = t_{j+1}-t_j$. In other words, this sequence of random variables almost surely converges to its $L^2$ limit.

(This is not guaranteed in general. Consider an independent sequence where $X_n=1$ with probability $1/n$ and $X_n=0$ otherwise. Then $\mathbb E(X^n_2)=1/n^2$ so $X_n$ converges to 0 in $L^2$. But $X_n$ does not converge to 0 almost surely, by the second Borel-Cantelli lemma.)

Instead of giving the details on that*, we will give details on something Mörters and Peres do not give as much detail on but which will also reveal how to do the above: what condition on the mesh is required for almost sure convergence of
$$
\sum g(B_{t_j})(\Delta B_{t_j})^2 \rightarrow \int g(B_s)ds,
$$
or equivalently, since $\int g(B_s)ds$ is a non-stochastic (but random) integral where the mesh does not matter,
$$
b_n:= \sum g(B_{t_j})[(\Delta B_{t_j})^2 - \Delta t_j] \rightarrow 0.
$$
They show that such convergence happens in $L^2$ without any assumption on the mesh (as most authors do) but then they are a little vague about the rest in Theorem 7.13. So as an exercise we can go back to Remark 7.7 and see if a similar condition can be obtained in this case.

By second-semester calculus it suffices to show that $\sum b_n^2<\infty$ almost surely:
$$
\{\omega: \sum b_n(\omega)^2<\infty\}\subseteq\{\omega: b_n(\omega)\rightarrow 0\}.
$$
For this it suffices to show that $\sum \mathbb E (b_n^2)<\infty$.

(Indeed, if $\sum \mathbb E (b_n^2)<\infty$ then [by Monotone Convergence] we have $\sum \mathbb E (b_n^2) = \mathbb E (\sum b_n^2)$ and so $\sum b_n^2$ is a nonnegative random variable with finite mean and hence $\sum b_n^2$ must be finite almost surely.

To prove that a nonnegative random variable with finite mean must be finite almost surely, note that for any random variable $X$, if $\mathbb P(X^2\ge M) \ge N/M$ then
$$
\mathbb E(X^2)\ge \mathbb E(X^2 | X^2\ge M) \mathbb P(X^2\ge M) \ge M \mathbb P(X^2\ge M) \ge N.
$$
Thus, if $\mathbb E(X^2) < N$, where $N$ is a given constant, then $\mathbb P(X^2\ge M) < N/M$.
Consequently, $\mathbb P(X^2=\infty)=0$.

Let $U_M = \{\omega: \sum b_n(\omega)^2\ge M\}$; then $\mathbb P(U_M) < N/M$. We may note that $U_M$ is computably enumerable only relative to the mesh. And indeed by an exercise in MP, the mesh must be computable in order to work for all ML-random $\omega$. With a nonstandard analysis construction of the Ito integral, however, we can sidestep this requirement.)

Calculating exactly as on the bottom of page 47 in Øksendal (6th edition)’s proof sketch for Ito’s lemma, with his $a_j$ being our $g(B_{t_j})$, we have
$$
\mathbb E(b_n^2) = 2\sum_{j=1}^{k(n)} \mathbb E(g(B_{t_j})^2) (\Delta t_j)^2.
$$
Here the factors $\mathbb E(g(B_{t_j})^2)$ will be bounded by a constant $c$, assuming we are integrating over a bounded domain $\int_S^T$.

So our condition is
$$
\sum_{n=1}^\infty 2\sum_{j=1}^{k(n)} \mathbb E(g(B_{t_j})^2) (\Delta t_j)^2 \le
\sum_{n=1}^\infty 2\sum_{j=1}^{k(n)} c (\Delta t_j)^2
<\infty
$$
which is exactly the same as condition (\ref{1}).

*Actually here are some details. We seek to relate the condition
$$
\sum_{n=1}^\infty \mathbb E \int_0^t (H_n(s)-H(s))^2 ds <\infty
$$
to the mesh. Here $H(s)=g(B_s)$ and $H_n(s)$ is an elementary approximation. The $n$th term is a sum over $j$ of:
$$
\mathbb E \int_{t_j}^{t_{j+1}} \left(g(B_s) – \sum_{j=1}^{k(n)} g(B_{t_j}) \chi_{[t_j,t_{j+1})}(s)\right)^2 ds
$$
$$
= \mathbb E\int_{t_j}^{t_{j+1}} (g(B_s) - g(B_{t_j}))^2 ds
= \int_{t_j}^{t_{j+1}} \mathbb E ((g(B_s) - g(B_{t_j}))^2) ds
$$
(by dominated convergence)
$$
\le c \int_{t_j}^{t_{j+1}} \mathbb E ((B_s) - B_{t_j})^2) ds
$$
(since $\mathbb E[(g(B_s)-g(B_{t_j})^2]
=\mathbb E\left[\left(\frac{g(B_s)-g(B_{t_j})}{B_s-B_{t_j}}\right)^2(B_s-B_{t_j})^2\right]
\le\mathbb E\left[\left(\frac{g(B_s)-g(B_{t_j})}{B_s-B_{t_j}}\right)^2\right]\mathbb E[(B_s-B_{t_j})^2]$)

$$
= c \int_{t_j}^{t_{j+1}} \mathbb E ((B_{s – t_j})^2) ds
= c \int_{t_j}^{t_{j+1}} s – t_j ds
$$
$$
= c \int_0^{\Delta t_j} u du = c (\Delta t_j)^2/2
$$
so we get
$$
\frac{1}{2} \sum_{n=1}^\infty \sum_{j=1}^{k(n)} (t_j^{(n)}-t_{j+1}^{(n)})^2 < \infty
$$
which is the same as condition (\ref{1}).

In the case where each $t_{j+1}^{(n)}-t_{j}^{(n)} = 1/k(n)$, this becomes
$$
\sum_{n=1}^\infty 1/k(n) < \infty
$$
so we can take $k(n)=n^2$ or, like Paley-Wiener, $k(n)=2^n$.

Øksendal (sixth edition) example 3.1.9: almost surely,
\[
B_t^2 - t = \int_0^t 2B_s dB_s
\]

This has a discrete version which holds everywhere: let \(X_n=\pm 1\) and \(S_n=\sum_{i=1}^n X_i\), then
\[
S^2_n-n = 2\sum_{i=0}^{n-1} S_i X_{i+1}
\]
To verify just note that both sides increase by \(2S_{n-1}X_n\) when going from $n-1$ to $n$.

Øksendal’s exercise 4.2:
\[
B_t^3 = \int_0^t 3B_s ds + \int_0^t 3B_s^2 dB_s
\]

Here the discrete version is not a perfect analogue:
\[
S_n^3 - S_n = 3\sum_{i=0}^{n-1} (S_i + S_i^2 X_{i+1})
\]
The extra term $S_n$ seems related to the fact that $(dB_t)^3 = 0$.

Let us consider the first example in Ch. 15 of Hull: Options, Futures, and other derivatives (6th edition).

$S_0=49$, $K=50$, $r=0.05$, $\sigma=0.2$, $T=0.3846 = 20/52$.

The price of the option per share should come out to $2.4$ according to Hull. We will check this against the Black-Scholes pricing formula for a call option $c$ on page 295.

We first calculate by hand that $\sigma^2/2=1/50$, $r\pm \sigma^2/2=(5\pm 2)/100$, and
$(r\pm\sigma^2/2)T = \frac{5\pm 2}{(13)(20)}$.

Moreover $S_0/K = 49/50$. A good approximation to the natural logarithm for inputs near 1 is $\ln x \approx x-1$.

Thus $\ln (S_0/K)\approx -\frac{1}{50}$.

Hence $(r+\sigma^2/2)T + \ln(S_0/K) \approx \frac{5+2}{(13)(20)} – \frac{1}{50} = \frac{9}{1300}$. This is the numerator of $d_1$. The denominator is $\sigma\sqrt{T} = \sqrt{5/13}/5 = 1/\sqrt{65} \approx 1/\sqrt{64} = 1/8 = .125$. So we have

$d_1 \approx \frac{9/1300}{1/8} = \frac{18}{325}$. It is tempting to say this is approximately $6\%$; a little calculator help shows it is approximately $5.5\%$. Then we also have $d_2 = d_1-\sigma\sqrt{T}\approx 5.5\% – 12.5\% = -7\%$.

Next we need to approximate the cumulative distribution function of the standard normal distribution, $N(d)$. Note that $N(0)=1/2$ and that $N’(0)=e^{-0^2/2}/\sqrt{2\pi} = 1/\sqrt{2\pi}$, so a linear approximation for $d\approx 0$ is $N(d)\approx \frac{1}{2} + \frac{d}{\sqrt{2\pi}}$.
A quite good approximation for $\sqrt{2\pi}$ is 2.5 (a little calculator help helps here).
This gives $N(d_1)\approx \frac{1}{2} + \frac{11}{500}$ and $N(d_2)\approx \frac{1}{2} – \frac{14}{500}$.

Next we approximate the exponential function near zero: $e^{-rT}\approx 1-rT = 1-\frac{1}{52}$ and $Ke^{-rT}\approx 50(1-\frac{1}{52})$.

Finally, $c = S_0 N(d_1) – K e^{-rT} N(d_2) = \frac{1}{2}(-\frac{1}{26}) + \frac{1}{500}(49\cdot 11 + 50(\frac{51}{52})(14))$. To get a good approximation we replace $50(51/52)$ by $49$, and get

$c \approx \frac{1225}{500} – \frac{1}{52} \approx \frac{2450}{1000} – \frac{1}{50} = 2.43$ which agrees to the precision given with our desired answer $2.4$.

For a nontrivial example of the use of Kolmogorov’s forward equation (the Fokker-Planck equation) where it can be contrasted with Kolmogorov’s backward equation, we look beyond Brownian motion, since the generator of the Brownian motion process is its own adjoint and applied to the transition density is the same in both the $x$ and $y$ variables. So we consider the next simplest example, the geometric Brownian motion process, which is given by
$$
dX_t = \mu X_t dt + \sigma X_t dW_t
$$
where we will assume $\sigma=1$ and $\mu=0$.

Generators and their adjoints

The generator for the GBM process in the $x$ variable is
$$
A = \frac{1}{2}x^2\frac{\partial^2}{\partial x^2}
$$
Following Øksendal’s book we need to find the adjoint $A^*$, satisfying $\langle Af,g\rangle = \langle f,A^* g\rangle$. The adjoint of $\partial/\partial x$ is $-\partial/\partial x$; this follows by integration by parts and the theory of distributions (where test functions are $C^\infty$ with compact support and hence the boundary terms vanish). Thus the adjoint of $A$, in the $y$ variable, is
$$
f\mapsto \frac{1}{2}\frac{\partial^2}{\partial y^2} (y^2 f)
$$
Using the product rule several times, we have
$$
A^*_y p = \frac{1}{2}\frac{\partial^2}{\partial y^2} (y^2 p) = p + 2y\frac{\partial p}{\partial y} + \frac{1}{2}y^2\frac{\partial^2 p}{\partial y^2}
$$

The pdf

By Shreve volume II page 119 the transition density for GBM is
$$
p(t,x,y)=\frac{1}{y\sqrt{2\pi T}}\exp\left(-\frac{(\log (y/x)-\nu T)^2}{2T}\right)
$$
where $\nu=\mu-\sigma^2/2=-1/2$, and where it is important to keep $x$ vs. $y$ straight.

Backward equation

If we wanted the backward equation we would replace $T$ by $T-t$ where $T>t$ and we are looking at the transition from $(t,x)$ to $(T,y)$ (instead of from $(0,x)$ to $(T,y)$). The resulting $p=p(t,x;T,y)$ gives the pdf of $X(T)$, as a function of $y$, given that $X(t)=x$. As $t\rightarrow T$ and with $x:=X(t)$ this pdf in $y$ will converge toward a Dirac delta function at the point $X(T)$. The backward equation is an expression of the fact that $p(t,X(t);T,y)$ is a martingale for fixed $T$ and $y$. It is the kind of martingale that keeps betting too much and hence will almost surely lose all its money. For a while the pdf at $y$ may be going up because $X$ approaches $y$, but this will only last for so long.

The backward equation can be verified and states that
$$
\frac{\partial p}{\partial t} = -A_x p
$$

Not-famous versions

On the other hand, since $p$ is a function of $t$ and $T$ only through $T-t$, it also follows that
$$
\frac{\partial p}{\partial T} = A_x p
$$
although this latter equation is not as famous as the Forward and Backward ones. It connects moving the future distribution forward in time, with space derivatives at the past time. Nor is it famous that
$$
\frac{\partial p}{\partial t} = – A^*_y p
$$

Verification of the Kolmogorov forward equation

In order to verify Kolmogorov’s forward equation we will have to take $\partial/\partial T$ of the transition density expression, which turns out to be less tedious if we use the functions $Q=\frac{1}{T}\log(y/x)$ and $R=Q+\frac{1}{2}$. If we ignore the trivial $\sqrt{2\pi}$ factor, after a while we obtain:
\begin{eqnarray}
p&=&\frac{1}{y\sqrt{T}}\exp(-R^2T/2) \\
\frac{\partial p}{\partial t} &=& p\cdot \left(\frac{-1}{2T} +RQ-R^2/2\right) \\
2y\frac{\partial p}{\partial y} &=& 2p(-1-R)\\
\frac{1}{2}y^2\frac{\partial^2 p}{\partial y^2} &=& \frac{p}{2}\left((-1-R)^2 +1+R-\frac{1}{T}\right)
\end{eqnarray}
By adding lines we get
$$
A^*_y p = \frac{\partial p}{\partial T}
$$
which is Kolmogorov’s forward equation.