Why $\frac{d~B_t}{dt}$ does not exist?
Difference quitient of B over the interval $[t,t_h]$: $$D_{t,h} = \frac{B_{t+h} - B_t}{h}$$
Now let's find expectaion and variance of that.
$$\mathbf{E}[D_{t,h}] = \frac{0-0}{h} = 0$$$$\mathbf{Var}[D_{t,h}] = \frac{h}{h^2} = \frac{1}{h}$$So when $h\to 0$ then $\mathbf{Var}[D_{t,h}]\to \infty$
So since if $D_{t,h}$ had a limit, it would have to be Gaussian, thus, $D_{t,h}$ does not converge. So $\frac{d~B_t}{dt}$ does not exist.
However, check this out:
So, in particular, $\displaystyle \sqrt{\mathbf{Var}[dB_t]} = \sqrt{dt}$.
To understand this, we will need "stochastic calculus".
One more leap of faith,
$$\left(dB_t\right)^2 = dt \ \ \text{Ito's rule}$$So, we say that $dB_t \sim \mathcal{N}(0, dt)$
where $\sigma=\text{a constant called volatility}$
This is a good model for stock proces (but not for interest rates, ...)
To see why the BSM is good for stock prices, let's first prove the Ito's formula.
Let $f:\mathbb{R}\to\mathbb{R}$ and $f$ is differentiable.
Order 1: $f(b) \approx f(a) + (b-a)f'(a)$
Order 2: $f(b) \approx f(a) + (b-a)f'(a) + \frac{(b-a)^2}{2}f''(a)$
Differntial motivation: $$f(x+dx) \approx f(x) + f'(x) dx + \frac{1}{2}f''(x)(dx)^2$$
Also, if $f$ also depends on another parameter $t$, then,
$$f(t+dt, x+dx) \approx f(t,x) + \frac{\partial~f}{\partial t}dt + \frac{\partial~f}{\partial x}dx + \frac{\partial^2~f}{\partial x^2} (dx)^2$$
In the above formula, we ignored the $\frac{\partial^2~f}{\partial t \partial x}dtdx$ because in pur stochastic calculus ...
For the function $f(t,x)$ on $\mathbb{R}_t\times\mathbb{R}$ ($f$ has $t-$derivatives of order 1, and $x-$derivatives of order 2).
then, $$f(t+dt,B_{t+dt}) = f(t,B_t) + \frac{\partial~f}{\partial x}(t,B_t)dB_t + \frac{1}{2}\frac{\partial^2~f}{\partial x^2}(t,B_t)dt + \frac{\partial~f}{\partial t}(t,B_t)dt$$
Apply the $(t,x)-$ dependent version of Taylor's formula we just saw with $x=B_t$ and $x+dx = B_{t+dt}$.
Simply, notice the notation:
$$dB_t := B_{t+dt} - B_t\ \ \ \ \ \ \ \text{ note }:=\text{ means definition}$$
We see that $dx = dB_t$ we just defined.
Indeed, $\displaystyle (dx)^2 = \left(B_{t+dt} - B_t\right)^2 = (dB_t)^2 = dt$ where the last equality is Ito's rule.
say $S_0=1$ just to simplify.
Thus,
$S_t = f(t,B_t)$ where $$f(t,x) =\displaystyle e^{\sigma x -\frac{1}{2} \sigma^2t}$$
Now, we apply Ito's formula:
$\frac{\partial~f}{\partial x}(t,x) = \sigma e^{\sigma x - \frac{1}{2}\sigma^2t}$
$\frac{\partial^2~f}{\partial x^2}(t,x) = \sigma^2 e^{\sigma x - \frac{1}{2}\sigma^2t}$
$\frac{\partial~f}{\partial t}(t,x) = -\frac{1}{2}\sigma^2 e^{\sigma x - \frac{1}{2}\sigma^2t}$
Let's subsititue $$f(t+dt,B_{t+dt}) = f(t,B_t) + \frac{\partial~f}{\partial x}(t,B_t)dB_t + \frac{1}{2}\frac{\partial^2~f}{\partial x^2}(t,B_t)dt + \frac{\partial~f}{\partial t}(t,B_t)dt$$
Ito's formula:
$$\begin{array}{lc}S_{t+dt} - S_t &= \left(\sigma e^{\sigma B_t - \frac{1}{2}\sigma^2t}\right)dB_t + \\ &\frac{1}{2}\left(\sigma^2 e^{\sigma B_t - \frac{1}{2}\sigma^2t}\right)dt + \\ &\left(-\frac{1}{2}\sigma^2 e^{\sigma B_t - \frac{1}{2}\sigma^2t}\right)dt \\ &= \sigma e^{\sigma x - \frac{1}{2}\sigma^2t} = \sigma S_t dB_t\end{array}$$Call options: you buy the right at time zero to purchase stock x at time $T$ (pre-detemined) inf future at a pre-determined price $K$.
If $S_T<K$ then just buy
If $S_T\ge K$ use the option and buy at price $K$ abd sell at price $S_t$ so you make profit $S_T-K$
The insurance contract gives you
$$\left\{\begin{array}{lrr}0&\text{if}&S_T< K\\ S_T - K &\text{if}&S_T \ge K \end{array}\right.$$The price seeling this insurance contract must find the right proce for it. It turns out the fair price is the expected value of the payoff. In fact, $PAYOFF = \Phi(S_T)$ and fair price is $\mathbf{E}[\Phi(S_T)] = \mathbf{E}[(S_T-K)_t]$
We know everything about the distrinution of $S_T$:
$$S_T = e^{\sigma B_T - \frac{1}{2}\sigma^2 t}$$$B_T \sim \mathcal{N}(0,T)$ so we write $S_T = e^{\sigma \sqrt{T} Z - \frac{1}{2}\sigma^2 T}$
So, $$\mathbf{E}[\Phi(S_T)] = \int_0^\infty \frac{1}{\sqrt{2\pi}} e^{-Z^2/2}\left(e^{\sigma \sqrt{T} Z - \frac{1}{2}\sigma^2 T} - K\right)_t dZ$$
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