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Why \(dW(t)^2\) is \(dt\)
This version: 2024-03-30
First version: 2024-03-30
Summary
In this short note, the key identity of stochastic calculus \(dW{(t)^2} \equiv dt\) is proved using intuitive way.
1 Non-rigorous proof
In the large domain of stochastic calculus, there is one rule which truly stands out and is fundamental for basically all results. That rule simply states that
\(\seteqnumber{0}{1.}{0}\)
\begin{equation*}
dW{(t)^2} = dt.
\end{equation*}
We all know this means that while Wiener process \(W\) itself is a random process, the squares of its increments are equal to a constant. Often, I see people to take this key result as ’God given rule’ (even saw such kind of statement on
quant.stackexchange), perhaps also because they expect there is some deep result behind it. But the fact is that it is actually quite easy to derive this result, using simple tools we readily have.
Let’s start from a discrete setting where
\(\seteqnumber{0}{1.}{0}\)
\begin{eqnarray*}
\Delta W(t) = W(t + \Delta t) - W(t),
\end{eqnarray*}
and it is known that that Wiener increments are normally distributed with zero mean and variance equal to time step, so
\(\seteqnumber{0}{1.}{0}\)
\begin{equation*}
\Delta W(t) = \sqrt {\Delta t} X, \text { where } X{\sim } N(0,1).
\end{equation*}
Taking square gives
\(\seteqnumber{0}{1.}{0}\)
\begin{equation*}
{\left ( {\Delta W(t)} \right )^2} = \Delta t{X^2} = \Delta tQ, \text { where } Q{\sim }{\chi ^2}(1),
\end{equation*}
so \(\left ( {\Delta W(t)} \right )^2\) can be expressed as Chi-squared distributed variable with \(\nu =1\) degree of freedom, multiplied by \(\Delta t\), simply because it is known that square of standardized normal has Chi-squared distribution with 1 degree of
freedom.
From known properties of Chi-square distribution we know that \({\chi ^2}(\nu )\)–distributed random variable has expected value \(\nu \) and variance \(2\nu \).
So for our case with \(\nu =1\) it turns out that
\(\seteqnumber{0}{1.}{0}\)
\begin{eqnarray}
\mathbb {E}\left [ {{{\left ( {\Delta W(t)} \right )}^2}} \right ] &=& \mathbb {E}\left [ {\Delta tQ} \right ] = \Delta t\mathbb {E}\left [ Q \right ] = \Delta t \nonumber \\ \mathbb {V}\left [ {{{\left ( {\Delta W(t)} \right
)}^2}} \right ] &=& \mathbb {V}\left [ {\Delta tQ} \right ] = {(\Delta t)^2}\mathbb {V}\left [ Q \right ] = 2{(\Delta t)^2}. \label {eq:E_V_W2}
\end{eqnarray}
If we replace \(\Delta \) by its infinitesimal version, i.e. we let \(\Delta W(t) \to dW(t),\Delta t \to dt\), we can apply the standard calculus principle that squares of deterministic differentials are considered 0.
So it turns out that \(dW{(t)^2}\) can be first seen as a ’random variable’ which has according to (1.1) mean \(dt\) and variance \(2dt^2=0\). Since such a ’random variable’ \(dW{(t)^2}\) has
no variance, it has lost its randomness and its value is always trivially equal to the expected value \(dt\).