Brownian Motion Squared Increments and Quadratic Variation

# Itos Lemma Derivation notation


# Itos Lemma Derivation notation












So in Hull (2012) the main point is that $\Delta x^2 = b^2 \epsilon ^2 \Delta t + $higher order terms$ $ has a term of order $\Delta t$ and can not be ignored as the Brownian motion exhibits the quadratic variation of $\Delta t$. My question is now what does $\epsilon ^2$ correspond to. Cochrane (2005) notes that $dz^2 = dt$, so I was confused since Hull defines $dz$ as $\epsilon \sqrt dt $. Hence, $dz^2$ would imply $\epsilon^2 dt $. As $\epsilon$ is standard normally distributed the mean would be zero and the variance one this would imply in $\Delta x^2 = b^2 \epsilon ^2 \Delta t$ that $b^2 \epsilon ^2 \Delta t$ would in the limit as $\Delta t$ goes to zero equal to $b^2 \Delta t$ as $E(\epsilon^2)$ =1. Hull argues that the variance of $\epsilon \Delta t $ would become too small and hence, lose its stochastic component and then equal to its expected value in the limit, but I didn't quite understand that. My only explanation would be that $\epsilon^2$ equals to one, but isn't it that $E(\epsilon^2) = 1$?

## Answer by Magic is in the chain (score 6)

https://quant.stackexchange.com/a/55138

The theory behind the actual reasoning is a bit complicated than the coverage in Hull's, but staying within the simple reasoning, the difference comes down to the following:

The Brownian increments over the interval $dt$ are normally distributed with mean zero and variance $dt$, so in terms of distribution, you can express the increments in terms of a standard normal: $dw_t \sim \epsilon \, \sqrt{dt}$. You can easily verify this: a constant times a normal is normal, the mean of $\sqrt{dt}$ times a standard normal is equal to zero, and the variance is equal to $dt \times \mathrm{variance \, of\, standard \, normal} =dt\times 1=dt$.

$dw_t$ and $\epsilon$ are random variables, so $dw_t^2=dt$ means this equality in some probabilistic/limiting sense. You can take that to mean variance, or $E\left[dw_t^2\right]$ because means of $dw_t$ is zero. But actually this equality holds in a much stronger sense - think of a simulated brownian path, and if you let the number of intervals become very large, you will see the sum of squared of brownian increments become equal to $dt$.

But for everyday use, you can assume $dw_t \sim \epsilon \, \sqrt{dt}$ and $dw_t^2 =dt$, thinking of $dw_t^2$ as variance or sum of the squares of the increments of brownians when the interval is divided into a very large number of sub-intervals.

## Answer by Jan Stuller (score 3)

https://quant.stackexchange.com/a/55142

I think the question also brings up a common confusion with notation. I think it is incredibly unfortunate to use notation such as $dW(t)$ (unless it's part of a stochastic integral), and I get upset when I see it being used in textbooks.

The definition of Brownian Motion is implicit and goes like this:

(i) $W(t=0) = 0$

(ii) $W(t)$ is (almost surely) continuous

(iii) $W(t)$ has independent increment

(iv) The increments $W(t) - W(s): t\geq s \geq0$ are normally distributed with mean zero and variance = (t-s).

What variance does $dW(t)$ have? In my opinion it's difficult to discuss that. Do we actually mean $W(dt)$ (so the variance is infinitesimal?)? Or more like $W(\delta t)$, so the variance is $\delta t$, i.e. very tiny? I have never seen a serious lecturer use the notation $dW(t)$ (aside from Stochastic integrals). I think discussing the quantity $dW(t)$ outside of Stochastic integrals doesn't make sense. Instead let's use $W(\delta t)$, in which case we can discuss its distribution.

Back to the question: In Hull, $Z$ confusingly refers to $W$ and $\epsilon$ refers to the Standard Normal random variable.

So when Hull writes $dZ = \epsilon \sqrt(dt)$, he really means to say that $Z(\delta t)$ equals in distribution to $\epsilon \sqrt(\delta t)$. Now:

$$ \mathbb{E}\left[\epsilon \sqrt{\delta t}\right]=0$$

$$\mathbb{E}[(\epsilon \sqrt{\delta t})^2]=Var(\epsilon \sqrt{\delta t})=\delta t Var(\epsilon)= \delta t$$

$$Var\left((\epsilon \sqrt{\delta t})^2\right) = Var \left( \epsilon^2 \delta t\right)= \delta t^2 Var \left( \epsilon^2 \right)$$

Above, the first equality is true because trivially $\mathbb{E}[\epsilon]=0$ by definition of standard normal variable. The second equality is true because trivially $Var(\epsilon)=1$, again by definition of standard normal variable. The third equality is true because for any random variable $X$, $Var(aX)=a^2Var(X)$.

In the third equality, one can see that irrespective of what $Var \left( \epsilon^2 \right)$ actually is, the term $Var \left( \epsilon^2 \delta t\right)$ is going to be of order $\delta t^2$.

So really, when someone writes $dz^2 = dt$, they actually mean to say that $Z(\delta t)^2$ converges to a non-stochastic quantity when $\delta t$ gets really small, because the Variance is of order $\delta t^2$, so the variance quickly converges to zero (and Random Variable with no variance is no longer random). The expected value of $Z(\delta t)^2$ is $\delta t$ as shown above, so in conclusion, $Z(\delta t)^2$ converges fast to non-random variable $\delta t$ when $\delta t$ gets arbitrarily close to zero.