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 $E(\epsilon^2) = 1$?


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.

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    $\begingroup$ The explanation of Magic is in the chain was great but, if you think of $\frac{t}{n}$ as $dt$ in your question, then this may also help. quant.stackexchange.com/questions/43195/… $\endgroup$ – mark leeds Jun 23 at 0:05
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    $\begingroup$ @Magic is in the chain: what is your view on the notation $dW(t)$ that I mention below? I do hold Quantpie in very high regard, so I am actually curious what your take is on that. Maybe it's not as much of a problem as I make it to be :) $\endgroup$ – Jan Stuller Jun 24 at 7:18
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    $\begingroup$ Thanks @Jan Stuller! I think the notation probably serves a purpose - to make the methods accessible to a much wider audience. Reason being most of us start learning maths the applied way - probably to do with the courses being designed to prepare us for engineering degrees. But agree the textbooks should make the meaning of the notation clear at the very beginning. Not making it clear has undesirable consequences later on- one has to go to basics at some stage to advance, and it is not easy (even though it should be the easiest!)- ask the professors who teach analysis- hard to fill the seats! $\endgroup$ – Magic is in the chain Jun 24 at 20:17

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.

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    $\begingroup$ Brilliant explanation. I think it's a pity that in none of the textbooks I mentioned actually explain this in terms that $dz^2 = dt$ holds in a distributional sense, which is the absolute central point. $\endgroup$ – Question Anxiety Jun 23 at 20:41
  • $\begingroup$ @QuestionAnxiety: I hear you. I used to get confused frequently when learning this stuff for the first time just because the notation used in the various textbooks tends to be so "non-rigorous". It's unfortunate and doesn't make the task of learning the subject any easier. $\endgroup$ – Jan Stuller Jun 24 at 11:46

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