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MR,Bernoullis and Risk-4

February 26, 2010

The Story of Mean Reversion

In the last blog post, I discussed briefly and touched upon some basics of mean reversion. And it is just that, the tendency to revert to mean.

Let me now, bring up a practical aspect to it. Imagine, an utter drunkard(almost on the verge of tipping) walking down the street. What do you think, will be his motion like?

Though, I have seen village drunkards, who betray serious alcohol content through their gaits, yet walk over 8 inch wide, 3 feet long plank over the village drain; still I am going to talk about drunkards who are not as skilled.

They will be apparently going nowhere.Returning back, making detours to their original point.Given long enough time, they might get somewhere, but its inherently very slow. It might resemble something like this:

Random Walk?

Here, L stands for the stride length and D is the distance to be covered. Green is the starting point, Red is the destination.

Anyway, so here was a physical representation of random walk. Now this random walk can be modelled as, something like this.By the way, the one I showed above was a 2 dimenional random walk. If you consider a similar condition, with only one choice, you flip a coin, depending on that you decide you move ahead or back.

Hence, this entire process, that is, your next tiny, incremental step, is going to be decided on a random variable(a continuous one for that matter and not a discrete one, like coin toss)  and a scaling factor, say \sigma . What this scaling factor creates, is , it scales a generic random variable to create an appropriate stride length.

Now, the above equation was with this reference that, where you stand right now, is zero. If we it was not, then the equation just changes to accomodate the offset.

dr= \mu + \sigma dW

Now, here, dW is the random process which keeps on spewing values which are random in themselves(hence called random process) , but with a mean of zero and standard deviation of 1. Moreover, here for our own general purpose we are considering it to be a normal process (i.e. it generates random values which follow normal distribution: remember the normal dist we talked about in the last blog post?)

Now, not me, but people say, then stock market returns in general are stationary processes. Stationary?

What the eff is stationary?

Well, a stationary process is simply a process which doesnt stay away from mean for a long time. Now, this is how you should see and think, but it has a few internal properties, which is not necessarily needed to work upon as of now. We might visit this, with time, but not necessarily.

So, I ask you, tell me, if this is stationary or not?

HINT: Its a mean reverting process, with mean zero, and standard deviation 1, and scaler 2;

So coming back to the previous point, people claim that stock market returns are random walk. I dont believe its that easy to justify. And of course it isnt. But the way, it is used as a crutch to say, you can’t really predict it, or profit from it, just irks traders and I am one of them. So my bottomline is:

a. Its not that simple to off handedly attribute random walk principle for financial market returns.

b. If its not possible to profit, then what do you make of hedge fund managers and traders.

So, alright, I will put this part off, because to be fair, random walk model for stock market returns is thoroughly debunked.

Anyway, moving ahead, we are faced with a crossroad.

by Paraflyer

One of the ways, take us straight into the heart of stochastic calculus, option pricing theory etc and the other makes a direct attack onto the fortes of econometric tools like AR process,ARMA, GARCH etc.

So, its a very interesting prospect, and when times comes,I will pick it up from where I left. But now the time is to move to a third direction.

I understand that these four blog posts can be a bit of information overload for some body just getting started, I would recommend to them to do a slight reading on any book on basic statistics.

The coin tossing problem, Bernoulli and Probabilities

Feymann once said, if we can’t explain some theory to a freshman with simplicity, then we probably dont know much about it anyway (or something to that effect).

Anyway, the point being, I have always thoroughly enjoyed building concepts from elemental concepts, those built on further elemental and so on. Like almost a smooth Chardonnay, flowing like the questions come out of the mind of a 6 year ol’.

And hence, I enjoy the process of addressing serious problems using simple understandings.

Its possible that, my repeated reference to the same very example bore the hell out of you, but well,its worth it, isnt it?

So, off we go back to our old companion, which we brought up on the first blog post- the coin toss.

Now, the question which rises is, if you are given a fair coin, and I conduct 10 trials of coin toss, what is the probability that I will end up with exactly 6 heads?

So the logic goes, to this point, that out of the 10 trials, you select any 6, and impose the condition that heads turn up.


p^{h}_{5} = ^{10}\textrm{C}_{6}(0.5)^{6} * (0.5)^{4}

So, it reads simple. Out of 10 trials, you select 6, impose the condition that heads turned up on those trials and the rest 4 tails turned up. Now the probability of heads turning up on 6 tosses is obviously, 0.5^{6} and so on and so forth.

So, moving onwards what if the coin is biased, with probability of heads turning up is p and for tails its (1-p) .


p^{H}_{k}= ^{N}\textrm{C}_{k}(p)^{k} * (1-p)^{N-k}

N =number of trials;k=number of times you want heads to turn up

So in effect, this will give you the probability of such an even occurring where out of N trials, k tosses yield heads.

Now lets bet on this coin toss. Let me say, I win 1 buck each time, heads come up and lose 1 buck each time, tails come up.

So, if my one experiment consists of 30 tosses, and generate my probability distribution, its going to look like this:


So, if I repeat this experiment(each experiment will have 30 coin tosses), a large number of times, there is approximately, 14% chance of having at least 14 losses. Hmm…

Makes sense?

Now, you might think that, this looks very close to normal distribution…


Absolutely. When Bernoullis Trials are repeated with huge number of trials, then the result very closely resembles a gaussian distribution.

Anyway, so having said and done all this, there is a problem at hand.

Calculate the probability of a trader having a losing streak of k length?

Leave comments for solutions.

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