## Monday, November 24, 2014

### Unizor - Probability - Normal Distribution - Using Sigma

Consider a game of winning against a house a unit of currency 1 bitcoin or losing to a house 1 bitcoin with equal probabilities of 1/2.
Now you decided to play a series of 100 games. How much money do you have to have in reserve to be able to pay the balance in case you are at loss after this series of games?

The simplest answer is to have 100 bitcoins in reserve to be able to pay for 100 losses in a row. In theory, if you want to assure with the probability of 100% that you will be able to pay, this is the only answer. But, at the same time, you might think that it would be good enough to assure the ability to pay with a relatively high probability, accepting a small risk to lose your good standing with the house. How much do you have to have in reserve then? Could you substantially reduce the amount of money required then?

The normal distribution and sigma limits can be of help in this case.

An individual game can be modeled as a Bernoulli random variable ξ taking values +1 and −1 with equal probabilities of 1/2.
The result of a series of 100 games is a sum of 100 such variables:
η = ξ1+ξ2+...+ξ100.
All members of this sum are independent and identically distributed according to Bernoulli distribution taking values +1 and −1 with equal probabilities of 1/2.

To evaluate the number of bitcoins you have to have to be able to pay off your losses with a relatively high probability, you have to establish your risk limits. For instance, you decide that if the probability of losing greater than S bitcoins is less than 0.025, it's good enough for you.
To find S (positive amount of bitcoins to have in reserve), we have to satisfy the inequality P(η ≤ −S) ≤ 0.025.

This is not a simple task. You have to find the probabilities for K=−100, K=−98, K=−96 etc. Summarize them one by one until their sum exceeds 0.025. The last value of K will be the maximum loss you can pay, ignoring all greater losses since their accumulated probability does not exceed 0.025.

As you see, this process is tedious, cumbersome, long, time consuming etc.

There is better way!

As we know, the distribution of a sum of many independent identically distributed random variables resembles the normal distribution. Which one? The one with an expectation equal to a sum of expectations of each component of a sum and a variance equal to a sum of their variances.

In our case the expectation of each Bernoulli random variable equals to
E(ξ)=(1)·(1/2)+(−1)·(1/2)=0
Its variance is
Var(ξ) =
=(1−0)2·(1/2)+(−1−0)2·(1/2)=1
Therefore, expectation and variance of a sum of 100 such variables are:
E(η) = 100·E(ξ) = 0
Var(η) = 100·Var(ξ) = 100

Let's assume now that random variable η is truly normal. Obviously, this assumption must be quantified, but we are looking for approximate values, so, for our purposes, it's quite valid assumption. Then we can use the sigma limits for a normal random variable. Knowing variance, we calculate the standard deviation as a square root from variance:
σ(η) = 10
And now we are ready to evaluate the probability of the sum of 100 results of the Bernoulli experiments based on sigma limit.

Recall the sigma limits for a normal random variable with expectation μ and standard deviation σ:
1. For a normal random variable with mean μ and standard deviation σ the probability of having a value in the interval [μ−σ, μ+σ] (the narrowest interval of these three) approximately equals to 0.6827.

2. For a normal random variable with mean μ and standard deviation σ the probability of having a value in the interval [μ−2σ, μ+2σ] (the wider interval) approximately equals to 0.9545.

3. For a normal random variable with mean μ and standard deviation σ the probability of having a value in the interval [μ−3σ, μ+3σ] (the widest interval of these three) approximately equals to 0.9973.

Consider the 2σ limit and apply it for our case of μ=0 and σ=10.
Since the probability of a random variable to be in the interval 2σ, that is in the interval [−20,20], equals to 0.9545, the probability to fall outside of this interval (less than −20 or greater than 20) equals to 0.0455. Since normal distribution is symmetrical, the probability to be less than −20 equals to the probability to be greater than 20 and equals to half of that value, that is, approximately, 0.0227.

Therefore, we can say that with the probability of about 0.0227 we will not lose more than 20 bitcoins. As you see, the calculations are simple and, accepting a little risk (with probability 0.0227), we can reduce the required amount of reserve from 100 bitcoins for non-risk strategy, to only 20 bitcoins.