# 3.15 Central Limit Theorem

The normal distribution is useful for modeling various random quantities, such as people’s heights, asset returns, and test scores. This is no coincidence. If a process is additive—reflecting the combined influence of multiple random occurrences—the result is likely to be approximately normal. This follows from the **central limit theorem**.

Let ** X** be an

*n*-dimensional random vector with independent and identically distributed (IID) components

*X*. It doesn’t matter what their common distribution is as long as its mean μ and standard deviation σ exist. Let be the random variable for the average of the

_{i}*X*. By [3.27] and [3.28], has mean

_{i}*m*and standard deviation . Accordingly, the normalized average

[3.210]

has mean 0 and standard deviation 1. The central limit theorem tells us is approximately *N*(0,1). Specifically, it states that, for any constant *x*,

[3.211]

where Φ(*x*) is the CDF of the standard normal distribution.

Exhibit 3.28 illustrates the PDF for the profit and loss (P&L) that will be realized by purchasing and holding for 1 month EUR 30,000 of a particular at-the-money 3-month call option on Euribor futures. The limited downside risk of the options strategy is evident in the skewed P&L distribution. It has skewness 0.96 and kurtosis 3.90.

Suppose random changes in Euribor are independent from one month to the next. We repeat our options strategy every month for 18 months. At the start of every month, we purchase EUR 30,000 of at-the-money 3-month options and liquidate them at the end of the month. Repeating this process for 18 consecutive months yields a total P&L for the 18 months whose PDF is graphed in Exhibit 3.29.12

The P&L distribution for the 18-month strategy is not skewed like that of the 1-month strategy. It does not afford the same protection against downside risk. With skewness of just 0.23, kurtosis of 3.05, and a familiar “bell” shape, it is almost normal. Rolling options for 18 months has a P&L distribution little different from that of just holding the underlying futures.

Our example illustrates the central limit theorem. With the 1-month strategy, we randomly draw a P&L from the probability distribution of Exhibit 3.28. With our 18-month strategy, we independently draw from that distribution 18 times. The 18-month P&L is the sum of these.

There are many versions of the central limit theorem.13 Several of these place additional restrictions on the *X _{i}* but do not require that they be identically distributed. The additional restrictions vary, but are generally designed to prevent one or a handful of random variables from dominating the average, which might happen if one random variable has a standard deviation far greater than the rest.

In Exhibit 3.30, probability distributions are illustrated for five independent random variables *X _{i}*. All five distributions have mean 0 and standard deviation 1 and are dramatically non-normal. They were selected arbitrarily, but their normalized average is approximately normal. Exhibit 3.31 provides summary information on the distributions of Exhibit 3.30.

*X*

_{1},

*X*

_{2},

*X*

_{3}, and

*X*

_{5}are continuous, so their PDFs are shown;

*X*

_{4}is discrete, so its PF is shown. The normalized average is approximately

*N*(0,1). All graphs indicate the interval [–3,3] on the

*x*-axis.

Other versions of the central limit theorem modestly weaken the independence assumption for the *X _{i}*. The central limit theorem generalizes to multiple dimensions.

###### Exercises

Let be the normalized average of *n* independent *U*(–1,1) random variables. Based upon your intuition:

- How large do you think
*n*must be for the PDF of to have the same general shape as the PDF of a standard normal distribution? - How large do you think
*n*must be for the kurtosis of to match the 3.0 kurtosis of a normal distribution to one decimal place?

Suppose *Y* equals a sum of 20 independent random variables. One is *U*(–10,10). The rest are *U*(–1,1). Is *Y* approximately normal?

Solution

For insight into the last exercise, use your intuition to sketch the PDFs of:

- a
*U*(–10,10) random variable; - a sum of a
*U*(–10,10) and 1 independent*U*(–1,1) random variables; - a sum of a
*U*(–10,10) and 2 independent*U*(–1,1) random variables; - a sum of a
*U*(–10,10) and 3 independent*U*(–1,1) random variables.