5.4.3 Other Applications of the Central Limit Theorem

The CLT can be used to justify the normal approximation to the binomial distribution discussed in Chapter 4. Recall that a binomial variable $X$ is the number of successes in a binomial experiment consisting of $n$ independent success/failure trials with $p=P\left(S\right)$ for any particular trial. Define a new $\mathrm{rv}{X}_1$ by

$$X_1=\{\begin{array}{ll}1& \text{ if the 1st trial results in a success }\\ 0& \text{ if the 1st trial results in a failure }\end{array}$$

and define $X_2,X_3,\dots ,X_n$ analogously for the other $n-1$ trials. Each $X_i$ indicates whether or not there is a success on the corresponding trial.

Because the trials are independent and $P\left(S\right)$ is constant from trial to trial, the $X_i$’s are iid (a random sample from a Bernoulli distribution). The CLT then implies that if $n$ is sufficiently large, both the sum and the average of the $X_i$’s have approximately normal distributions. When the $X_i$’s are summed, a 1 is added for every $S$ that occurs and a 0 for every $F$, so $X_1+\cdots +X_n=X$. The sample mean of the $X_i$ ’s is $X/n$, the sample proportion of successes. That is, both $X$ and $X/n$ are approximately normal when $n$ is large. The necessary sample size for this approximation depends on the value of $p$ : When $p$ is close to.5, the distribution of each $X_i$ is reasonably symmetric (see Figure 5.20), whereas the distribution is quite skewed when $p$ is near 0 or 1. Using the approximation only if both $np\ge 10$ and $n\left(1-p\right)\ge 10$ ensures that $n$ is large enough to overcome any skewness in the underlying Bernoulli distribution.

Figure 5.20 Two Bernoulli distributions: (a) $p=.4$ (reasonably symmetric); (b) $p=.1$ (very skewed)

Consider $n$ independent Poisson rv’s $X_1,\dots ,X_n$, each having mean value $\mu /n$. It can be shown that $X=X_1+\cdots +X_n$ has a Poisson distribution with mean value $\mu$ (because in general a sum of independent Poisson rv’s has a Poisson distribution). The CLT then implies that a Poisson rv with sufficiently large $\mu$ has approximately a normal distribution. A common rule of thumb for this is $\mu >20$.

Lastly, recall from 4.5 Other Continuous Distributions that $X$ has a lognormal distribution if $\ln \left(X\right)$ has a normal distribution. Let $X_1,X_2,\dots ,X_n$ be a random sample from a distribution for which only positive values are possible $\left[P\left(X_i>0\right)=1\right]$. Then if $n$ is sufficiently large, the product $Y=X_1{X}_2\cdots \cdots X_n$ has approximately a lognormal distribution.

To verify this, note that

$$\ln \left(Y\right)=\ln \left(X_1\right)+\ln \left(X_2\right)+\cdots +\ln \left(X_n\right)$$

Since $\ln \left(Y\right)$ is a sum of independent and identically distributed rv’s (the $\ln \left(X_i\right)$ ’s), it is approximately normal when $n$ is large, so $Y$ itself has approximately a lognormal distribution. As an example of the applicability of this result, Bury (Statistical Models in Applied Science, Wiley, p. 590) argues that the damage process in plastic flow and crack propagation is a multiplicative process, so that variables such as percentage elongation and rupture strength have approximately lognormal distributions.