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# State and apply the Central Limit Theorem

For this learning objective, a certain knowledge of the normal distribution and knowing how to use the Z-table is assumed.

The central limit theorem is of the most important results in the probability theory. It states that the sum of a large number of independent random variables has an approximately normal distribution. It provides a simple method for computing approximate probabilities for sums of independent random variables and helps explain the remarkable fact that the empirical frequencies of so many natural populations exhibit bell-shaped (or normal) curves.

After analyzing the moment generating technique, we have found that the mean $$\bar{X}$$ of a random sample size n from a distribution with mean $$\mu$$ and variance $$\sigma^2 > 0$$ is a random variable with the properties:

$$E(\bar{X}) = \mu \quad\text{and}\quad Var(\bar{X}) = \frac{\sigma^2}{n}$$

As n increases, the variance of $$\bar{X}$$ decreases. Consequently, the distribution of $$\bar{X}$$ clearly depends on n, and we see that we are dealing with sequences of distributions.

If we consider n mutually independent normal variables with n means and n variances, each one belong to its n sub-indexes, then the linear function:

$$Y = \sum_{i=1}^{n}c_iX_i$$

has the normal distribution:

$$N\bigg(\sum_{i=1}^{n}c_i\mu_i,\sum_{i=1}^{n}c_{i}^2\sigma_{i}^2\bigg)$$

This can be proved by applying the moment generating technique to the linear function.

Having applied this we can note that as n increases, the probability becomes concentrated in a small interval centered at $$\mu$$. This means, as n increases, $$\bar{X}$$ tends to converge to $$\mu$$ or ($$\bar{X} – \mu$$) tends to converge to 0 in a probability sense.

For most cases, if we assume:

$$W=\frac{\sqrt n}{\sigma}\left(\ \bar{X}-\mu\right)=\frac{\ \bar{X}-\mu}{\sigma/\sqrt n}=\frac{Y-n\mu}{\sqrt n\sigma}$$

where $$Y$$ is the sum of a random sample of size n from some distribution with mean $$\mu$$ and variance $$\sigma^2$$, then, for each positive integer n,

$$E(W) = E\bigg[\frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \bigg]= \frac{E(\bar{X})-\mu}{\sigma/\sqrt{n}}=\frac{\mu – \mu}{\sqrt{n}\sigma}=0$$

and

$$Var(W)= E(W^2) = E\bigg[\frac{(\bar{X}-\mu)^2}{\sigma^2/n} \bigg]= \frac{E\big[(\bar{X}-\mu)^2\big]}{\sigma^2/n}=\frac{\sigma^2/n}{\sigma^2/n}=1$$

Then, when $$\bar{X}-\mu$$ tends to “reduce” to 0, the factor $$\sqrt{n}/\sigma$$ in $$\sqrt{n}(\bar{X}-\mu)/\sigma$$ starts making the probability enough to prevent this “reduction.”

But what happens to $$W$$ when n increases? If this sample comes from a normal distribution, then we know that $$\bar{X}$$ is $$N(\mu,\sigma^2/n)$$, and hence $$W$$ is $$N(0,1)$$ for each positive n. So in this limit, the distribution of $$W$$ necessarily will be $$N(0,1)$$. Circling back to the original question: if this does not depend on the underlying distribution, the answer must be $$N(0,1)$$.

Now, we can state the following theorem:

## The Central Limit Theorem

If $$\bar{X}$$ is the mean of a random sample $$X_1,X_2,\cdots,X_n$$ of size n from a distribution with a finite mean $$\mu$$ and a finite positive variance $$\sigma^2$$, then the distribution of:

$$W = \frac{\bar{X} – \mu}{\sigma/\sqrt{n}} = \frac{\sum_{i=1}^{n}X_i – n\mu}{\sqrt{n}\sigma}$$

is $$N(0,1)$$ in the limit as $$n \to \infty$$. When n is “sufficiently large”, a practical use of the central limit theorem is approximating the cdf of $$W$$:

$$P(W \leq w) \approx \int_{-\infty}^{w}\frac{1}{\sqrt{2\pi}}e^{-z^{2}/2}dz = \Phi(w).$$

An interesting thing about the Central Limit Theorem is that it does not matter what the distribution of the $$X_i\prime s$$ is; $$X_i\prime s$$ can be discrete, continuous, or mixed random variables.

For example, assume that $$X_i\prime s$$ are Bernoulli (p) random variables, then $$E[X_i]=p,\ Var\left(X_i\right)=p(1-p)$$.  Also, $$Y_n=X_1+X_2+\ldots+X_n$$ has a Binomial (n,p) distribution. Thus,

$$Z_n=\frac{Y-np}{\sqrt{np\left(1-p\right)}}$$

Where $$Y_n\sim Binomial\ \left(n,p\right)$$.

In the example, $$Z_n$$ is a discrete random variable; thus, mathematically, we refer to it as having a PMF and not a PDF. This is the reason why the Central Limit Theorem states that the CDF and not the PDF of $$Z_n$$ converge to the standard normal CDF.

A common question asked is how large n should be so that the normal approximation can be used. Using the normal approximation will generally depend on $$X_i$$’s distribution. However, a rule of thumb is often stated that if $$n\geq30$$, then a normal approximation applies.

### Steps on How to Apply the Central Limit Theorem (CLT)

Step 1: Write the random variable of interest, $$Y$$, as the sum of n independent random variables $$X_j^\prime s$$:

$$Y=X_1+X_2+\ldots+X_n$$

Step 2: Compute $$E(Y)$$ and $$Var(Y)$$ by noting that:

$$E\left(Y\right)=n\mu, \text{ and } Var\left(Y\right)=n\sigma^2$$

Where $$\mu=E(X_i)$$ and $$\sigma^2=Var(X_i)$$.

Step 3: As per the Central Limit Theorem, conclude that $$\frac{Y-E(Y)}{\sqrt{Var\left(Y\right)}}=\frac{Y-n\mu}{\sqrt n\sigma}$$ is approximately standard normal.

Hence, to find $$P\left(y_1\le Y\le y_2\right)$$, we can write,

$$P\left(y_1\le Y\le y_2\right)=P\left(\frac{y_1-n\mu}{\sqrt n\sigma}\le\frac{Y-n\mu}{\sqrt n\sigma}\le\frac{y_2-n\mu}{\sqrt n\sigma}\right)$$

Which is given by:

$$P\left(y_1\le Y\le y_2\right)=\Phi\left(\frac{y_2-n\mu}{\sqrt n\sigma}\right)-\Phi\left(\frac{y_1-n\mu}{\sqrt n\sigma}\right)$$

#### Example: Central Limit Theorem #1

Let $$\bar{X} = 18$$ and $$Var(X) = 3$$ for a random sample of $$n = 30$$. Approximate $$P(17.4 < \bar{X} < 18.5$$.

Solution

From the information given, $$\bar{X}$$ has an approximate $$N(18,3/30)$$ distribution. We can compute probabilities such as:

\begin{align} P(17.4 < \bar{X} < 18.5) & = P\bigg(\frac{17.4-18}{\sqrt{3/30}} < \frac{\bar{X}-18}{\sqrt{3/30}} < \frac{18.5-18}{\sqrt{3/30}}\bigg)\\ & \approx \Phi(0.158) – \Phi(-0.189) = 0.94295-0.02872 = 0.9142 \end{align}

#### Example: Central Limit Theorem #2

Let $$X_1,X_2,\cdots, X_{15}$$ be a random sample of size 15 from a joint random distribution. Let $$E(X_i) =\frac{1}{4}$$ and $$Var(X_i) = \frac{1}{24}$$ for $$i=1,2,\cdots, 20$$.

If $$Y$$ is a transformation $$Y = X_1 + X_2 + \cdots X_{15}$$, approximate $$P(Y \leq 4.11)$$.

Solution

\begin{align} P(Y \leq 4.11) & = P\bigg(\frac{Y – 15(1/4)}{\sqrt{15/24}}\leq \frac{4.11 – 3.75}{\sqrt{15/24}}\bigg) = P(W \leq 0.455)\\ & \approx \Phi(0.455) = 0.676 \end{align}

Notice how the formula in Example 1 is different from the formula in Example 2. In example 1, we are using a single variable with a single sample, so we are using the left side of the expression, whereas, in example 2, we are using a random sample from a random distribution with $$X_n$$ data points, so we need to weight the distribution, and we end up using the right side of the formula.

#### Example: Central Limit Theorem #3

A company offers payment for its employees; the amount paid is 10,000 for its 200 employees if they survive a set criterion. The probability of survival for each employee is 1.1%. The person who built this fund says there is a 99%  probability that the fund will handle the payouts.

Calculate the smallest amount of money that the company should put into the fund.

Solution

Let $$P$$ be the payments and $$X$$ the number of deaths, $$P = 10,000X$$, where $$X \sim Bin(200,0.011)$$.

$$E(P) = E(10,000X)= n\cdot p =10,000(200)(0.011)=22000$$

$$Var(P) = Var(10,000X) = n\cdot p \cdot (1-p)= 10,000^2(200)(0.011)(1-0.011)=217,580,000$$

$$Sd(P)=\sqrt{Var(P)}= 14,750.60$$

Since there is a probability of at least 0.99 that the fund will be able to handle the payout, then:

$$Pr\left(Z\le\frac{P-22,000}{14,750.60}\right)=0.99$$

Thus,

$$\Rightarrow\Phi\left(\frac{P-22,000}{14,750.60}\right)=0.99$$

Intuitively,

$$\frac{P-22,000}{14,750.60}=\Phi^{-1}(0.99)$$

$$\therefore P=22,000+14,750.60\left(2.326\right)=56,309.90$$

The value $$2.326$$ is nothing more than our application of the Central Limit Theorem ($$\Phi(0.99)$$).

Learning Outcome

Topic 3.i: Multivariate Random Variables – State and apply the Central Limit Theorem.

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