{"id":3028,"date":"2019-06-28T15:45:41","date_gmt":"2019-06-28T15:45:41","guid":{"rendered":"https:\/\/analystprep.com\/study-notes\/?p=3028"},"modified":"2022-01-25T07:46:12","modified_gmt":"2022-01-25T07:46:12","slug":"state-and-apply-the-central-limit-theorem","status":"publish","type":"post","link":"https:\/\/analystprep.com\/study-notes\/actuarial-exams\/soa\/p-probability\/multivariate-random-variables\/state-and-apply-the-central-limit-theorem\/","title":{"rendered":"State and apply the Central Limit Theorem"},"content":{"rendered":"

For this learning objective, a certain knowledge of the normal distribution and knowing how to use the Z-table is assumed.<\/em><\/p>\n

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.<\/p>\n

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

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

As n<\/em> increases, the variance of \\(\\bar{X}\\) decreases. Consequently, the distribution of \\(\\bar{X}\\) clearly depends on n<\/em>, and we see that we are dealing with sequences of distributions.<\/p>\n

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

$$ Y = \\sum_{i=1}^{n}c_iX_i$$<\/p>\n

has the normal distribution:<\/p>\n

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

This can be proved by applying the moment generating technique to the linear function.<\/p>\n

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

For most cases, if we assume:<\/p>\n

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

where \\(Y\\) is the sum of a random sample of size n<\/em> from some distribution with mean \\(\\mu\\) and variance \\(\\sigma^2\\), then, for each positive integer n<\/em>,<\/p>\n

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

and<\/p>\n

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

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.”<\/p>\n

But what happens to \\(W\\) when n<\/em> 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<\/em>. 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)\\).<\/p>\n

Now, we can state the following theorem:<\/p>\n

The Central Limit Theorem<\/strong><\/h2>\n

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

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

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

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

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.<\/p>\n

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,<\/p>\n

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

Where \\(Y_n\\sim Binomial\\ \\left(n,p\\right)\\).<\/p>\n

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.<\/p>\n

A common question asked is how large n<\/em> 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.<\/p>\n

Steps on How to Apply the Central Limit Theorem (CLT)<\/h3>\n

Step 1: <\/strong>Write the random variable of interest, \\(Y\\), as the sum of n<\/em> independent random variables \\(X_j^\\prime s\\):<\/p>\n

$$Y=X_1+X_2+\\ldots+X_n$$<\/p>\n

Step 2: <\/strong>Compute \\(E(Y)\\) and \\(Var(Y)\\) by noting that:<\/p>\n

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

Where \\(\\mu=E(X_i)\\) and \\(\\sigma^2=Var(X_i)\\).<\/p>\n

Step 3: <\/strong>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.<\/p>\n

Hence, to find \\(P\\left(y_1\\le Y\\le y_2\\right)\\), we can write,<\/p>\n

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

Which is given by:<\/p>\n

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

Example: Central Limit Theorem #1<\/strong><\/h4>\n

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

Solution<\/strong><\/p>\n

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

\\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}<\/p>\n

Example: Central Limit Theorem #2<\/strong><\/h4>\n

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\\).<\/p>\n

If \\(Y\\) is a transformation \\(Y = X_1 + X_2 + \\cdots X_{15}\\), approximate \\(P(Y \\leq 4.11)\\).<\/p>\n

Solution<\/strong><\/p>\n

\\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}<\/p>\n

Notice how the formula in Example 1<\/strong><\/em> is different from the formula in Example 2<\/strong><\/em>. In example 1<\/strong><\/em>, we are using a single variable with a single sample, so we are using the left side of the expression, whereas, in example 2<\/strong><\/em>, 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.<\/p>\n

Example: Central Limit Theorem<\/strong> #3<\/strong><\/h4>\n

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%\u00a0 probability that the fund will handle the payouts.<\/p>\n

Calculate the smallest amount of money that the company should put into the fund.<\/p>\n

Solution<\/strong><\/p>\n

Let \\(P\\) be the payments and \\(X\\) the number of deaths, \\(P = 10,000X\\), where \\( X \\sim Bin(200,0.011)\\).<\/p>\n

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

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

$$Sd(P)=\\sqrt{Var(P)}= 14,750.60$$<\/p>\n

Since there is a probability of at least 0.99 that the fund will be able to handle the payout, then:<\/p>\n

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

Thus,<\/p>\n

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

Intuitively,<\/p>\n

$$\\frac{P-22,000}{14,750.60}=\\Phi^{-1}(0.99)$$<\/p>\n

$$\\therefore P=22,000+14,750.60\\left(2.326\\right)=56,309.90$$<\/p>\n

The value \\(2.326\\) is nothing more than our application of the Central Limit Theorem (\\(\\Phi(0.99)\\)).<\/p>\n

Learning Outcome<\/strong><\/em><\/p>\n

Topic 3.i:\u00a0Multivariate Random Variables –\u00a0State and apply the Central Limit Theorem.<\/strong><\/em><\/p>\n","protected":false},"excerpt":{"rendered":"

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