Steven R. Dunbar
Department of Mathematics
203 Avery Hall
University of Nebraska-Lincoln
Lincoln, NE 68588-0130
http://www.math.unl.edu
Voice: 402-472-3731
Fax: 402-472-8466
Stochastic Processes and
Advanced Mathematical Finance
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The Central Limit Theorem
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Mathematically Mature: may contain mathematics beyond calculus with proofs.
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What is the most important probability distribution? Why do you choose that distribution as most important?
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and let $Z$ be the “standard” normally distributed random variable with mean $0$ and variance $1$. Then ${Z}_{n}$ converges in distribution to $Z$, that is:
In words, a shifted and rescaled sample distribution is approximately standard normal.
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Lemma 1. Let ${X}_{1},{X}_{2},\dots $ be a sequence of random variables having cumulative distribution functions ${F}_{{X}_{n}}$ and moment generating functions ${\varphi}_{{X}_{n}}$. Let $X$ be a random variable having cumulative distribution function ${F}_{X}$ and moment generating function ${\varphi}_{X}$. If ${\varphi}_{{X}_{n}}\left(t\right)\to {\varphi}_{X}\left(t\right)$, for all $t$, then ${F}_{{X}_{n}}\left(t\right)\to {F}_{X}\left(t\right)$ for all $t$ at which ${F}_{X}\left(t\right)$ is continuous.
We say that the sequence ${X}_{i}$ converges in distribution to $X$ and we write
Notice that $\mathbb{P}\left[a<{X}_{i}\le b\right]={F}_{{X}_{i}}\left(b\right)-{F}_{{X}_{i}}\left(a\right)\to F\left(b\right)-F\left(a\right)=\mathbb{P}\left[a<X\le b\right]$, so convergence in distribution implies convergence of probabilities of events. Likewise, convergence of probabilities of events implies convergence in distribution.
This lemma is useful because it is routine to determine the pointwise limit of a sequence of functions using ideas from calculus. It is usually much easier to check the pointwise convergence of the moment generating functions than it is to check the convergence in distribution of the corresponding sequence of random variables.
We won’t prove this lemma, since it would take us too far aﬁeld into the theory of moment generating functions and corresponding distribution theorems. However, the proof is a routine application of ideas from the mathematical theory of real analysis.
Here’s a simple representative example of using the convergence of the moment generating function to prove a useful result. We will prove a version of the Weak Law of Large numbers that does not require the ﬁnite variance of the sequence of independent, identically distributed random variables.
Theorem 2 (Weak Law of Large Numbers). Let ${X}_{1},\dots ,{X}_{n}$ be independent, identically distributed random variables each with mean $\mu $ and such that $\mathbb{E}\left[\left|X\right|\right]$ is ﬁnite. Let ${S}_{n}={X}_{1}+\cdots +{X}_{n}$. Then ${S}_{n}\u2215n$ converges in probability to $\mu $. That is:
Proof. If we denote the moment generating function of $X$ by $\varphi \left(t\right)$, then the moment generating function of
is ${\left(\varphi \left(t\u2215n\right)\right)}^{n}$. The existence of the ﬁrst moment assures us that $\varphi \left(t\right)$ is diﬀerentiable at $0$ with a derivative equal to $\mu $. Therefore, by tangent-line approximation (ﬁrst-degree Taylor polynomial approximation)
where ${r}_{2}\left(t\u2215n\right)$ is a error term such that
This is equivalent to $\left(1\u2215t\right)\underset{n\to \infty}{lim}n{r}_{2}\left(t\u2215n\right)=0$ or just $\underset{n\to \infty}{lim}n{r}_{2}\left(t\u2215n\right)=0$, needed for taking the limit in (1). Then we need to consider
$$\varphi {\left(\frac{t}{n}\right)}^{n}={\left(1+\mu \frac{t}{n}+{r}_{2}\left(t\u2215n\right)\right)}^{n}.$$ | (1) |
Taking the logarithm of ${\left(1+\mu \left(t\u2215n\right)+r\left(t\u2215n\right)\right)}^{n}$ and using L’Hospital’s Rule, we see that
But this last expression is the moment generating function of the (degenerate) point mass distribution concentrated at $\mu $. Hence,
□
Theorem 3 (Central Limit Theorem). Let random variables ${X}_{1},\dots {X}_{n}$
Consider ${S}_{n}={\sum}_{i=1}^{n}{X}_{i}.$ Let
Then ${Z}_{n}$ converges in distribution to $Z$, that is:
Remark. The Central Limit Theorem is true even under the slightly weaker assumptions that ${X}_{1},\dots {X}_{n}$ only are independent and identically distributed with ﬁnite mean $\mu $ and ﬁnite variance ${\sigma}^{2}$ without the assumption that moment generating function exists. However, the proof below using moment generating functions is simple and direct enough to justify using the additional hypothesis.
Proof. Assume at ﬁrst that $\mu =0$ and ${\sigma}^{2}=1$. Assume also that the moment generating function of the ${X}_{i}$, (which are identically distributed, so there is only one m.g.f) is ${\varphi}_{X}\left(t\right)$, exists and is everywhere ﬁnite. Then the m.g.f of ${X}_{i}\u2215\sqrt{n}$ is
Recall that the m.g.f of a sum of independent r.v.s is the product of the m.g.f.s. Thus the m.g.f of ${S}_{n}\u2215\sqrt{n}$ is (note that here we used $\mu =0$ and ${\sigma}^{2}=1$)
The quadratic approximation (second-degree Taylor polynomial expansion) of ${\varphi}_{X}\left(t\right)$ at $0$ is by calculus:
again since the hypotheses assume $\mathbb{E}\left[X\right]={\varphi}^{\prime}\left(0\right)=0$ and $Var\left[X\right]=\mathbb{E}\left[{X}^{2}\right]-{\left(\mathbb{E}\left[X\right]\right)}^{2}={\varphi}^{\u2033}\left(0\right)-{\left({\varphi}^{\prime}\left(0\right)\right)}^{2}={\varphi}^{\u2033}\left(0\right)=1$. Here ${r}_{3}\left(t\right)$ is an error term such that $\underset{t\to 0}{lim}{r}_{3}\left(t\right)\u2215{t}^{2}=0$. Thus,
implying that
Now by some standard results from calculus,
as $n\to \infty $. (If the reader needs convincing, it’s easy to show that
using L’Hospital’s Rule to account for the ${r}_{3}\left(t\right)$ term.)
To handle the general case, consider the standardized random variables $\left({X}_{i}-\mu \right)\u2215\sigma $, each of which now has mean $0$ and variance $1$ and apply the result. □
Abraham de Moivre proved the ﬁrst version of the central limit theorem around 1733 in the special case when the ${X}_{i}$ are binomial random variables with $p=1\u22152=q$. Pierre-Simon Laplace subsequently extended the proof to the case of arbitrary $p\ne q$. Laplace also discovered the more general form of the Central Limit Theorem presented here. His proof however was not completely rigorous, and in fact, cannot be made completely rigorous. A truly rigorous proof of the Central Limit Theorem was ﬁrst presented by the Russian mathematician Aleksandr Liapunov in 1901-1902. As a result, the Central Limit Theorem (or a slightly stronger version of the Central Limit Theorem) is occasionally referred to as Liapunov’s theorem. A theorem with weaker hypotheses but with equally strong conclusion is Lindeberg’s Theorem of 1922. It says that the sequence of random variables need not be identically distributed, but instead need only have zero means, and the individual variances are small compared to their sum.
The statement of the Central Limit Theorem does not say how good the approximation is. One rule of thumb is that the approximation given by the Central Limit Theorem applied to a sequence of Bernoulli random trials or equivalently to a binomial random variable is acceptable when $np\left(1-p\right)>18$ [2, page 34], [3, page 134]. The normal approximation to a binomial deteriorates as the interval $\left(a,b\right)$ over which the probability is computed moves away from the binomial’s mean value $np$. Another rule of thumb is that the normal approximation is acceptable when $n\ge 30$ for all “reasonable” probability distributions.
The Berry-Esséen Theorem gives an explicit bound: For independent, identically distributed random variables ${X}_{i}$ with $\mu =\mathbb{E}\left[{X}_{i}\right]=0$, ${\sigma}^{2}=\mathbb{E}\left[{X}_{i}^{2}\right]$, and $\rho =\mathbb{E}\left[\left|{X}^{3}\right|\right]$, then
Figure 1 is a graphical illustration of the Central Limit Theorem. More precisely, this is an illustration of the de Moivre-Laplace version, the approximation of the binomial distribution with the normal distribution.
The ﬁgure is actually an non-centered and unscaled illustration since the binomial random variable ${S}_{n}$ is not shifted by the mean, nor normalized to unit variance. Therefore, the binomial and the corresponding approximating normal are both centered at $\mathbb{E}\left[{S}_{n}\right]=np$. The variance of the approximating normal is ${\sigma}^{2}=\sqrt{npq}$ and the widths of the bars denoting the binomial probabilities are all unit width, and the heights of the bars are the actual binomial probabilities.
From the Central Limit Theorem we expect the normal distribution applies whenever an outcome results from numerous small additive eﬀects with no single or small group of eﬀects dominant. Here is a standard illustration of that principle.
Consider the following data from the National Longitudinal Survey of Youth (NLSY). This study started with 12,000 respondents aged 14-21 years in 1979. By 1994, the respondents were 29-36 years old and had 15,000 children among them. Of the respondents 2,444 had exactly two children. In these 2,444 families, the distribution of children was boy-boy: 582; girl-girl 530, boy-girl 666, and girl-boy 666. It appears that the distribution of girl-girl family sequences is low compared to the other combinations. Our intuition tells us that all combinations should be equally likely and should appear in roughly equal proportions. We will assess this intuition with the Central Limit Theorem.
Consider a sequence of 2,444 trials with each of the two-child families. Let ${X}_{i}=1$ (success) if the two-child family is girl-girl, and ${X}_{i}=0$ (failure) if the two-child family is otherwise. We are interested in the probability distribution of
In particular, we are interested in the probability ${\mathbb{P}}_{n}\left[{S}_{2444}\le 530\right]$, that is, what is the probability of seeing as few as 530 girl-girl families or even fewer in a sample of 2444 families? We can use the Central Limit Theorem to estimate this probability.
We are assuming the family “success” variables ${X}_{i}$ are independent, and identically distributed, a reasonable but arguable assumption. Nevertheless, without this assumption, we cannot justify the use of the Central Limit Theorem, so we adopt the assumption. Then $\mu =\mathbb{E}\left[{X}_{i}\right]=\left(1\u22154\right)\cdot 1+\left(3\u22154\right)\cdot 0=1\u22154$ and $Var\left[{X}_{i}\right]=\left(1\u22154\right)\left(3\u22154\right)=3\u221516$ so $\sigma =\sqrt{3}\u22154$ Note that $2444\cdot \left(1\u22154\right)\cdot \left(3\u22154\right)=45.75>18$ so the rule of thumb justiﬁes the use of the Central Limit Theorem. Hence
$$\begin{array}{llll}\hfill {\mathbb{P}}_{n}\left[{S}_{2444}\le 530\right]& ={\mathbb{P}}_{n}\left[\frac{{S}_{2444}-2444\cdot \left(1\u22154\right)}{\left(\sqrt{3}\u22154\cdot \sqrt{2444}\right)}\le \frac{530-2444\cdot \left(1\u22154\right)}{\left(\sqrt{3}\u22154\cdot \sqrt{2444}\right)}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx \mathbb{P}\left[Z\le -3.7838\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 0.0000772\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$It is highly unlikely that under our assumptions such a proportion would have occurred. Therefore, we are justiﬁed in thinking that under our assumptions, the actual proportion of girl-girl families is low. We then begin to suspect our assumptions, one of which was the implicit assumption that the appearance of girls was equally likely as boys, leading to equal proportions of the four types of families. In fact, there is ample evidence that the birth of boys is more likely than the birth of girls.
We expect the normal distribution to apply whenever the numerical description of a state of a system results from numerous small additive eﬀects, with no single or small group of eﬀects dominant. Here is another illustration of that principle.
We can us the Central Limit Theorem to assess risk. Two large banks compete for customers to take out loans. The banks have comparable oﬀerings. Assume that each bank has a certain amount of funds available for loans to customers. Any customers seeking a loan beyond the available funds will cost the bank, either as a lost opportunity cost, or because the bank itself has to borrow to secure the funds to lend to the customer. If too few customers take out loans then that also costs the bank since now the bank has unused funds.
We create a simple mathematical model of this situation. We suppose that the loans are all of equal size and for deﬁniteness each bank has funds available for a certain number (to be determined) of these loans. Then suppose $n$ customers select a bank independently and at random. Let ${X}_{i}=1$ if customer $i$ selects bank H with probability $1\u22152$ and ${X}_{i}=0$ if customers select bank T, also with probability $1\u22152$. Then ${S}_{n}={\sum}_{i=1}^{n}{X}_{i}$ is the number of loans from bank H to customers. Now there is some positive probability that more customers will turn up than the bank can accommodate. We can approximate this probability with the Central Limit Theorem:
$$\begin{array}{llll}\hfill \mathbb{P}\left[{S}_{n}>s\right]& ={\mathbb{P}}_{n}\left[\left({S}_{n}-n\u22152\right)\u2215\left(\left(1\u22152\right)\sqrt{n}\right)>\left(s-n\u22152\right)\u2215\left(\left(1\u22152\right)\sqrt{n}\right)\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx \mathbb{P}\left[Z>\left(s-n\u22152\right)\u2215\left(\left(1\u22152\right)\sqrt{n}\right)\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathbb{P}\left[Z>\left(2s-n\right)\u2215\sqrt{n}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Now if $n$ is large enough that this probability is less than (say) $0.01$, then the number of loans will be suﬃcient in 99 of 100 cases. Looking up the value in a normal probability table,
so if $n=1000$, then $s=537$ will suﬃce. If both banks assume the same risk of sellout at $0.01$, then each will have $537$ for a total of 1074 loans, of which 74 will be unused. In the same way, if the bank is willing to assume a risk of $0.20$, i.e. having enough loans in 80 of 100 cases, then they would need funds for 514 loans, and if the bank wants to have suﬃcient loans in 999 out of 1000 cases, the bank should have 549 loans available.
Now the possibilities for generalization and extension are apparent. A ﬁrst generalization would be allow the loan amounts to be random with some distribution. Still we could apply the Central Limit Theorem to approximate the demand on available funds. Second, the cost of either unused funds or lost business could be multiplied by the chance of occurring. The total of the products would be an expected cost, which could then be minimized.
The proofs in this section are adapted from Chapter 8, “Limit Theorems”, A First Course in Probability, by Sheldon Ross, Macmillan, 1976. Further examples and considerations come from Heads or Tails: An Introduction to Limit Theorems in Probability, by Emmanuel Lesigne, American Mathematical Society, Chapter 7, pages 29–74. Illustration 1 is adapted from Dicing with Death: Chance, Health, and Risk by Stephen Senn, Cambridge University Press, Cambridge, 2003. Illustration 2 is adapted from An Introduction to Probability Theory and Its Applications, Volume I, second edition, William Feller, J. Wiley and Sons, 1957, Chapter VII.
The experiment is ﬂipping a coin $n$ times, and repeat the experiment $k$ times. Then compute the proportion for which the deviation of the sample sum from $np$ by more than $\sqrt{p\left(1-p\right)n}$ is less than $a$. Compare this to the theoretical probability from the standard normal distribution.
R script for the Central Limit Theorem.
Octave script for the Central Limit Theorem.
Perl PDL script for the Central Limit Theorem.
Scientiﬁc Python script for the Central Limit Theorem.
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where ${X}_{1},{X}_{2},\dots $ are independent, identically distributed continuous random variables with mean $0$ and variance ${\sigma}^{2}$. (Note that this is an additive assumption about the change in a stock price. In the binomial tree models, we assumed that a stock’s price changes by a multiplicative factor up or down. We will have more to say about these two distinct models later.) Suppose that a stock’s price today is $100$. If ${\sigma}^{2}=1$, what can you say about the probability that after $10$ days, the stock’s price will be between $95$ and $105$ on the tenth day?
where ${X}_{1},{X}_{2},\dots $ are independent, identically distributed continuous random variables with mean $0$ and variance ${\sigma}^{2}$. Write an expression for the probability that you do not recover your purchase price.
Explain with a reason whether or not you expect the approximation to be a good approximation.
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[1] William Feller. An Introduction to Probability Theory and Its Applications, Volume I, volume I. John Wiley and Sons, third edition, 1973. QA 273 F3712.
[2] Emmanuel Lesigne. Heads or Tails: An Introduction to Limit Theorems in Probability, volume 28 of Student Mathematical Library. American Mathematical Society, 2005.
[3] Sheldon Ross. A First Course in Probability. Macmillan, 1976.
[4] Stephen Senn. Dicing with Death: Chance, Health and Risk. Cambridge University Press, 2003.
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