0.25
3
0.33
7
0.50
5
0.33
12
0.25
8
0.34

Performance of Two Investments

To rank these two investments under the Standard Dominance Approach in Finance, first we must compute the mean and standard deviation and then analyze the results. Using the Multinomial for calculation, we notice that the Investment I has mean = 6.75% and standard deviation = 3.9%, while the second investment has mean = 5.36% and standard deviation = 2.06%. First observe that under the usual mean-variance analysis, these two investments cannot be ranked. This is because the first investment has the greater mean; it also has the greater standard deviation; therefore, the Standard Dominance Approach is not a useful tool here. We have to resort to the coefficient of variation (C.V.) as a systematic basis of comparison. The C.V. for Investment I is 57.74% and for Investment II is 38.43%. Therefore, Investment II has preference over the Investment I. Clearly, this approach can be used to rank any number of alternative investments. Notice that less variation in return on investment implies less risk.

Expectation of a sum of a random number of random variables: Suppose that the number of people entering a department store on a given day is a random variable with mean 50. Suppose further that the amount of money spent by these customers is independent random variables having a common mean of $80. What is the expected amount of money spent in the store on a given day?.

E (sum of N random variables Xi) = E(N) . E(X)

Hence, the expected amount of money spent in the store is (50)(80) = $4000.

You might like to use this JavaScript in performing some numerical experimentation to:

  1. Show that E[aX + b] = aE(X) + b.
  2. Show that V[aX + b] = a2V(X).
  3. Show that: E(X2)= V(X) + (E(X))2.

Normal Density Function

In the Descriptive Statistic Section of this Web site, we have been concerned with how empirical scores are distributed and how best to describe their distribution. We have discussed several different measures, but the mean m will be the measure that we use to describe the center of the distribution, and the standard deviation s will be the measure we use to describe the spread of the distribution. Knowing these two facts gives us ample information to make statements about the probability of observing a certain value within that distribution. If I know, for example, that the average Intelligence Quotient (I.Q.) score is 100 with a standard deviation of s = 20, then I know that someone with an I.Q. of 140 is very smart. I know this because 140 deviates from the mean mby twice the average amount as the rest of the scores in the distribution. Thus, it is unlikely to see a score as extreme as 140 because most of the I.Q. scores are clustered around 100 and only deviate 20 points from the mean m .

Many applications arise from the central limit theorem (CLT). The CLT states that, average of values of n observations approaches normal distribution, irrespective of the form of original distribution under quite general conditions. Consequently, normal distribution is an appropriate model for many, but not all, physical phenomena, such as distribution of physical measurements on living organisms, intelligence test scores, product dimensions, average temperatures, and so on.

Know that the Normal distribution is to satisfy seven requirements: (1) the graph should be bell shaped curve; (2) mean, median and mode are all equal; (3) mean, median and mode are located at the center of the distribution; (4) it has only one mode, (5) it is symmetric about mean, (6) it is a continuous function; (6) it never touches x-axis; and (7) the area under curve equals one.

Many methods of statistical analysis presume normal distribution.

When we know the mean and variance of a Normal then it allows us to find probabilities. So, if, for example, you knew some things about the average height of women in the nation, including the fact that heights are distributed normally, you could measure all the women in your extended family and find the average height. This enables you to determine a probability associated with your result, if the probability of getting your result, given your knowledge of women nationwide, is high. Then your family's female height cannot be said to be different from average. If that probability is low, then your result is rare (given the knowledge about women nationwide), and you can say your family is different. You have just completed a test of the hypothesis that the average height of women in your family is different from the overall average.

The ratio of two independent observations from the standard normal is distributed as the Cauchy Distribution which has thicker tails than a normal distribution. It density function is f(x) = 1/[p(1+x2)], for all real value x.


An Application: A portfolio manager believes that the overnight loss of his portfolio is distributed normally with mean $0 and standard deviation of $10 000. Find the 5% one-day value at risk for this portfolio.

Let X denotes the random portfolio loss distributed as X ~ N (0, 10 0002). The value at risk v5% is defined by definition a number such that

P(X £ v5%) = 0.95.

To find v5% we standardize the random variable on the left-hand side:

X £ v5%     Û     X – 0 £ v5% – 0    Û    [X – 0] / [10 000] £ [v5% – 0] / [10 000].

The transformation is denoted by Z = (X - 0) / 10 000 which has standard normal distribution. Therefore,

P{} = 0.95.

If we denote by z95% the 95% quantile of a standard normal distribution, then

[v5%] / [10 000]   =   z95%

v5% can be found in normal statistical table:

z95%  = 1.645, v95%   = 10 000z95%  = 16 450

Therefore, the overnight 5% value at risk is $16450.

You might like to use Standard Normal JavaScript instead of using tabular values from your textbook, and the well-known Lilliefors' Test for Normality to assess the goodness-of-fit.

Poisson Probability Function

Life is good for only two things, discovering mathematics and teaching mathematics.
-- Simeon Poisson                   

An important class of decision problems under uncertainty is characterized by the small chance of the occurrence of a particular event, such as an accident. Poisson probability function computes the probability of exactly x independent occurrences during a given period of time, if events take place independently and at a constant rate. Poisson probability function also represent number of occurrences over constant areas or volumes:

Poisson probabilities are often used; for example in quality control, software and hardware reliability, insurance claim, number of incoming telephone calls, and queuing theory.

Application: Gives probability of exactly x independent occurrences during a given period of time if events take place independently and at a constant rate. May also represents number of occurrences over constant areas or volumes. It is used frequently in quality control, reliability, queuing theory, and so on.

Example: Used to represent distribution of number of defects in a piece of material, customer arrivals, insurance claims, incoming telephone calls, alpha particles emitted, and so on.

A process that creates fabric is monitored. If the number of defects (X) per meter of fabric exceeds 5 then the process is stopped for diagnosis. The random variable X follows a Poisson distribution with rate = number of defects per meter of fabric.

An Application: One of the most useful applications of the Poisson distribution is in the field of queuing theory. In many situations where queues occur it has been shown that the number of people joining the queue in a given time period follows the Poisson model. For example, if the rate of arrivals to an emergency room is l per unit of time period (say 1 hr), then:

P ( n arrivals) = ln  e-l / n!

The mean and variance of random variable n are both l . However if the mean and variance of a random variable have equal numerical values, then it is not necessary that its distribution is a Poisson. Its mode is within interval [l -1, l].

Applications:

P ( 0 arrival) = e-l
P ( 1 arrival) = l  e-l / 1!
P ( 2 arrival) = l2  e-l/ 2!

and so on. In general:

P ( n+1 arrivals ) = l P ( n arrivals ) / n.

Normal approximation for Poisson: All Poisson tables are limited in their scope; therefore, it is necessary to use standard normal distribution in computing the Poisson probabilities. The following numerical example illustrates how good the approximation could be.

Numerical Example: Emergency patients arrive at a large hospital at the rate of 0.033 per minute. What is the probability of exactly two arrivals during the next 30 minutes?

The arrival rate during 30 minutes is l = (30)(0.033) = 1. Therefore,

P (2 arrivals) = [12 /(2!)] e-1 = 18%

The mean and standard deviation of distribution are:

m = l = 1, and s = l 1/2 = 1,

respectively; therefore, the standardized observation for n = 2, by using the continuity factor (which always enlarges) are:

z1 = [(r-1/2) - m] / s = (1.5 -1)/1 = 0.5, and

z2 = [(r+1/2) - m] / s = (2.5 -1)/1 = 1.5.

Therefore, the approximated P (2 arrivals) is P (z being within the interval 0.5, 1.5). Now, by using the standard normal table, we obtain:

P (2 arrivals) = 0.43319 - 0.19146 = 24%

As you see the approximation is slightly overestimated, therefore the error is on the safe side. For large values of l, say over 20, one may use the Normal approximation to calculate Poisson probabilities.

Notice that by taking the square root of a Poisson random variable, the transformed variable is more symmetric. This is a useful transformation in regression analysis of Poisson observations.

Poisson approximation for binomial: