Raising the bar

Chapter 5. Discrete Random Variables and Probability Distributions

Unit 5.2. General properties of means

Exercise 5.1. With probability 0.001 a strong earthquake destroys your house this year and then your loss is $latex \$200,000$. Otherwise, your loss is just normal wear and tear, estimated at $latex \$4,000$ per year. What is your expected loss?

Sometimes I ask questions before giving formal definitions, to prompt you to guess. Guessing and inductive reasoning are important parts of our thinking process. In economics there is a French term "tâtonnement" (groping one's way in the dark). There are cases when you clearly see the idea and the solution is straightforward. In more complex situations there may be several competing ideas, and some of them may not be clear, looking like unfamiliar creatures in the dark. Looking for them and nurturing them until they shape up so that you can test them is a great skill. This is where you need to collect all the relevant information, mull it over again and again, until it becomes your second nature. This process is more time-consuming than getting a ready answer from your professor or classmate but the rewards are also greater.

The expected loss is $latex (-4{,}000) \times 0.999 + (-200{,}000) \times 0.001 = -4{,}196$. The general idea is that it is a weighted sum of the values of the variable of interest (the loss function). We attach a higher importance to the value that has a higher probability.

In research ideas go before definitions.

Expected value of a discrete random variable. Using the working definition of a random variable, we define the expected value of $latex X$ (also called a population mean, or an average) to be

$latex EX = {x_1}{p_1} + ... + {x_n}{p_n} = \sum\limits_{i = 1}^n {{x_i}{p_i}}. \qquad (\textrm{Eq. 5.1})$

Comments. (1) Expected value is a function whose argument is a complex object (it is described by Table 5.1, see Random variable) and the value is simple: $latex EX$ is just a number. And it is not a product of $latex E$ and $latex X$!

(2) Relation to the sample mean. A discrete random variable is said to be uniformly distributed if $latex {p_1} = ... = {p_n} = 1/n$. In this case the expected value is just the sample mean:

$latex EX = {x_1}\frac{1}{n} + ... + {x_n}\frac{1}{n} = \frac{{{x_1} + ... + {x_n}}}{n} = \bar X.$

(3) Grouped data formula. Recall the procedure for finding relative frequencies. Denote $latex {y_1},...,{y_n}$ the values in the sample. Equal values are joined in groups. Let $latex {x_1},...,{x_m}$ denote the distinct values and $latex {n_1},...,{n_m}$ their absolute frequencies. Their total is, clearly, $latex n$. The sample mean is

$latex \bar Y = ({y_1} + ... + {y_n})/n \qquad (\textrm{Eq. 5.2})$

(sorting out $latex y$'s into groups with equal values)

$latex = \left( \overbrace{ {x_1} + ... + {x_1} }^{{n_1}\textrm{ times}} + ... + \overbrace{ {x_m} + ... + {x_m} }^{{n_m}\textrm{ times}} \right)/n$

$latex = ({n_1}{x_1} + ... + {n_m}{x_m})/n$ (this is the grouped data formula)

(dividing through by $latex n$ and using relative frequencies)

$latex = {r_1}{x_1} + ... + {r_m}{x_m} = EX.$

In this context, the sample mean $latex \bar Y$, the grouped data formula (in the fourth line) and the expected value are the same thing. The weighted mean formula from NCT is the same thing as the grouped data formula with $latex {n_i} = {w_i}$ and $latex n = \sum {{w_i}}$. Note also that when data are grouped by intervals, in the grouped data formula the midpoints $latex {m_i}$ of the intervals are taken as the values of the variables.

(4) When the sample size becomes very large, the relative frequencies approach the true probabilities by the Law of Large Numbers. Therefore Eq. (5.2) gives rise to the following intuitive interpretation of the expected value: $latex EX$ is what $latex \bar X$ approaches when $latex n$ is very large.

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