It will be useful to note a couple of the properties

2.4 Examples of Discrete Random

of the variance.

Variables

First,

Var(X) = E (X ’ µ X )2 2.4.1 Uniform Distribution. A discrete random

variable X that takes the K different values in a set

= E X2 ’ 2Xµ X + µ2

X

= {x1 , . . . , x K } with equal likelihood is called

= E X2 ’ 2µ X E(X) + E µ2 a uniform random variable. Its probability function

X

= E X2 ’ µ2 . is given by

X

if x ∈

1/K

f X (x) =

The third step in this derivation, distributing

0 otherwise.

the expectation operator, is accomplished by

Note that the speci¬cation of this distribution

applying properties (2.4) and (2.5). The last step

depends upon K parameters, namely the K

is achieved by applying the expectation operator

different values that can be taken. We use the

and simplifying the third line.

shorthand notation

Second, if a random variable is shifted by a

X ∼ U( )

constant, its variance does not change. Adding a

constant shifts the realizations of X to the left

to indicate that X is uniformly distributed on . If

or right, but it does not change the dispersion of

the K values are given by

those realizations. On the other hand, multiplying

k’1

xk = a + (b ’ a), for k = 1, . . . , K

a random variable by a constant does change the

K ’1

dispersion of its realizations. Thus, if a and b are

for some a < b, then the parameters of the uniform

constants, then

distribution are the three numbers a, b, and K . It

Var(aX + b) = a 2 Var(X). (2.6) is readily shown that the mean and variance of a

discrete uniform random variable are given by

2.3.5 Random Vectors. Until now we have E U(a, b, K ) = (a + b)/2

considered the case in which a single random

variable is de¬ned on a sample space. However, Var U(a, b, K ) = (b ’ a) /12.

2

we are generally interested in situations in which Note that the mean and variance do not depend on

more than one random variable is de¬ned on K .

2: Probability Theory

24

2.4.2 Binomial Distribution. We have already occur in n ways, the probability of observing

h

discussed the binomial distribution in the coin

n

this event is h p h (1 ’ p)n’h .

tossing and model validation examples [2.2.2].

When an experiment consists of n independent Hence the binomial distribution is de¬ned by

tosses of a fair coin, the number of heads H that ±

n p h (1 ’ p)n’h for 0 ¤ h ¤ n

come up is a binomial random variable. Recall

h

f H (h) =

n

that the sample space for this experiment has 2

n 0 otherwise.

equally likely elements and that there are h

(2.7)

ways to observe the event {H = h}. This random

variable H has probability function We can readily verify that this is indeed a proper

probability distribution. First, the condition that

n 1 n.

f H (h) = h f H ≥ 0 is clearly satis¬ed. Second,

2

n n

nh

f H (h) = h p (1 ’ p)

n’h

In general, the ˜coin™ is not fair. For example,

consider sequences of n independent daily h=0 h=0

observations of West Glacier rainfall [2.1.2] and n

= p + (1 ’ p) = 1.

classify each observation into two categories

Thus, the probabilities sum to 1 as required.

depending upon whether the rainfall exceeds the

The shorthand H ∼ B(n, p) is used to

0.1 inch threshold. This natural experiment has

indicate that H has a binomial distribution with

the same number of possible outcomes as the coin

tossing experiment (i.e., 2n ), but all outcomes are two parameters: the number of trials n and the

probability of success p. The mean and variance

not equally likely.

of H are given by

The coin tossing and West Glacier experiments

are both examples of binomial experiments. That

E(H) = np (2.8)

is, they are experiments that:

Var(H) = np(1 ’ p). (2.9)

• consist of n independent Bernoulli trials, and

2.4.3 Example: Rainfall Forecast. Consider

• have the same probability of success on every

again the daily rainfall at West Glacier, Wash-

trial.

ington. Let R be the event that the daily rainfall

A binomial random variable is de¬ned as the exceeds the 0.1 inch threshold and let ¬R be

number of successes obtained in a binomial the complement (i.e., rain does not exceed the

threshold).

experiment.

Let us now suppose that a forecast scheme has

The probability distribution of a binomial f

random variable H is derived as follows. Let S been devised with two outcomes: R = there will

denote a ˜success™ and assume that there are n be more than 0.1 inch of precipitation and ¬R .

f

trials and that P (S) = p on any trial. What is The binomial distribution can be used to assess the

the probability of observing H = h? One way to skill of categorical forecasts of this type.

The probability of threshold exceedance at West

obtain {H = h} is to observe

Glacier is 0.38 (i.e., P (R) = 0.38). Suppose that

the forecasting procedure has been tuned so that

h times

P R f = P (R).

SSS · · · S F F F · · · F .

Assume ¬rst that the forecast has no skill, that is,

n’h times

that it is statistically independent of nature. Let C

Since the trials are independent, we may apply denote a correct forecast. Using (2.1) and (2.3) we

see that the probability of a correct forecast when

(2.3) repeatedly to show that

there is ˜no skill™ is

P (SSS · · · S F F F · · · F) = p (1 ’ p) .

h n’h

P (C) = P R f — P (R)

+ P ¬R f — P (¬R)

Also, because of independence, we get the

same result regardless of the order in which = 0.382 + 0.622 ≈ 0.53.

the successes and failures occur. Therefore all

outcomes with exactly h successes have the same The forecasting scheme is allowed to operate

probability of occurrence. Since {H = h} can for 30 days and a total of 19 correct forecasts

2.4: Examples of Discrete Random Variables 25

since 23 hits are unlikely under the no-skill

are recorded. The forecasters claim that they have