Maths9

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Special probability distributions
9.1 Discrete probability distributions
Example 9.1

(Bernoulli random variable) Let (S, A, P ) be a probability space and consider an event A A, for which P (A) = p is given, where 0 < p < 1. Then the random variable 1 if A occurs X= 0 if A does not occur is called a Bernoulli random variable (corresponding to the event A). In this case X takes the value 1 with probability p, and take s the value 0 with probability 1 - p. This probability distribution has two elements and its only parameter is p.

(Binomial distribution) We carry out a probability experiment consisting of n repeated trials. The repeated trials are independent. We observe at each trial if a given event A occurs. Assume that the probability P (A) = p, 0 < p < 1 is constant from trial to trial. Let X mean the number of occurrences of the event A. As it was seen in the problem of sampling with replacement the probability distribution of the random variable X is
Example 9.2

P (X = k) = Bin(n, p).

n k p (1 - p)n-k k

k = 0, 1, 2, . . . , n

The probability distribution of X is called binomial distribution and denoted by This distribution has n + 1 elements and two parameters, namely n and p.

If Xk denotes the Bernoulli random variable corresponding to the event A at the k th trial, then X = X1 + . . . + Xn that is, the random variable X with binomial distribution Bin(n, p) is the sum of n Bernoulli variables of parameter p.

(Hypergeometric distribution) Let the experiment be a sampling without replacement. Take an N element set, in which m element is defective. A random sample of size n is selected without replacement (n m). Let X mean the number of defective elements in the selected sample. Then the probability distribution of X is
Example 9.3

P (X = k) =

m k

·

N -m n-k N n

k = 0, 1, 2, . . . , n

This distribution is called hypergeometric distribution and denoted by Hip(n, m, N ). The hypergeometric distribution has n + 1 elements and three parameters n, m and N .
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9.2.

CONTINUOUS PROBABILITY DISTRIBUTIONS

(Geometric distribution) Repeated independent trials are performed until a given event A rst occurs. Assume, that P (A) = p is constant from trial to trial, where 0 < p < 1. Let X mean the number of trials. Then the probability distribution of X is
Example 9.4

P (X = k) = (1 - p)k-1 p

k = 1, 2, . . .

This distribution is called geometric distribution and denoted by Geo(p). The geometric distribution has innite elements and its only parameter is p. It is easily seen that we really dened a probability distribution, because


(1 - p)k-1 p = p ·
k=1
Example 9.5

1 =1 1 - (1 - p)

X is {0} N and its distribution is P (X = k) =

(Poisson distribution) Assume that the range of the random variable
k - e k! k = 0, 1, 2, . . .

where > 0 is a given real number. By the power series of the exponential function we have that


k=0

k - e = e- · e = 1 k!

verifying that a probability distribution is obtained.
9.2 Continuous probability distributions

Example 9.6

(Uniform distribution) Let [a, b] be a given interval. Consider the random variable X having the density function
f (t) =
1 b-a

0

if a < t < b otherwise

Then the random variable X is of uniform distribution on the interval [a, b]. This distribution has two parameters: a and b and denoted by U ni(a, b). In this case the probability that the value taken by X is in a subinterval of [a, b] is proportional to the length of the subinterval. It is an easy calculation to check, that the cumulative distribution function of X is if x a 0 x-a if a < x b F (x) = b-a 1 if x > b
Example 9.7

(Exponential distribution) Let > 0 be a given real number. Consider the random variable X having density function
f (t) = e-t 0

if t > 0 otherwise

Then the distribution of the random variable X is called exponential distribution. The exponential distribution has one parameter: , and denoted by Exp(). Calculating the cumulative distribution function of X having distribution Exp() we obtain that 0 if x 0 F (x) = 1 - e-x if x > 0
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9.3.

HOMEWORK

Example 9.8 (Normal distribution) Let m and be given real numbers, where > 0. Consider the random variable X having density function

f (t) =

(t-m)2 1 e- 22 2

- < t < +

Then X is called a normal random variable. The probability distribution of the normal random variable has two parameters: m and > 0, and denoted by N (m, ). In particular, the distribution of the normal random variable X is called standard normal distribution, if m = 0 and = 1 (that is, an N (0, 1) distribution), thus its density function is
t2 1 (t) = e- 2 2

- < t < +

Using the Gaussian integral it can be checked, that



-

(t-m)2 1 e- 22 dt = 2

-

t2 1 e- 2 dt = 1 2

Analyzing the properties of the density function we nd that the function has a global maximum at t = m and inexion points at points t = m ± . The density function is continuous, therefore its denite integral exists, but this function can not be given using elementary functions. The function
x

(x) =
-

t2 1 e- 2 dt 2

is called the cumulative distribution function of the standard normal distribution.
9.3 Homework

· CH: 5.1 5.6 + CH: 6.1 6.7 · Exercises: pp.185-187, and 205-208 partly

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