Idea Transcript
CHAPTER 3: Random Variables and Probability Distributions
Concept of a Random Variable: 3.1
• The outcome of a random experiment need not be a number. • However, we are usually interested not in the outcome itself, but rather in some measurement of the outcome.
Example: Consider the experiment in which batteries coming off an assembly line were examined until a good one (S) was obtained. S = {S, FS, FFS, . . .}. We may be interested in the number of batteries examined before the experiment terminates. A random variable is a function that associate a real number with each element in the sample space. Example: Tossing two coins S = {HH, TT, HT, TH} Let X = # of heads observed.
Example: A group of 4 components is known to contain 2 defectives. An inspector tests the components one at the time until the 2 defectives are located. Let X denote the number of the test on which the second defective is found.
Two types of random variables • A discrete random variable is a random variable whose possible values either constitute a finite set or else can be listed in an infinite sequence. • A random variable is continuous if its set of possible values consists of an entire interval on the number line. Many random variables, such as weight of an item, length of life of a motor etc., can assume any value in certain intervals.
1
Discrete Probability Distributions: 3.2 Probability mass function of a discrete random variable X is defined by f (x) = P (X = x)
Example: tossing two coins X = # of heads. f (0) = P (X = 0) = P (TT) = 1/4 f (1) = P (X = 1) = P (HT, TH) = 1/2 f (2) = P (X = 2) = P (HH) = 1/4 Example: An information source produces symbols at random from a five-letter alphabet: S = {a, b, c, d, e}. The probabilities of the symbols are p(a) =
1 1 1 1 , p(b) = , p(c) = , p(d) = p(e) = . 2 4 8 16
A data compression system encodes the letters into binary strings as follows: a b c d e
1 01 001 0001 0000
Let the random variable Y be equal to the length of the binary string output by the system. f (1) = P (Y = 1) = f (2) = p(Y = 2) = f (3) = p(Y = 3) = f (4) = p(Y = 4) =
f (x) = P (X = x) satisfies the following conditions: 1. fP(x) ≥ 0 2. f (x) = 1
Example: A box contains 5 balls numbered 1, 2, 3, 4, and 5. Three balls are drawn at random and without replacement from the box. If X is the median of the numbers on the 3 chosen balls, then what is the probability function for X, where nonzero? Solution
2
Example: Determine c so that the function f (x) can serve as the probability mass function of a random variable X: f (x) = cx for x = 1, 2, 3, 4, 5 Solution:
The cumulative distribution function: F (x) of a discrete random variable X with probability mass function f (x) is defined for every number x by X f (t) F (x) = P (X ≤ x) = t≤x
Example: Assume that f (2) = p(X = 2) = 1/6 f (3) = p(X = 3) = 1/3 f (4) = p(X = 4) = 1/2 Then F (2) = F (3) = F (4) = F (x) =
Example: A mail order computer business has six telephone lines. Let X denote the number of lines in use at a specified time. Suppose that the probability mass function of X is given below x p(x)
0 0.10
1 0.15
2 0.20
3 0.25
4 0.20
5 0.06
6 0.04
a. Find the cumulative distribution function b. Find the probability that {at most 3 lines are in use}. c. Find the probability that {at least 4 lines are in use}.
3
Example: If X has the cumulative distribution function:
find the probability mass function.
0 1/3 1/2 F (x) = 5/6 1
if if if if if
x Y );
(x + y) , for x = 0, 1, 2, 3 y = 0, 1, 2. 30
(b) P (X > 2, Y ≤ 1)
(d) P (X + Y = 4)
Solution:
14
Definition 3.10: The marginal distribution of X alone and of Y alone are X X g(x) = f (x, y) and h(y) = f (x, y) y
x
Example x f (x, y) 0 y 1 2
0 0 1 30 2 30
1
2
3
1 30 2 30 3 30
2 30 3 30 4 30
3 30 4 30 5 30
P (X = 0) = P (X = 1) = P (X + 2) = P (Y = 0) =. Thus x g(x)
0
1
2
y h(y)
0
1
2
3
Definition 3.11: The conditional distribution of the random variable Y , given that X = x, is f (y|x) =
f (x, y) , g(x)
g(x) > 0
Similarly, the conditional distribution of the random variable X, given that Y = y, is f (x|y) =
f (x, y) , h(y)
h(y) > 0
Statistical Independence: Definition: X and Y are said to be statistically independent if and only if f (x|y) = g(x). Result: X and Y are statistically independent if and only if f (x, y) = g(x)h(y) for all (x, y). proof: f (x|y) =
f (x, y) = g(x), h(y)
implies that f (x, y) = g(x)h(y) for all (x, y). 15
Example: An investor owns two assets. He is interested in the value of his investments in one year. The value of the first asset in one year is a random variable X, and the value of the second asset in one year is a random variable Y . The joint probability mass function of X and Y is given in the following table:
f (x, y) y 0 10
90 0.05 0.15
x 100 0.27 0.33
110 0.18 0.02
(a) Are X and Y independent?
(b) Find P (X ≥ 100)
(c) Find P (Y = 0|X = 100) Example: A computer program can make calls to two subroutines, A and B. In a randomly chosen run, let X be the number of calls to subroutine A and let Y denote the number of calls to subroutine B. The joint probability mass function of X and Y is given in the following table. y f (x, y) 1 2 3 x 1 0.15 0.10 0.10 2 0.10 0.20 0.15 3 0.05 0.05 0.10 a. Find the marginal probability mass function of X. b. Find the marginal probability mass function of Y . c. Are X and Y independent? Explain. d. Find the probability that the number of calls to the two subroutines in 4. e. Find P (Y = 1|X = 2). f. Assume that each execution of subroutine A takes 100 ms, and that each execution of subroutine B takes 200 ms. Express the number of milliseconds used in all the calls to the two subroutines in terms of X and Y.
16