Download RANDOM VARIABLES Definition: A random variable X is a rule that

RANDOM VARIABLES
Definition: A random variable X is a rule that assigns a numerical value to each outcome in a sample
space S.
We shall let RXdenote the set of numbers assigned by a random variable X, and we shall refer to RXas
the
range space.
EXAMPLE
A pair of fair dice is tossed. The sample space S consists of the 36 ordered
pairs (a, b) where a and b can be any of the integers from 1 to 6.
Let X assign to each point in S the sum of the numbers; then X is a random variable with range space
RX= {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Let Y assign to each point the maximum of the two numbers; then Y is a random variable with range space
RY= {1, 2, 3, 4, 5, 6}
Probability Distribution of a Random Variable
Let X be a random variable on a finite sample space S with range space Rx= {x1, x2, . . . , xt}. Then X
induces a function f which assigns probabilities pkto the points xkin Rxas follows:
f (xk) = pk= P (X = xk) = sum of probabilities of points in S whose image is xk
The set of ordered pairs (x1, f (x1)), (x2, f (x2)), . . . , (xt, f (xt)) is called the distribution of the random variable
X; it is usually given by a table as in Fig. 7-4. This function f has the following two properties:
(i) f (xk) ≥ 0 and (ii)
f (xk) = 1
k
Thus RXwith the above assignments of probabilities is a probability space. (Sometimes we will use the pair
notation [xk, pk] to denote the distribution of X instead of the functional notation [x, f (x)]).
Fig. 7-4 Distribution f of a random variable X
Theorem:
Let S be an equiprobable space, and let f
be the distribution of a random variable X on S with
the range space RX= {x1, x2, . . . , xt}. Then
pi= f (xi) = number of points inSwhose image is xi
number of points in S
EXAMPLE Let X be the random variable in which assigns the sum to the toss of a pair of
dice. Note n(S) = 36, and Rx= {2, 3, . . . , 12}. Using Theorem 7.8, we obtain the distribution f of X as
follows:
f (2) = 1/36, since there is one outcome (1, 1) whose sum is 2.
f (3) = 2/36, since there are two outcomes, (1, 2) and (2,1), whose sum is 3.
f (4) = 3/36, since there are three outcomes, (1, 3), (2, 2) and (3, 1), whose sum is 4.
Similarly, f (5) = 4/36, f (6) = 5/36, . . ., f (12) = 1/36. Thus the distribution of X follows
x
2
3
4
5
6
7
8
9
10
11
12
f (x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
Expectation of a Random Variable
Let X be a random variable on a probability space S = {s1, s2, . . . , sm} Then the mean or expectation of
X
is denoted and defined by:
µ = E(X) = X(s1)P (s1) + X(s2)P (sa2) + · · · + X(sm)P (sm) =
X(sk)P (sk)
In particular, if X is given by the distribution f
in Fig. 7-4, then the expectation of X is:
µ = E(X) = x1f (x1) + x2f (x2) + · · · + xtf (xt)
Alternately, when the notation [xk, pk] is used instead of [xk, f (xk)],
µ = E(X) = x1p1+ x2p2+ · · · + xtpt
(For notational convenience, we have omitted the limits in the summation symbol .)
EXAMPLE
(a) Suppose a fair coin is tossed six times. The number of heads which can occur with their respective
probabilities follows:
xi
0
1
2
3
4
5
6
winning are
p1
1/64 6/64 15/64 20/64 15/64 6/64 1/64
Then the mean or expectation (or expected number of heads) is:
µ = E(X) =0(1/64)+1(6/64)+2(15/64)+3(20/64)+4(15/64)+5(6/64)+6(1/64)=
(b) Three horses a , b, and c are in a race; suppose their respective probabilities a,b,c
2
1/2
6
1/3
9
1/6
according as a, b, or c
E(X) = X(a)P (a) + X(b)P (b) + X(c)P (c)
=2(1/2)+6(1/3)+ 9(1/6)=
Variance and Standard Deviation of a Random Variable
Let X be a random variable with mean µ and distribution f Then the variance of X, denoted
by V ar(X), is defined by:
V ar(X) = (x1− µ)2f (x1) + (x2− µ)2f (x2) + · · · + (xt− µ)2f (xt) = (xk− µ)2f (xk) = E((X − µ)2
V ar(X) = (x1− µ)2p1+ (x2− µ)2p2+ · · · + (xt− µ)2pt= (xk− µ)2pk= E((X − µ)2)
The standard deviation of X, denoted by σ
:
V ar(X
"
)
V ar(X )= E(X2) − µ2
V ar(X) = x12p1+x22p2 + · · · + xt2pt
"2
EXAMPLE
Let X denote the number of times heads occurs when a fair coin is tossed six times. The
distribution of X ), where its mean µ = 3 is computed. The variance of X is computed
as follows:
V ar(X) = (0 − 3)2 ( 1/ 64)+(1 − 3)2 (6/64) +(2 − 3)2 (15/64) + · · · + (6 − 3)2 1/64=1.5
Alternatively:
Thus the standard deviation
is(1.5)1/2
σ =(1.
Binomial Distributio
Consider a binomial experiment B(n, p). That is, B(n, p) consists of n independent repeated trials with
two
outcomes, success or failure, and p is the probability of success (and q
= (1 − p) is the probability of
failure).
The number X of k successes is a random variable with distribution appearing in Fig. 7-5.
Fig. 7-5
The following theorem applies.
Theorem 7.9: Consider the binomial distribution B(n, p). Then:
(i) Expected value E(X) = µ = np.
(ii) Variance V ar(X) = σ2= npq.
(iii) Standard deviation σ = √npq.
EXAMPLE 7.17
(a) The probability that a man hits a target is p = 1/5. He fires 100 times. Find the expected number µ of
times
he will hit the target and the standard deviation σ .
Here p = 1/5and so q = 4/5. Hence
µ = np = 100 (1/5) = 20
and
σ = √npq=
5
(b) Find the expected number E(X) of correct answers obtained by guessing in a five-question
true–false test.
Here p = 1/2. Hence E(X) = np = 5 ·1/2= 2.5.