Download X = x

Random Variable (RV)
A function that assigns a numerical value to
each outcome of an experiment.
Notation: X, Y, Z, etc
Observed values: x, y, z, etc
Example 1
Two fair coins tossed. Let X = No of Heads
Outcomes
HH
HT
TH
TT
Probability
¼
¼
¼
¼
Value of X
2
1
1
0
Random Variable

Discrete RV – finite or countable number of
values

Continuous RV – taking values in an interval
Probability Distribution

Probability distribution of a discrete RV
described by what is known as a Probability
Mass Function (PMF).

Probability distribution of a continuous RV
described by what is known as a Probability
Density Function (PDF).
Probability Mass Function (PMF)
p(x) = Pr (X = x) satisfying

p(x) ≥ 0 for all x

∑p(x) = 1
Example 1 (Contd)
X = No of Heads. The PMF of X is:
x
0
1
2
Pr (X = x)
1/4
2/4=1/2
1/4
Example 1 (Contd)
Probability Density Function (PDF)
f(x) satisfying

f(x) ≥ 0 for all x

∫f(x) dx = 1

∫ab f(x) dx = Pr (a < X < b)

Pr (X = a) = 0
Example 2
Example 3
Expectation
E (X) = ∑x p (x) for a Discrete RV
E (X) = ∫x f (x) dx for a Continuous RV
Expectation
E (X2)= ∑x2 p (x) for a Discrete RV
E (X2) = ∫x2 f (x) dx for a Continuous RV
Expectation
E (g(X)) = ∑g(x) p (x) for a Discrete RV
E (g(X)) = ∫g(x) f (x) dx for a Continuous RV
Variance
Var (X) = E (X2) – (E(X))2
Standard Deviation
SD (X) = √Var (X)
Properties of Expectation
E (c) = c for a constant c
E (c X) = c E (X) for a constant c
E (c X + d) = c E (X) + d for constants c & d
Properties of Variance
Var (c) = 0 for a constant c
Var (c X) = c2 Var (X) for a constant c
Var (c X + d) = c2 Var (X) for constants c & d
Example 1 (Contd)
X = No of Heads. Find the following:
(a)
(b)
(c)
(d)
(e)
(f)
E (X)
E (X2)
E ((X+10)2)
E (2X)
Var (X)
SD (X)
Ans: 1
Ans: 1.5
Ans: 121.5
Ans: 2.25
Ans: 0.5
Ans: 1/√2
Example 4
If X is a random variable with the probability density function
f (x) = 2 (1 - x) for 0 < x < 1
find the following:
(a) E (X)
(b) E (X2)
(c) E ((X+10)2)
(d) Var (X)
(e) SD (X)
Ans:
Ans:
Ans:
Ans:
Ans:
1/3
1/6
106.8333
1/18
1/(3√2)
Example 5
An urn contains 4 balls numbered 1, 2, 3 & 4.
Let X denote the number that occurs if one
ball is drawn at random from the urn. What
is the PMF of X?
Example 5 (Contd)
Two balls are drawn from the urn without
replacement. Let X be the sum of the two
numbers that occur. Derive the PMF of X.
Example 6
The church lottery is going to give away a
£3,000 car and 10,000 tickets at £1 a piece.
(a) If you buy 1 ticket, what is your expected
gain. (Ans: -0.7)
(b) What is your expected gain if you buy 100
tickets? (Ans: -70)
(c) Compute the variance of your gain in these
two instances. (Ans: 899.91 & 89100)
Example 7
A box contains 20 items, 4 of them are
defective. Two items are chosen without
replacement. Let X = No of defective items
chosen. Find the PMF of X.
Example 8
You throw two fair dice, one green and one red.
Find the PMF of X if X is defined as:
A)
B)
C)
D)
Sum of the two numbers
Difference of the two numbers
Minimum of the two numbers
Maximum of the two numbers
Example 9
If X has the PMF p (x) = ¼ for x = 2, 4, 8, 16
compute the following:
(a)
E (X)
Ans: 7.5
(b)
E (X2)
Ans: 85
(c)
E (1/X)
Ans: 15/64
(d)
E (2X/2)
Ans: 139/2
(e)
Var (X)
Ans: 115/4
(f)
SD (X)
Ans: √115/2
Example 10
If X is a random variable with the probability density function
f (x) = 10 exp (-10 x) for x > 0
find the following:
(a) E (X)
(b) E (X2)
(c) E ((X+10)2)
(d) Var (X)
(e) SD (X)
Ans:
Ans:
Ans:
Ans:
Ans:
0.1
0.02
102.02
0.01
0.1
Example 11
If X is a random variable with the probability density function
f (x) = (1/√(2)) exp (-0.5 x2) for - < x < 
find the following:
(a) E (X)
(b) E (X2)
(c) E ((X+10)2)
(d) Var (X)
(e) SD (X)
Ans:
Ans:
Ans:
Ans:
Ans:
0
1
101
1
1
Example 12
A game is played where a person pays to roll
two fair six-sided dice. If exactly one six is
shown uppermost, the player wins £5. If
exactly 2 sixes are shown uppermost, then
the player wins £20. How much should be
charged to play this game is the player is to
break-even?
Example 13
Mr. Smith buys a £4000 insurance policy on his
son’s violin. The premium is £50 per year.
If the probability that the violin will need to
be replaced is 0.8%, what is the insurance
company’s gain (if any) on this policy?