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 Probability Value of X

HH ¼ 2

HT ¼ 1

TH ¼ 1

TT ¼ 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 Pr (X = x)

0 1/4

1 2/4=1/2

2 1/4

Example 1 (Contd)

Probability Density Function (PDF)

f(x) satisfying f(x) ≥ 0 for all x ∫f(x) dx = 1 ∫a

b 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) E (X) Ans: 1

(b) E (X2) Ans: 1.5

(c) E ((X+10)2) Ans: 121.5

(d) E (2X) Ans: 2.25

(e) Var (X) Ans: 0.5

(f) SD (X) 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) Ans: 1/3(b) E (X2) Ans: 1/6(c) E ((X+10)2) Ans: 106.8333(d) Var (X) Ans: 1/18(e) SD (X) Ans: 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) Sum of the two numbersB) Difference of the two numbersC) Minimum of the two numbersD) 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) Ans: 0.1(b) E (X2) Ans: 0.02(c) E ((X+10)2) Ans: 102.02(d) Var (X) Ans: 0.01(e) SD (X) Ans: 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) Ans: 0(b) E (X2) Ans: 1(c) E ((X+10)2) Ans: 101(d) Var (X) Ans: 1(e) SD (X) Ans: 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?