DISCRETE RANDOM VARIABLES
Achievement Standards: 90643 (3.3) part, external; 90646 (3.6) part, external
Key words: Sigma notation, Expected value, variance, winnings, profit, gain, return,
independent, standard deviation.
1. SIGMA NOTATION
This notation will be used at times during this topic.
5
a
Examples: A= { 1,3,5,7,9,11,……)
i =1
1
= 1 + 3 + 5 + 7 + 9 = 25
5
 (4i + 1) = 5 + 9 + 13 + 17 + 21 = 65
i=1
Nulake p 147
Sigma p121, Ex 7.01, Ex 7.02, 7.03
2. EXPECTATION
For a given probability function the mean () or
expected value of X (E[X]) is given by the formula: E[X] =

xip(xi)
Example 1: Spinner
A probability distribution for the result of spinning this spinner is:
x
P(X=x)
0
0.1
1
0.3
2
0.4
In this example the mean, or expected, value is:
3
0.2
3
0
1
2
E[X] = 0  0.1 + 1  0.3 + 2  0.4 + 3  0.2 = 1.7
Example 2: Raffle
A raffle has 100 tickets and has a first prize of $200, second prize of $100 and a third
prize of $50. Prizewinners are drawn, without replacement, in order (1st, 2nd, 3rd).
Let X be a random variable representing the winnings of a single ticket buyer.
x
P(X=x)
200
0.01
100
0.01
50
0.01
0
0.97
E[X] = 200  0.01 + 100  0.01 + 50  0.01 + 0  0.9 = $ 3.50
(Note that this means that if the raffle is to be fair tickets should sell at $3.50)
Example 3: Car Insurance
A racing car valued at $200 000 has the probability of being a total loss estimated at
0.002, a 50% loss with probability 0.01, and a 25% loss with probability 0.1.
What should the insurance company charge if it wants to make an average profit of
$1 000 per car that it insures?
Let X be a random variable representing the amount that the company has to pay out.
x
P(X=x)
200 000
0.002
100 000
0.01
50 000
0.1
0
0.888
E[X] = $6 400, so the company would have to charge $7 400.
3. EXPECTED VALUE OF A LINEAR FUNCTION OF A RANDOM VARIABLE
E[aX+b] = aE[X] + b
Nulake p 156
Sigma p130, Ex 7.04
 pi(axi +b)
=  (pi (axi) + pi (b))
=  (pi axi) +  ( pi b)
= a (pi xi) + b ( pi),  ( pi) = 1, axi = axi
Proof: E[aX+b] =
= aE[X] + b
Example:
The random variable X is defined by the following probability distribution:
x
P(X=x)
0
0.1
1
0.3
2
0.4
3
0.2
giving E[X] = 1.7
Then the probability distribution for 3X+2 is:
x
P(X=x)
2
0.1
5
0.3
8
0.4
Note that 3E[X]+2 = 3  1.7 + 2
= 7.1
= E[3X+2]
11
0.2
giving E[3X+2] = 7.1
4. VARIANCE OF A RANDOM VARIABLE
Var[X] = E[(X-)2]
= E[X2] - 2
Proof:
Nulake p 152
Sigma p135, Ex 7.05, 7.06
Var[X] = E[(X-)2]
= E[X2 – 2X + 2]
= E[X2] – E[2X] + E[2]
= E[X2] – 2E[X] + E[2]
= E[X2] – 22 + 2
= E[X2] – 2
Example:
X
P(X=x)
1
0.1
2
0.4
3
0.2
4
0.3
giving  = E[X] = 2.7
(X-)2
P(X=x)
2.89
0.1
0.49
0.4
0.09
0.2
1.69
0.3
giving Var[X] = E[(X-)2]
= 1.01
X2
P(X=x)
1
0.1
4
0.4
9
0.2
16
0.3
giving E[X2] = 8.3
Note that E[X2] – 2 = 8.3 – 2.72
= 1.01
= Var[X]
The standard deviation () of a random variable X is defined as the square root of the
variance.
   Var[X]
5. VARIANCE OF A FUNCTION OF A RANDOM VARIABLE
Var[aX + b] = a2Var[X]
Proof:
Example:
Nulake p 160
Sigma p138, Ex 7.07
Var[aX+b] = E[(aX+b)2] - aX+b2
= E[ a2X 2 +2abX+b2] – E[aX+b] 2
= (a2E[X 2] + 2abE[X] + b2) – (aE[X]+b)2
= (a2E[X 2] + 2abE[X] + b2) – (a2E[X] 2+2abE[X]+b2)
= a2E[X 2] - a2E[X] 2
= a2(E[X 2] - E[X] 2)
= a2Var[X]
X
P(X=x)
0
0.1
1
0.2
2
0.3
3
0.4
E[X] = 2, E[X2] = 5; so Var[X] = 1
Then for Y = 3X – 1
E[Y] = E[3X- 1]
= 3E[X] – 1
=32-1
=5
Var[Y] = Var[3X- 1]
= 9Var[X]
=91
=9
6. SUMS AND DIFFERENCES OF RANDOM
VARIABLES
(a) E[ X + Y ] = E[X] + E[Y],
Nulake p 160
Sigma p143, Ex 7.09, 7.10
E[ X – Y ] = E[X] - E[Y]
(b) If X and Y are independent then Var[ X  Y ] = Var[X] + Var[Y]
(c) This means that for independent random variables X and Y we have:
E[ aX + bY + c] = aE[X] + bE[Y] + c,
Var[aX + bY + c] = a2Var[X] + b2Var[Y]
Example: Empty jam containers have a mean weight of 250g with a standard
deviation of 20g.
The contents have a mean weight of 900g with a standard
deviation of 15g.
(a) Find the mean and standard deviation of a filled container
Answer: If X represents the weight of an empty container then:
E[X] = 250, Var[X] = 202
= 400
If Y represents the weight of the contents then:
E[Y] = 900, Var[Y] = 152
= 225
If T = X + Y (T represents the total weight)
Then E[T] = E[X + Y]
Var[T] = Var[X + Y]
= E[X] + E[Y]
= Var[X] + Var[Y] (assuming X,Y indpt)
= 250 + 900
= 400 + 225
= 1150g
= 625
So then mean weight of a full container is 1150g, with a standard
deviation of
625 = 25g.
(b) If the price of making the contents is 0.3 cents per gram plus a fixed cost of
25 cents, find the mean and standard deviation of the cost of making the
contents of a jar of jam.
Answer: If C represents the cost of making the contents then C = 0.3Y + 25 cents.
E[C] = E[0.3Y + 25]
Var[C] = Var[ 0.3Y + 25]
= 0.3E[Y] + 25
= 0.09Var[Y]
= 0.3  900 + 25
= 0.09  225
= 295c
= 20.25c
So the mean cost of making the contents of a jar of jam is $2.95
and the standard deviation is 20·25 = 4.5cents.
(d) For identically distributed independent random variables X1, X2, ……….Xn,
each with mean , and standard deviation :
E[X1, + X2, ……….+ Xn] = E[X1] + E[X2] ……….+ E[Xn]
=  +  +…………………..+ 
= n
Var[X1, + X2, ……….+ Xn] = Var[X1] + Var[X2] ……….+ Var[Xn]
= 2 + 2 +……………,……..+ 2
= n2
Example 1: Blocks of cheese have mean weight 5kg with a standard deviation of
0.5kg.
A carton consists of 25 such blocks.
What is the mean and standard deviation of the contents of a carton?
Answer: Let X1, X2, ……….X25 be random variables representing the weights of
blocks 1 to 25. Then each Xi is a random variable with  = 5, 2 = 0.25.
Take Y be a random variable representing the weight of the contents of a
carton. Then Y = X1, + X2, ……….+ X25.
E[Y] = E[X1, + X2, ……….+ X25]
= E[X1] + E[X2] ……….+ E[X25]
= 25  5 (assuming identical)
= 125 Kg
Var[Y] = Var[X1, + X2, ……….+ X25]
= Var[X1] + Var[X2] ……….+ Var[X25] (assuming independent)
= 25  0.25 (assuming identical)
= 6.25
So y = 2.5
Example 2: Packets of 12 matchboxes have mean weight 125g and a standard of
10g.
What is the mean and standard deviation of a single box?
Answer: If X1, X2, ……….X12 represent the weights of the boxes then:
E[X1, + X2, ……….+ X25] = 125
E[X1] + E[X2] ……….+ E[X25] = 125
12  = 125
 = 10.4g, where  is the mean weight of a single box.
Var[X1, + X2, ……….+ X12] = 100
Var[X1] + Var[X2] ……….+ Var[X12] = 100 (assuming independent)
12 2 = 100
2 = 8.333
 = 2.89g (2DP)
where  is the s.d. of the weight of a single box