THEOREMS ON EXPECTATION - Quantitative Techniques for management

Expectation Theorem

Theorem : Expected value of a constant is the constant itself, i.e., E(b) = b, where b is a constant.

Proof: The given situation can be regarded as a probability distribution in which the random variable takes a value b with probability 1 and takes some other real value, say a, with probability 0.
Thus, we can write E(b) = b × 1 + a × 0 = b

Theorem : E(aX) = aE(X), where X is a random variable and a is constant.

Proof: For a discrete random variablze X with probability function p(X), we have:

probability function

Combining the results of theorems 1 and 2, we can write

E(aX + b) = aE(X) + b

Remarks : Using the above result, we can write an alternative expression for the variance of X, as given below :

σ2 = E(X - m)2 = E(X2 - 2mX + m2)
= E(X2) - 2mE(X) + m2 = E(X2) - 2m2 + m2
= E(X2) - m2 = E(X2) - [E(X)]2
= Mean of Squares - Square of the Mean

We note that the above expression is identical to the expression for the variance of a frequency distribution.

Theorems on Variance

Theorem : The variance of a constant is zero.

Proof: Let b be the given constant. We can write the expression for the variance of b as:

Var(b) = E[b - E(b)]2 = E[b - b]2 = 0.

Theorem : Var(X + b) = Var(X).

Proof: We can write Var(X + b) = E[X + b - E(X + b)]2 = E[X + b - E(X) - b]2

= E[X - E(X)]2 = Var(X)

Similarly, it can be shown that Var(X - b) = Var(X)

Remarks: The above theorem shows that variance is independent of change of origin.

Theorem : Var(aX) = a2Var(X)

Proof: We can write Var(aX) = E[aX - E(aX]2 = E[aX - aE(X)]2= a2E[X - E(X)]2 = a2Var(X).

Combining the results of theorems 2 and 3, we can write
Var(aX + b) = a2Var(X).

This result shows that the variance is independent of change origin but not of change of scale.

Remarks:

  1. On the basis of the theorems on expectation and variance, we can say that if X is a random variable, then its linear combination, aX + b, is also a random variable with mean aE(X) + b and Variance equal to a2Var(X).
  2. The above theorems can also be proved for a continuous random variable.

Example: Compute mean and variance of the probability distributions obtained in examples 1, 2 and 3.

Solution:

  • The probability distribution of X in example 1 was obtained as

The probability distribution

From the above distribution, we can write

The probability distribution

(b) The probability distribution of X in example 2 was obtained as

probability distribution of X

∴ E(X)= 2*1/36+3*2/36+4*3/36+5*4/36+6*5/36+7*6/36+
8*5/36+9*4/36+10*3/36+11*2/36+12*1/36 =252/36

Further, E(X2) = 4*1/36+9*2/36+16*3/36+25*4/36+36*5/36+
49*6/36+64*5/36+81*4/36+100*3/36+121*2/36+144*1/36 =1974/36=54.8

thus var(x)= 54.8-49= 5.8

(c) The probability distribution of X in example 3 was obtained as

probability distribution of X

E(x)= 1*4/20+2*12/20+3*4/20=2

and E(x2)= 1*4/20+4*12/20+9*4/20= 4.4

∴ var(x)= 4.4-4= 0.4

Expected Monetary Value (EMV)

When a random variable is expressed in monetary units, its expected value is often termed as expected monetary value and symbolised by EMV.

Example : If it rains, an umbrella salesman earns Rs 100 per day. If it is fair, he loses Rs 15 per day. What is his expectation if the probability of rain is 0.3?

Solution: Here the random variable X takes only two values, X1 = 100 with probability 0.3 and X2 = – 15 with probability 0.7.

Thus, the expectation of the umbrella salesman

= 100 *0.3 – 15 * 0.7 = 19.5

The above result implies that his average earning in the long run would be Rs 19.5 per day.

Example: A person plays a game of throwing an unbiased die under the condition that he could get as many rupees as the number of points obtained on the die. Find the expectation and variance of his winning. How much should he pay to play in order that it is a fair game?

Solution: The probability distribution of the number of rupees won by the person is given below:

probability distribution

Thus, E(X)= 1*1/6+2*1/6+3*1/6+4*1/6+5*1/6+6*1/6= rs 7/2

and E(X2)= 1*1/6+4*1/6+9*1/6+16*1/6+25*1/6+36*1/6=91/6

&there4 σ2= 91/6-(7/2)2= 35/12=2.82.Note that the unit of s2 will be (Rs)2.

Since E(X) is positive, the player would win Rs 3.5 per game in the long run. Such a game is said to be favourable to the player. In order that the game is fair, the expectation of the player should be zero. Thus, he should pay Rs 3.5 before the start of the game so that the possible values of the random variable become 1 - 3.5 = - 2.5, 2 - 3.5 = - 1.5, 3 - 3.5 = - 0.5, 4 - 3.5 = 0.5, etc. and their expected value is zero.

Example: Two persons A and B throw, alternatively, a six faced die for a prize of Rs 55 which is to be won by the person who first throws 6. If A has the first throw, what are their respective expectations?

Solution: Let A be the event that A gets a 6 and B be the event that B gets a 6. Thus,

P(A) = 1/6 and P(B) = 1/6

If A starts the game, the probability of his winning is given by:

P(A wins )= P(A)+P(Ā).P(B).P(A)+P(Ā).P(B)P(Ā).P(B).P(A)+....

= 1/6+5/6*5/6*1/6+5/6*5/6*5/6*5/6*1/6+....

=1/6[1+(5/6)2+(5/6)4+....]= 1/6[1/1-(25/36)]= 1/6*36/11= 6/11

SIMILARLY, P(B WINS)= P(Ā).P(B)+P(Ā).P(B).P(Ā).P(B)+....

= 5/6*1/6+5/6*5/6*5/6*1/6+....

=5/6*1/6[1+(5/6)2+(5/6)4+...]= 5/6*1/6*36/11= 5/11

Expectation of A and B for the prize of Rs 55

Since the probability that A wins is 6/11, therefore, the random variable takes a value 55 with probability
6/11 and value 0 with probability 5/11. Hence,

Example : An unbiased die is thrown until a four is obtained. Find the expected value and variance of the number of throws.

Solution: Let X denote the number of throws required to get a four. Thus, X will take values 1, 2, 3, 4, ...... with respective probabilities.

1/6,5/6*1/6,(5/6)2*1/6.(5/6)3*1/6...ETC.

∴ E(X)= 1.1/6+2.5/6.1/6+3.(5/6)2*1/6+4(5/6)3.1/6....

=1/6[1+2.5/6+3.(5/6)2+4.(5/6)3+...] let S = 1+2.5/6+3.(5/6)2+4.(5/6)3+...

MULTIPLYING BOTH SIDES BY 5/6, 5/6S= 5/6+2.(5/6)2+3.(5/6)3+4(5/6)4+...

∴ S-5/6S= 1+(2-1)5/6+(3-2)(5/6)2+(4-3)(5/6)3+...

1/6S= 1+5/6+(5/6)2+(5/6)3+...= 1/1-(5/6)=6

Thus S=36 AND Hence E(X)= 1/6*36=6

Further, to find variance we first find E(X2)

E(X2)= 1.1/6+4.5/6.1/6+9(5/6)2.1/6+16(5/6)3.1/6....

= 1/6[1+4(5/6)+9.(5/6)2+16(5/6)3+..]

Let S= 1+4.(5/6)+9.(5/6)2+16(5/6)3+...

Multiply both sides by (5/6) and subtract from S, to get

1/6S=1+(22-1)(5/6)+(32-22)(5/6)2+(42-32)(5/6)3+...

Hence. Variance= E(X2)-[E(X)]2= 66-36= 30


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