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Statistics and Mathematics for Economics

Statistics and Mathematics for Economics. Statistics Component: Lecture Three. Objectives of the Lecture. To acquaint you with a further rule relating to the expectations operator To provide an example of an application of this rule

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Statistics and Mathematics for Economics

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  1. Statistics and Mathematics for Economics Statistics Component: Lecture Three

  2. Objectives of the Lecture • To acquaint you with a further rule relating to the expectations operator • To provide an example of an application of this rule • To show different methods for calculating the value of the variance of a random variable • To present rules relating to the variance

  3. A Further Rule relating to the Expectations Operator The expected value of a linear combination of random variables is equal to the same linear combination of the expected values of the random variables. a, b, c are parameters, X and Y are random variables. Hence, E[a + bX + cY] = a + bE[X] + cE[Y].

  4. A Non-linear Combination of Random Variables In general, the expected value of a non-linear combination of random variables is not equal to the same non-linear combination of the expected values of the random variables. For example, in general: E[X.Y]  E[X].E[Y]. The exception, though, is when the variables are statistically independent.

  5. An Example of an Application of this Rule • Three equal-size balls are contained in a black velvet bag • In order to be able to distinguish between them, the balls have the different numbers marked on them, 1, 2 and 3 • In sequence and without replacement, two of the balls are drawn from the bag • X denotes the number which is marked on the first ball • Y denotes the number which is marked on the second ball

  6. Further Information • It is necessary to pay a price of five pounds for the privilege of drawing the balls from the bag • In return, there is guaranteed to be received an amount which is equal to the sum of the number marked on the second ball and twice the number that is marked on the first ball • Calculate the expected net gain

  7. Joint Probability Distribution of X and Y X and Y are discrete random variables. Possible pairs of values of X and Y are: (X = 1, Y = 2); (X = 1, Y = 3); (X = 2, Y = 1); (X = 2, Y = 3); (X = 3, Y = 1); (X = 3, Y = 2). On the basis that the game is fair then every one of the possible pairs of values is associated with the same joint probability of 1/6.

  8. Mathematical Presentation of the Joint Probability Distribution P(X = x, Y = y) = 1/6, x, y = 1, 2, 3, x  y.

  9. Presentation of the Joint Probability Distribution in the form of a Table

  10. Calculation of the Expected Net Gain Net Gain = W = -5 + 2X + Y Expected Net Gain = E[W] = E[-5 + 2X + Y]. Upon applying the rule relating to the expectations operator: E[W] = -5 + 2E[X] + E[Y].

  11. Marginal Probability Distributions of X and Y The marginal probabilities of the different values of X can be obtained by adding together the joint probabilities in each of the rows of the table. The marginal probabilities of the different values of Y can be obtained by adding together the joint probabilities in each of the columns of the table. x P(X = x) y P(Y = y) 1 2/6 1 2/6 2 2/6 2 2/6 3 2/6 3 2/6

  12. The Expected Values of X and Y x P(X = x) x.P(X = x) 1 1/3 1/3 2 1/3 2/3 3 1/3 3/3 ---- 6/3 Therefore, E[X] = 2. Because the probability distribution of Y is identical to the probability distribution of X then E[Y] = 2. Consequently, the expected net gain is: E[W] = -5 + 2(2) + 2 = £1.

  13. Alteration of the Conditions of the Game Again, a price of £5 has to be paid for the privilege of drawing the balls out of the bag. In return, an amount is guaranteed which is, in pounds, equal to the product of the numbers which are marked on the two balls. What is the expected net gain?

  14. Calculating the Expected Net Gain If W denotes the net gain then W = -5 + X.Y The expected net gain: E[W] = E[-5 + X.Y] = -5 + E[X.Y] However, for all possible pairs of values of x and y, P(X = x, Y = y)  P(X = x).P(Y = y), hence, X and Y are not statistically independent, and so E[X.Y]  E[X].E[Y].

  15. Calculation of E[X.Y] E[X.Y] = x.y.P(X = x, Y = y) = (1)(1)(0) + (1)(2)(1/6) + (1)(3)(1/6) + (2)(1)(1/6) + (2)(2)(0) + (2)(3)(1/6) + (3)(1)(1/6) + (3)(2)(1/6) + (3)(3)(0) = (1/6)(2 + 3 + 2 + 6 + 3 + 6) = 22/6 So, E[W] = -5 + 11/3 = -£1.33

  16. Erroneous Calculation Let us suppose that we had believed that: E[W] = -5 + E[X].E[Y], such that E[W] = -5 + (2)(2) = -£1.00. The extent of the expected loss would have been underrecorded

  17. The Second Central Moment of a Probability Distribution • The second central moment of a probability distribution provides a measure of dispersion • Its value indicates the extent to which the population of values of the random variable are distributed about its expected value • The second central moment equates with the variance of the random variable

  18. Formula for the Variance of a Random Variable If X is a discrete random variable then var.(X) = E[(X – E[X])2] Equivalently, var.(X) = (x – E[X])2.P(X = x)

  19. Demonstration of the Calculation of the Value of the Variance of a Discrete Random Variable X is the number which is obtained following a single throw of a six-sided dice. x 1 2 3 4 5 6 P(X = x) 1/6 1/6 1/6 1/6 1/6 1/6 E[X] = 7/2 var.(X) = E[(X – E[X])2] = (x – E[X])2.P(X = x)

  20. Calculation of the Value of the Variance var.(X) = 35/12 units squared

  21. Alternative Formula for the Variance of X var.(X) = E[X2] - (E[X])2, where E[X2] = x2.P(X = x)

  22. Alternative Approach towards the Calculation of the Value of the Variance var.(X) = E[X2] – (E[X])2, where E[X2] = x2.P(X = x). Thus, var.(X) = 91/6 - (7/2)2 = (182 – 147)/12 = 35/12 units2

  23. A Limitation of the Variance • The variance is limited as a measure of dispersion as its value is not expressed in terms of the same units as the random variable, itself • The value of a variance is expressed in units squared • A related measure of dispersion is the standard deviation • The standard deviation is the positive square root of the variance • Hence, the value of the standard deviation is expressed in terms of the same units as the random variable, itself

  24. Rules relating to the variance of a random variable • The variance of a constant term is equal to zero • The variance of the sum of a constant and a random variable is equal to the variance of the random variable • The variance of the product of a constant and a random variable is equal to the product of the square of the constant and the variance of the random variable • var.(a + bX) = b2.var.(X), where a and b are constants

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