Chapter 4 Discrete random variables 4.1

Definition, Mean and Variance

Let X be a (discrete) r.v. taking on the values xi with corresponding probabilities pi = f (xi ), i = 1, ..., n. Function f (x) is called probability density function (p.d.f). It is convenient to present the values of a (discrete) r.v. and the corresponding probabilities in a tabular form as follows. x x1 f (x) p1

... ...

xn pn

Total 1

The cumulative distribution function of a discrete random variable X, denoted as F (x) is defined as X F (x) = P (X ≤ x) = f (xi ) xi ≤x

For a discrete random variable X, F (X) satisfies the following properties: 1. 0 ≤ F (x) ≤ 1 2. If x ≤ y, then F (x) ≤ F (y) The cumulative distribution function can be used to calculate the probabilities: P (a < x ≤ b) = P (x ≤ b) − P (x ≤ a) = F (b) − F (a) The mathematical expectation of r.v. X (or just expectation or mean value of X or just mean of X) is denoted by E[X] and is defined by: E[X] =

n X

xi f (xi ).

i=1

The alternative notations µ(X) or µ are also often used. 1

2

CHAPTER 4. DISCRETE RANDOM VARIABLES Example 4.1

Suppose an insurance company pays the amount of £1,000 for lost luggage on an airplane trip. From past experience, it is known that the company pays this amount in 1 out of 200 policies it sells. What premium should the company charge? Solution Define the r.v. X as follows: X = 0 if no loss occurs, which happens with 1 = 0.005. probability 1 − (1/200) = 0.995, and X = −1, 000 with probability 200 x 0 −1, 000 Total f (x) 0.995 0.005 1 Then the expected loss to the company is: E[X] = 0 × 0.995 + (−1, 000 × 0.005) = −5. Thus, the company must charge £5 to break even. △ Properties of mean: 1. E[cX] = cE[X] Proof

E[cX] =

n X

cxi f (xi ) = c

i=1

n X

xi f (xi ) = cE[X]

i=1

△ 2. E[cX + d] = cE[X] + d Proof

E[cX + d] =

n X

(cxi + d)f (xi ) =

i=1 n X

= c

i=1

n X i=1

xi f (xi ) + d

n X

cxi f (xi ) +

n X

f (xi )

i=1

f (xi ) = cE[X] + d

i=1

△ The variance of a r.v. X is denoted by Var(X) and is defined by: V ar[X] = E[X − E[X]]2 2 . The alternative notations σ 2 (X) and σX are also often used for the V ar(X).

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CHAPTER 4. DISCRETE RANDOM VARIABLES

The variance account for the difference in size of the range of the distributions of r.v. More generally, for a r.v. X taking on finitely many values x1 , ..., xn with respective probabilities f (x1 ), ..., f (xn ), the variance is: Pn V ar[X] = i=1 (xi − E[X])2 f (xi ) and represents the sum of the weighted squared distances of the points xi , i = 1, ..., n from the center of location of the distribution, E[X]. Thus, the further from E[X] the xi s are located, the larger the variance, and vice versa. Because of this characteristic property of the variance, the variance is referred to as a measure of dispersion of the underlying distribution. The positive square root of the V ar[X] is called the standard deviation (s.d.) of X. Unlike the variance, the s.d. is measured in the same units as X. Properties of variance: 1. V ar[X] = E[X]2 − (E[X])2 Proof

2

V ar[X] = E[X − E[X]] =

n X i=1

(xi − E[X])2 f (xi )

=

n X

(x2i − 2xi E[X] + E[X]2 )f (xi )

=

n X

x2i f (xi )

i=1

i=1

− 2E[X]

n X i=1

2

xi f (xi ) + E[X]

n X

f (xi )

i=1

= E[X 2 ] − 2E[X]E[X] + E[X]2 = E[X 2 ] − E[X]2 △ 2. V ar[cX] = c2 V ar[X] Proof

V ar[cX] = E[(cX)2 ] − E[cX]2 = E[c2 X 2 ] − (cE[X])2 = c2 E[X 2 ] − c2 E[X]2 = c2 (E[X]2 − (E[X])2 ) = c2 V ar[X] △ 3. V ar[cX + d] = c2 V ar[X] Proof

CHAPTER 4. DISCRETE RANDOM VARIABLES

V ar[cX + d] = = = =

4

E[(cX + d)2 ] − E[cX + d]2 E[c2 X 2 + 2cdX + d2 ] − (cE[X] + d)2 c2 E[X 2 ] + 2cdE[X] + d2 − c2 E[X]2 − 2cdE[X] − d2 c2 (E[X]2 − (E[X])2 ) = c2 V ar[X] △

4. V ar[X] = E[X(X − 1)] + E[X] − E[X]2 Proof We know that V ar(X) = E[X 2 ] − E[X]2 E[X(X − 1)] = E[X 2 − X] = E[X 2 ] − E[X] = V ar[X] + E[X]2 − E[X], so that V ar[X] = E[X(X − 1)] + E[X] − E[X]2 △ Example 4.2 The r.v. X has p.d.f. f given by:  0, x < 4     0.1, x = 4      0.3, x = 5 0.3, x = 6 f (x) =   0.2, x = 8     0.1, x = 9    0, x > 9. 1. Draw the graph of p.d.f f .

2. Calculate the probabilities P (X ≤ 6.5), P (X > 8.1), P (5 < X < 8). 3. Calculate and draw the graph of c.d.f F . 4. Calculate mean E[X]. 5. Calculate variance V ar[X]. 6. If we define new r.v. Y = 5 − 2X, calculate E[Y ] and V ar[Y ]. Solution 1.

CHAPTER 4. DISCRETE RANDOM VARIABLES

5

2. P (X ≤ 6.5) = P (X = 4) + P (X = 5) + P (X = 6) = 0.1 + 0.3 + 0.3 = 0.7 P (X > 8.1) = P (X = 9) = 0.1 (P (5 < X < 8) = P (X = 6) = 0.3 3.  0,     0.1,    0.4, F (x) = 0.7,     0.9,    1.0,

x