15 2. Probability Distributions 2.1. Random Variables, Expected Values and Variance Definition (random variable): A random variable is a numerical description of the outcome of an experiment. There are two common types of random variables. They are: Discrete random variable: a quantity assumes either a finite number of values or an infinite sequence of values, such as 0, 1, 2, Continuous random variable: a quantity assumes any numerical value in an interval or collection of intervals, such as time, weight, distance, and temperature. Definition (probability distribution (or density) function (p.d.f)): a function describes how probabilities are distributed over the values of the random variable. Required conditions for a discrete probability distribution function: Let a1 , a 2 ,, a n , be all the possible values of the discrete random variable X. Then, the required conditions for f (x ) to be the discrete probability distribution for X are (a) P X ai f (ai ) 0, for every i. (b) f a f a f a f a 1 i 1 i 1 2 n Required conditions for a continuous probability density: Let the continuous random variable Z taking values in subsets of , . Then, the required conditions for f (x) to be the continuous probability density function for Z are (a) f ( x) 0, x . (b) f ( x )dx 1 Note: As X is discrete, P(c X d ) f a . cai d 15 i 16 As X is continuous, d P (c X d ) f ( x)dx. c Expected Value and Variance: As X is discrete, E ( X ) ai f ai a1 f ( a1 ) a2 f ( a2 ) an f (an ) i 1 and Var ( X ) 2 E X E ( X ) (ai ) 2 f (ai ) 2 i (a1 ) f (a1 ) (a2 ) f (a2 ) (an ) 2 f (an ) 2 2 As X is continuous, E( X ) xf ( x) dx and Var ( X ) 2 E X E ( X ) 2 ( x u) 2 f ( x) dx Important Properties of Expected Value and Variance: a, b are constants and X is discrete or continuous. 1. Var X E X 2 2 . 2. EaX b aE X b . 3. Var aX b a 2Var X Example 1: The probability distribution functions (discrete random variable) or probability density functions (continuous random variable) for a random variable X are (a) c exp 6 x , x 0 f ( x ) cx, 1 x 0 0, otherwise 16 17 (b) f x cx 2 exp x 3 , x 0 (c) x 1 f x c , x 0, 1, 2, 3 Find c . [solution:] (a) cx 2 c exp 6 x cxdx c exp 6 x dx 1 1 0 1 2 6 0 1 c c 3c c 3 1 1 c 2 6 6 2 0 0 (b) c exp x 3 1 3 3 cx exp x dx 1 c exp x dx 1 1 0 0 3 3 0 c 1 c 3 3 2 3 (c) x 1 1 1 2 1 1 3c 1 c 1 c 1 1 c 2 1 1 x 0 3 3 3 3 c 2 3 Example 2: The probability distribution function for a discrete random variable X is f ( x ) 2k , x 10 k , x 20 k 0.2, x 30 0, otherwise where k is some constant. Please find (a) k (b) P( X 15 or X 40 ) (c) E X and Var X (d) E5 X 2 (e) Var 3 X 7 17 18 [solution:] (a) f ( x) f (10) f (20) f (30) 2k k k 0.2 1 x k 0.3 . (b) P( X 15 or X 40) P( X 10 or X 20 or X 30) 1 . (c) u E ( X ) xf ( x) 10 f (10) 20 f (20) 30 f (30) x 10 0.6 20 0.3 30 0.1 15 and E ( X 2 ) x 2 f ( x) 10 2 f 10 20 2 f 20 30 2 f 30 x 100 0.6 400 0.3 900 0.1 270 Var X E X 2 2 270 225 45 (d) E5 X 2 5E X 2 5 15 2 77. (e) Var (3 X 7) 32 Var X 9 45 405 . Example 3: Let X be a discrete random variable representing the number of hours a college student spending on reading novels per week. The following probability distribution has been proposed. i3 f i , i 1, 2, 3 , 9c where k is some constant. (a) Compute c. (b) Compute P X 1.2 and P X 2.2 . (c) Compute E X a nd Var X . [solution:] (a) f (i) f (1) f (2) f (3) i 13 23 33 36 4 1 9c 9c 9c 9c c c 4. (b) P ( X 1.2) P ( X 1) f 1 1 1 .and 9 4 36 P( X 2.2) P( X 3) f 3 (c) 18 33 27 . 9 4 36 19 u E ( X ) if (i ) 1 f (1) 2 f (2) 3 f (3) i 1 1 8 27 98 49 2 3 36 36 36 36 18 and E ( X 2 ) i 2 f (i ) 12 f 1 2 2 f 2 32 f 3 i 1 1 8 27 276 46 4 9 36 36 36 36 6 Var X E X 2 2 2 46 49 0.2561 6 18 Example 4: The probability density function for a continuous random variable X is f ( x ) a bx 2 , 0 x 1 0, otherwise. where a, b are some constants. Please find (a) a, b if E ( X ) 3 (b) Var( X ) . 5 [solution:] (a) a bx dx 1 1 1 f ( x)dx 1 2 0 ax 0 a b 3 1 x |0 1 3 b 1 3 and 1 1 0 0 E ( X ) xf ( x)dx x a bx 2 dx Solve for the two equations, we have a a 2 b 4 1 a b 3 x x |0 2 4 2 4 5 3 6 . , b 5 5 (b) f ( x) 3 6 x2 , 0 x 1 5 5 0, otherwise. Thus, 19 20 Var ( X ) E X E ( X ) E ( X ) E ( X ) 2 2 2 3 E( X ) 5 2 9 3 6 2 x dx 25 5 5 0 1 6 5 1 9 1 6 9 2 x3 x |0 5 25 25 5 25 25 25 1 1 9 x f ( x) dx 25 0 x 2 2 Example 5: The probability density function for a continuous random variable X is x 2 ,2 x 4 f ( x ) 18 0, otherwise Please find (a) P X 1 (b) P X 2 9 (c) E X and Var X (d) E 9 X 2 8 X 2 (e) Var6 X 8 . [solution:] 1 x2 x x2 1 1 1 1 2 P X 1 P 1 X 1 dx 18 36 9 1 36 9 36 9 9 1 1 (a) (b) P X 2 9 P 3 X 3 x2 3 18 dx 3 2 0dx 3 3 x2 x x2 25 2 18 dx 36 9 36 2 3 (c) 4 EX 2 x 2 dx x 18 4 x2 x x3 x2 18 9 dx 54 18 2 . 2 2 4 Since x 4 E X 2 2 2 x 2 dx 18 4 x3 x2 x4 x3 18 9 dx 72 27 6 , 2 2 4 Var X E X E X 2 2 6 2 2 2 . 2 (d) E 9 X 2 8 X 2 9E X 2 8E X 2 9 6 8 2 2 72 . (e) Var 6 X 8 6 2 Var X 36 2 72 20 2 21 Definition (cumulative distribution function (c.d.f)): F x P X x. Important Properties of CDF: 1. CDF is right continuous. 2. As X is discrete, F x f a ; ai x i As X is continuous and f is continuous, f x F ' x . Example 2 (continuous): The probability distribution function for a discrete random variable X is f ( x ) 0.6, x 10 0.3, x 20 0.1, x 30 0, otherwise Thus, the CDF is F ( x ) 0, x 10 0.6, 10 x 20 0.3, 20 x 30 1, x 30 Example 1 (a): The probability distribution functions (discrete random variable) or probability density functions (continuous random variable) for a random variable X are 3 2 exp 6 x , x 0 3 f ( x) x, 1 x 0 2 0, otherwise Thus, as 1 x 0 , x 3x 2 3x 3 2 F x dx 1 x 2 4 1 4 1 x as 0 x , 3x 3 3 3 exp 6 x 3 1 F x dx exp 6 x dx 1 exp 6 x . 2 2 4 2 6 0 4 4 1 0 x x x 21 22 Thus, the CDF is F ( x ) 0, x 1 3 1 x2 , 1 x 0 4 3 1 1 exp 6 x , 0 x 4 4 22