Math438 Actuarial Probability

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1 Math438 Actuarial Probability Jinguo Lian Department of Math and Stats Jan. 22, 2016

2 Continuous Random Variables-Part I: Definition A random variable X is continuous if its set of possible values is an entire interval of numbers A probability distribution or probability density function (pdf) of X is a function f(x) such that for any two numbers a and b P(a X b) = b a f (x)dx

3 Uniform Distribution A continuous rv X is said to have a uniform distribution on the interval [a, b] if the pdf of X is { 1 b a, a x b; 0, otherwise. E(x) = a + b 2 Var(X ) = (b a)2 12

4 Continuous Random Variables-Part I: C.D.F. If X is a continuous rv, then for any number c, P(x = c) = 0. For any two numbers a and b with a < b, P(a X b) = P(a < X b) = P(a X < b) = P(a < X < b) The cumulative distribution function,f(x) for a continuous rv X is defined for every number x by F (x) = P(X x) = x f (y)dy P(X > a) = 1 F (a), P(a X b) = F (b) F (a) F (x) = f (x)

5 Percentile, Median, Mean and Variance The (100p)th percentile of the distribution of a continuous rv X denoted by x p, is defined by p = F (x p ) The median of a continuous distribution, denoted by µ, is the 50th percentile, µ = x 0.5. Expected value µ X = E(X ) = x f (x)dx Expected Value of h(x ), µ h(x ) = E(h(X )) = h(x) f (x)dx Variance σx 2 = Var(X ) = (x µ)2 f (x)dx = E((X µ) 2 ) Standard Deviation σ X = Var(X ), Var(X ) = E(X 2 ) [E(X )] 2

6 Example 1 The loss due to a fire in a commercial building is modeled by a random variable X with density function { 0.005(20 x), for 0 < x < 20; 0, otherwise. Given that a fire loss exceeds 8, what is the probability that it exceeds 16? (A) 1/25 (B) 1/9 (C) 1/8 (D) 1/3 (E) 3/7

7 Solution of Example 1 Solution:B Note that P(X > x) = 20 where 0 < x < 20. Therefore, x (0.005(20 t)dt = 0.005(200 20x x 2 ) P(X > 16 X > 8) = P(X > 16) P(X > 8) = 1 9

8 Example 2 The lifetime of a machine part has a continuous distribution on the interval (0, 40) with probability density function f, where f (x) is proportional to (10 + x) 2. What is the probability that the lifetime of the machine part is less than 5? (A) 0.03 (B) 0.13 (C) 0.42 (D) 0.58 (E) 0.97

9 Solution of Example 2 Solution: C Let the random variable T be the future lifetime of a 30-year-old. We know that the density of T has the form f (x) = C(10 + x) 2 for 0 < x < 40 (and it is equal to zero otherwise). First, determine the proportionality constant C from the condition 40 0 f (x)dx = 1 C = 12.5 Then, calculate P(T < 5) by integrating f (x) = 12.5(10 + x) 2 over the interval (0, 5).

10 Example 3 A group insurance policy covers the medical claims of the employees of a small company. The value, V, of the claims made in one year is described by V = 100, 000Y where Y is a random variable with density function { k(1 y 4 ), for 0 < y < 1; 0, otherwise. where k is a constant. What is the conditional probability that V exceeds 40,000, given that V exceeds 10,000? (A) 0.08 (B) 0.13 (C) 0.17 (D) 0.20 (E) 0.51

11 Solution of Example 3 Solution: B To determine k, note that 1 = We next need to find 1 0 k(1 y) 4 dy k = 5 P[V > 10, 000] = P[100, 000Y > 10, 000] = P[Y > 0.1] = and P[V > 40, 000] = P[100, 000Y > 40, 000] = P[Y > 0.4] = It now follows that P[V > 40, 000 V > 10, 000] = P[V > 40, 000] P[V > 10, 000] = (1 y) 4 d 5(1 y) 4 d

12 Example 4: Mode The mode is the value that appears most often in a set of data. The mode of a discrete probability distribution is the value x at which its probability mass function takes its maximum value. In other words, it is the value that is most likely to be sampled. The mode of a continuous probability distribution is the value x at which its probability density function has its maximum value, so the mode is at the peak. The loss amount, X, for a medical insurance policy has cumulative distribution function 0, x < 0; F (x) = 1 9 (2x 2 x3 3 ), 0 x 3; 1, x > 3. Calculate the mode of the distribution. (A) 2/3 (B) 1 (C) 3/2 (D) 2 (E) 3

13 Solution of Example 4 Solution: D The mode of a distribution is the point where f(x) is maximal. With the given distribution we have f (x) = F (x) = (4/9)x (1/9)x 2 for 0 < x < 3, and f (x) = 0 outside the interval [0, 3]. Differentiating, we get f (x) = (4/9) (2/9)x, and setting this equal to 0, we see that x = 2 is the only critical point of f. Hence the mode must occur at x = 2.

14 Example 5 An insurance company insures a large number of homes. The insured value, X, of a randomly selected home is assumed to follow a distribution with density function { 3x f (x) = 4, for x > 1; 0, otherwise. Given that a randomly selected home is insured for at least 1.5, what is the probability that it is insured for less than 2? (A) (B) (C) (D) (E) 0.875

15 Solution of Example 5 Solution: A Let F denote the distribution function of f. Then F (x) = P(X x) = Using this result, we see x 1 f (x)dx = 1 x 3 P(X < 2 X 1.5) = P(X < 2 X 1.5) P(X 1.5) = P(1.5 X < 2) P(X 1.5) = F (2) F (1.5) 1 F (1.5) = 0.578

16 Example 6 An insurance policy pays for a random loss X subject to a deductible of C, where 0 < C < 1. The loss amount is modeled as a continuous random variable with density function { 2x, for 0 < x < 1; f (x) = 0, otherwise. Given a random loss X, the probability that the insurance payment is less than 0.5 is equal to Calculate C. (A) 0.1 (B) 0.3 (C) 0.4 (D) 0.6 (E) 0.8

17 Solution of Example 6 Solution: B Denote the insurance payment by the random variable { 0, if 0 < x C; Y = X C, if C < x < 1. Then we are given that 0.64 = P(Y < 0.5) = P(0 < X < 0.5+C) = 0.5+C 0 2xdx = (0.5+C) 2 Therefore, solving for C, we find C = ± Finally, since 0 < C < 1, we conclude that C = 0.3.

Notes for Math 324, Part 20

Notes for Math 324, Part 20 7 Notes for Math 34, Part Chapter Conditional epectations, variances, etc.. Conditional probability Given two events, the conditional probability of A given B is defined by P[A B] = P[A B]. P[B] P[A B]