Section 5: Expected Value; Percentiles, Median, and Mode

Transcription

1 Section 5: Expected Value; Percentiles, Median, and Mode September 30th, 2014

2 In this lesson we define a basic statistical function of a random variable, the expected value. The expected value will become our key tool in analyzing a random variable X, and it will be the basis for the other important functions that we will discuss in the next several lessons.

3 Say you are tossing a fair six-sided die, with sides numbered from 1 through 6. We know that for any one toss, we have an equal chance of coming up with the result 1, 2, 3, 4, 5, or 6. However, the more often we toss the die (say as the number of tosses tends to infinity), we expect the outcomes to approach a certain value. We call this average value the mean, or expected value, of the associated random variable. If X is our random variable, we denote the expected value of X by E[X] or µ X.

4 Suppose X is a discrete random variable. Then we define E[X] as follows: E[X] = x p(x) = x 1 p(x 1 ) + x 2 p(x 2 ) + where x ranges over all points where X has nonzero probability and p(x) is the probability function of X. For our previous example, the possible outcomes are i = 1, 2,..., 6, and p(i) = 1 for each i. Therefore, if X is the 6 random variable denoting the outcome of tossing a fair six-sided die, we have E[X] = 1( 1 6 ) + 2( 1 6 ) + + 6( 1 6 ) = 7 2 Thus as we continue to throw the die, we expect the outcomes to approach the value 7 (note X can never actually take this 2 value).

5 For a continuous random variable X, we define the expected value as For the pdf we have E[X] = f (x) = x f (x) dx { 2x 0 < x < 1 0 otherwise E[X] = 1 0 x 2x dx = 1 0 2x 2 dx = 2 3 x = 2 3 We think of the expected value as giving us the weighted center of the distribution of X.

6 Given a function h(x), we can define the expected value of h(x): E[h(X)] = h(x) p(x) x h(x) f (x) dx So in the previous (continuous) example, E[ X] = 1 0 X is discrete X is continuous 1 x 2x dx = 2x 3/2 dx = We call E[X n ] the n-th moment of X. If µ is the mean of X, then the n-th central moment of X is E[(X µ) n ].

7 Example (5.1) Let X equal the number of tosses of a fair die until the first 1 appears. Find E[X].

8 Example (5.2) A fair die is tossed until the first 1 appears. Let x equal the number of tosses required. You are to receive (.5) x dollars if the first 1 appears on the x-th roll. What is the expected amount that you will receive?

9 Suppose X is a continuous random variable with pdf f (x), and suppose that there exists a value c such that f (c + t) = f (c t) for all t > 0. In this case, we say that X has a symmetric distribution about the point x = c. If X has a symmetric distribution about x = c, then E[X] = c. While it will likely never be necessary for you to recognize and utilize this fact of a symmetric distribution, it can save you significant time in your work.

10 For 0 < p < 1, the 100p-th percentile of the distribution of X is the number c p which satisfies both P[X c p ] p and P[X c p ] 1 p. If X is continuous, it is enough to find the value c p for which P[X c p ] = p. We will generally see percentiles for continuous distributions. If p =.5, then c.5 = M is the 50-th percentile of the distribution, and is called the median of the distribution. Thus M is the value where half of the distribution is to the left and half of the distribution is to the right. The mode of a distribution is any point m at which the probability or density function f (x) is maximized.

11 Example (5.3) An insurer s annual weather related loss, X, is a random variable with density function f (x) = Find the median of X. { 2.5(200) 2.5 x 3.5 for x otherwise

12 One important result on standard deviations is the following: For any constants a 1, a 2, b and functions h 1, h 2, E[a 1 h 1 (X) + a 2 h 2 (X) + b] = a 1 E[h 1 (X)] + a 2 E[h 2 (X)] + b. Note that this means that E[X 2 + 4X + 4] = E[X 2 ] + 4E[X] + 4, or similar for any polynomial in X.

13 Example (5.4) Smith finds the carnival game over-under-seven irresistible. The game involves the random toss of two fair dice. If a player bets 1 on over and the total on the dice is over 7, then the player wins 1 (otherwise he loses the 1 he bet). If he bets 1 on under and the total on the dice is under 7 then he wins 1 (otherwise he loses). If he bets 1 on seven and the total on the dice is 7 then he wins 4 (otherwise he loses 1). Smith devises the following strategy. His first bet is 1 on under. If he wins, he walks away with his net gain of 1. If he loses, he doubles his bet to 2 on under. If he wins, he walks away with his net gain of 1. If he loses, he doubles his bet to 4 and bets on under, etc. Smith walks away as soon as he wins. Find his expected gain, and the number of bets it will take to win, on average.