Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Regression analysis wikipedia , lookup
Information theory wikipedia , lookup
Birthday problem wikipedia , lookup
Simplex algorithm wikipedia , lookup
Generalized linear model wikipedia , lookup
Probability box wikipedia , lookup
Least squares wikipedia , lookup
Uniform random variable • A random variable is said to be uniformly distributed over the interval (a, b) if its probability density function is 1 given by , axb ba f(x) 0, otherwise • Theorem. If U is uniformly distributed over (0, 1) and a and b are real numbers with a < b, then X = (b – a)U + a is uniformly distributed over (a, b). • The latter theorem will allow us to use Maple’s random number generator to generate any uniformly distributed random variable that we want. Uniform random variable, continued • Theorem. If X is uniformly distributed over (a, b), then (b a) 2 ab E[X] , Var(X) . 2 12 • Problem. Buses arrive at a specified stop at 7:15am and 7:30am. If a passenger arrives at the stop at a time that is uniformly distributed between 7 and 7:30, find the probability that she waits less than 5 minutes for the bus. Solution. We use a uniform r. v. on (0, 30). The desired probability is given by: P{10 X 15} P{25 X 30} 15 1 10 30 dx 30 1 25 30 dx 13 . Normal random variables • X is a normal random variable (or normally distributed) with parameters (, 2) if the density of X is 1 ( x ) 2 / 2 2 f(x) e , x . 2 • We often write X ~ N( , 2 ). • Theorem. If X is normally distributed (, 2), then 1. Z = (X – )/ is normally distributed (0, 1), that is, Z is a standard normal random variable. 2. E[X] = . 3. Var(X) = 2. Cumulative distribution function of a standard normal r. v. • The c. d. f. of a standard normal random variable is traditionally denoted by (x). That is, x 1 y2 / 2 (x) e dy. 2 y2 / 2 • Theorem. It follows from the symmetry of e that (x) 1 (x). • (x) is tabulated in the textbook in Tables 1 and 2. • Example. Let X be a normal random variable with parameters (3, 9). We “standardize” to get the following probability: 23 X3 53 P{2 X 5} P{ } 3 3 3 (2 / 3) (1 / 3) 0.3779 68-95-99.7 Rule • If X~ N( , 2 ), then P( - X ) 0.68 P( - 2 X 2 ) 0.95 P( - 3 X 3 ) 0.997. Scottish Soldier’s Chest Size is Normally Distributed • Let X be the chest measurement in inches of the chest size of a Scottish soldier. Belgian scholar Quetelet has determined that X ~ N(39.8, 2.052) . • What is the probability that a randomly selected Scottish soldier has a chest size of 40 or more inches? X 39.8 40 39.8 P(X 40) P( ) 2.05 2.05 P(Z 0.10) 1 (0.1) 0.46 • Note the use of Z to denote a standard normal r. v. Scores on an Achievement Test • The scores on an achievement test given to 100,000 students are normally distributed, N(500, 1002). What should the score of a student be to place him among the top 10% of all students? • We want to find the score t such that P(X t) 0.10 or P(X t) 0.90. • This becomes X 500 t 500 t 500 P( ) 0.90, or ( ) 0.90. 100 100 100 • From Table 2, t 500 1.28 so that t 628. 100 Normal approximation to the binomial distribution • Theorem (DeMoivre-Laplace) If Sn denotes the number of successes in n independent trials, each resulting in a success with probability p, then for any a < b, Sn np P{a b} (b) (a), as n . np(1 p) • Problem. Let X be the number of times a coin, flipped 40 times, lands heads. Find P{X=20} using the normal approximation with the continuity correction. Solution. P{X 20} P{19.5 X 20.5} P{.16 X 20 .16} 10 (.16) (.16) 0.1272 Exponential random variable • A continuous random variable whose density is f(x) e x , x 0 0 , x0 for some > 0 is an exponential random variable with parameter . • Theorem. For an exponential r. v. with parameter , F(a) 1 e a , a 0 0 , a0 Also, E[X] = 1/ and Var(X) = 1/ 2. Exponential random variable continued • The exponential random variable is often the amount of time until some specific event occurs. For example, it could be the amount of time (starting now) until an earthquake occurs. • Problem. Suppose that the length of an ATM session in minutes is an exponential random variable with = 0.1. If someone arrives immediately ahead of you at an ATM, find the probability that you will wait between 10 and 20 minutes. Solution. P{10 X 20} F(20) F(10) e 1 e 2 0.233 • A nonnegative random variable X is memoryless if P{X s t | X t} P{X s} for all s, t 0. • Theorem. Exponentially distributed random variables are memoryless. Proof of memoryless property for exponential random variable • We want to show that P(X s t, X t) P(X s). P(X t) • This is equivalent to P(X s t) P(X s)P(X t). • The result follows using the following: P(X s t) 1 [1 e (s t) ] e (s t) P(X s) 1 [1 e s ] e s P(X t) 1 [1 et ] et The Cauchy distribution • A random variable has a Cauchy distribution with parameter if its density is given by 1 1 f( x) , x . 2 1 (x ) • Theorem. If the angle in the figure below is uniformly distributed from –/2 to /2, then X has the Cauchy distribution with = 0. flashlight 1 x-axis X Flashlight Interpretation • Suppose that a narrow beam flashlight is spun around its center, which is located a unit distance from the x-axis (see the figure on the previous slide). When the flashlight has stopped spinning, consider the point X at which the beam intersects the x-axis. (If the beam is not pointing toward the x-axis, repeat the experiment.) • F(x) = P(X x) P(tan x) P( arctan x) 1 1 arctan x, 2 where the latter equality follows since α is uniformly distributed over (–/2, /2). Upon differentiation of F(x), we see that the density function of X is the density function of the Cauchy distribution with = 0.