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STAT 430/510 Lecture 10 STAT 430/510 Probability Lecture 10: Continuous Random Variable Pengyuan (Penelope) Wang June 13, 2011 STAT 430/510 Lecture 10 Introduction The set of possible values for discrete random variable is discrete However, there also exist random variables who can take values on a whole interval STAT 430/510 Lecture 10 Definition of Continuous Random Variable X is a continuous random variable if it take continuous values STAT 430/510 Lecture 10 Definition of Continuous Random Variable X is a continuous random variable if it take continuous values There exists a nonnegative function f , having the property that, for any set B of real numbers P(X ∈ B) = f (x)dx B for example, for set B = (a, b), P(X ∈ B) = b a f (x)dx The function f is called the probability density function of random variable X STAT 430/510 Lecture 10 Definition of Continuous Random Variable X is a continuous random variable if it take continuous values There exists a nonnegative function f , having the property that, for any set B of real numbers P(X ∈ B) = f (x)dx B for example, for set B = (a, b), P(X ∈ B) = b a f (x)dx The function f is called the probability density function of random variable X f must satisfy: f ≥ 0, f (x)dx = Important: how to interpret f ? STAT 430/510 Lecture 10 How to represent the probability distribution of such random variables? Think of PMF It is sort of like the continuous version of PMF For any set B of real numbers P(X ∈ B) = f (x)dx B for example, for set B = (a, b), P(X ∈ B) = b a f (x)dx STAT 430/510 Lecture 10 Example The amount of time in hours that a computer functions before breaking down is a continuous random variable with probability density function given by f (x) = −x/100 , 100 e x ≥0 0, x < Confirm that it is a density function STAT 430/510 Lecture 10 Probability on an interval P(a ≤ X ≤ b) = b a f (x)dx The amount of time in hours that a computer functions before breaking down is a continuous random variable with probability density function given by f (x) = −x/100 , 100 e x ≥0 0, x < What is the probability that (a) it will function for 100 - 150 hours? STAT 430/510 Lecture 10 The probability that a continuous random variable would take exactly one number is P(a ≤ X ≤ b) = b a f (x)dx Then for any value u, P(X = u) = (u ≤ X ≤ u) = u u f (x)dx = The probability that the computer would function for exactly 100 hours is STAT 430/510 Lecture 10 Cumulative Probability Function Cumulative Probability Function: F (x) = P(X < x) = P(X ≤ x) = x ∞ f (u)du The amount of time in hours that a computer functions before breaking down is a continuous random variable with probability density function given by f (x) = −x/100 , 100 e x ≥0 0, x < What is the probability that (b) it will function for fewer than 100 hours? STAT 430/510 Lecture 10 Summary of Properties of Continuous Random Variable = P(X ∈ (−∞, ∞)) = P(a ≤ X ≤ b) = b a ∞ −∞ f (x)dx f (x)dx P(X = u) = Cumulative Probability Function: F (x) = P(X < x) = P(X ≤ x) = x −∞ f (u)du STAT 430/510 Lecture 10 Expected Value If X is a continuous random variable with probability density function f (x), then its expected value is, ∞ E[X ] = xf (x)dx −∞ And, for any function g, ∞ E[g(X )] = g(x)f (x)dx −∞ It is totally analogous to the discrete case STAT 430/510 Lecture 10 Expected Value and Variance of Continuous R.V For continuous random variable X with pdf f (x), ∞ −∞ xf (x)dx ∞ E[X ] = −∞ x f (x)dx ∞ Var (X ) = −∞ (x − µ)2 f (x)dx E[X ] = µ = EX SD(X ) = Var (X ) = E[X ] − (E[X ])2 , where STAT 430/510 Lecture 10 Properties of Expected Value and Variance: totally the same as the discrete case E[aX + b] = aE[X ] + b, where a and b are constants Var (aX + b) = a2 Var (X ) , where a and b are constants E[X + Y ] = E[X ] + E[Y ], where X and Y are random variables Var [aX + bY ] = a2 Var [X ] + b2 Var [Y ], if X and Y are independent They are the same as the discrete case STAT 430/510 Lecture 10 Example Find E[X ] and Var(X) when the density function X is f (x) = x2xdx = 2/3 E[X ] = x 2xdx = 1/2 Var (X ) = E[X ] − (E[X ])2 2x, ≤ x ≤ 0, otherwise E[X ] = = 1/18 STAT 430/510 Lecture 10 Example The density function X is given by f (x) = Find E[eX ] E[eX ] = x e 1dx =e−1 1, ≤ x ≤ 0, otherwise STAT 430/510 Lecture 10 Uniform Random Variable A continuous random variable X is said to have a uniform distribution on the interval [A, B], if X can take any value in [A, B] and the distribution is flat on every points Probability density function: f (x) = B−A , A≤x ≤B 0, otherwise STAT 430/510 Lecture 10 cdf For uniform r.v X on [A, B], the cdf is  0, m < A  x−A , A≤x ≤B F (x) =  B−A 1, x > B STAT 430/510 Lecture 10 Expected Value and Variance X is uniform random variable on [A, B] E[X ] = Var (X ) A+B 2 = (B−A) 12 STAT 430/510 Lecture 10 Example If X is uniformly distributed over (0,10), calculate the probability that (a) X < (b) X > (c) < X < P(X < 3) = 10 dx = 10 STAT 430/510 Lecture 10 Example If X is uniformly distributed over (0,10), calculate the probability that (a) X < (b) X > (c) < X < P(X < 3) = P(X > 6) = 3 10 dx = 10 10 10 dx = STAT 430/510 Lecture 10 Example If X is uniformly distributed over (0,10), calculate the probability that (a) X < (b) X > (c) < X < 3 10 dx = 10 10 dx = 25 P(X > 6) = 10 P(3 < X < 8) = 10 dx = 12 P(X < 3) = STAT 430/510 Lecture 10 Example Buses arrive at a specific stop at 15-minute interval starting at A.M That is, they arrive at 7, 7:15, 7:30, 7:45, and so on If a passenger arrives at the stop at a time that is uniformly distributed between and 7:30, find the probability that he waits (a) less than minutes for a bus (b) more than 10 minutes for a bus STAT 430/510 Lecture 10 Example: Solution Let X denote the number of minutes past that the passenger arrives at the stop {waiting for < 5min} = {10 < X < 15} ∪ {25 < X < 30} P(10 < X < 15) + P(25 < X < 30) = 15 30 1 10 30 dx + 25 30 dx = {waiting for > 10min} = {0 < X < 5} ∪ {15 < X < 20} 20 P(0 < X < 5) + P(15 < X < 20) = 30 dx + 15 30 dx = STAT 430/510 Lecture 10 last Example A stick of length is split at a point U that is uniformly distributed over (0,1) Determine the expected length of the piece that contains the mid point