2 Continuous Random Variables and their Distributions

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1 Name: Discussion-5 1 Introduction - Continuous random variables have a range in the form of Interval on the real number line. Union of non-overlapping intervals on real line. - We also know that for any x R,P (X = x) = 0. - Analogous to the theory of discrete random variables. 2 Continuous Random Variables and their Distributions - From the definition of CDF, F X (x) = P (X x) we get the definition of a continuous random variable Definition:A random variable X with CDF F X (x) is said to be continuous if F X (x) is a continuous function for all x R. - The CDF is a continuous function with no jumps. - No jumps is consistent with the fact that P (X = x) = 0 for all x. - CDF of a continuous random variable is differentiable almost everywhere in R. 2.1 Probability Density Function - For continuous random variables, CDF works but PMF does not since P (X = x) = 0. - For a continuous random variable X, we can intuitively define the Probability Density Function f X (x) as P (x < X x + ) f X (x) = lim. 0 +

2 FX(x) 1 a b x Figure 1: CDF for a continuous random variable uniformly distributed over [a, b]. - f X (x) is limit of the probability density of an interval as the length of the interval goes to 0. - Formally, we define the PDF of random variable X as Definition: Consider a continuous random variable X with CDF F X (x). The function f X (x) is the probability density function (PDF) of X, defined by f X (x) = df X(x) dx = F X(x), if F X (x) is differentiable at x. - If f X (x 1 ) > f X (x 2 ), we can say the value of X is more likely to be around x 1 than x 2. - The CDF can be obtained from PDF by integration F X (x) = - We also have x f X (u)du. P (a < X b) = F X (b) F X (a) = - Properties of PDF- b a f X (u)du. 2

3 Consider a continuous random variable X with PDF f X (x). We have 1. f X (x) 0 for all x R. 2. f X(u)du = P (a < X b) = F X (b) F X (a) = b a f X(u)du. 4. More generally, for a set A, P (X A) = A f X(u)du. 2.2 Range - Range of a random variable X is the set of all possible values of the random variable. - For a continuous random variable, we can define it as the set of all real numbers with non-zero PDF. R X = {x f X (x) > 0} 2.3 Expected Value and Variance - The definition of expected value of a continuous random variable is EX = xf X (x)dx - Expectation is a linear operation. E[aX + b] = aex + b for all a, b R E[X 1 + X X n ] = EX 1 + EX EX n for any set of random variables X 1, X 2,.., X n. 2.4 Expected Value of a Function of a Continuous Random Variable - We can use LOTUS to determine the expected value of a function of a continuous random variable Law of the unconscious statistician (LOTUS) for continuous random variables: E[g(X)] = g(x)f X(x)dx 3

4 2.4.1 Variance - Variance of a random variable is defined as V ar(x) = E[(X µ X ) 2 ] = EX 2 (EX) 2 - For a continuous random variable we can write Var(X) = E [ (X µ X ) 2] = = EX 2 (EX) 2 = (x µ X ) 2 f X (x)dx x 2 f X (x)dx µ 2 X - For a, b R, we have V ar(ax + b) = a 2 V ar(x) 2.5 Discrete vs Continuous Random Variables - Differences between discrete and continuous random variables. Discrete PMF P X (x) = P (X = x) EX = x k R X x k P X (x k ) LOTUS E[g(x)] = g(x k )P X (x k ) x R X Continuous PDF f X (x) = df X(x) dx EX = xf X(x)dx LOTUS E[g(X)] = g(x)f X(x)dx 4

5 3 Special Distributions 3.1 Uniform Distribution A continuous random variable X is said to have a Uniform distribution over the interval [a, b], shown as X Uniform(a, b), if its PDF is given by f X (x) = { 1 b a a < x < b 0 x < a or x > b EX = a + b 2 (b a)2, Var(X) = Exponential Distribution A continuous random variable X is said to have an exponential distribution with parameter λ > 0, shown as X Exponential(λ), if its PDF is given by { λe λx x > 0 f X (x) = 0 otherwise If X Exponential(λ), then EX = 1 λ and Var(X) = 1 λ 2. If X is exponential with parameter λ > 0, then X is a memoryless random variable, that is P (X > x + a X > a) = P (X > x), for a, x 0. 5

6 4 Problems 1. Let X be a random variable with PDF given by { 3 f X (x) = 2 x2 x 1 0 otherwise (a) Find P (0 X 1 2 ). (b) Find EX and Var(X). (c) Find EX 4. Solution: (a) To find P (0 X 1 ), we can write 2 P (0 X 1 2 ) = 3 2 (b) To find EX, we can write 2 0 x 2 dx = EX = = 3 2 = 0. uf X (u)du u 3 du In fact, we could have guessed EX = 0 because the PDF is symmetric around x = 0. To find Var(X), we have Var(X) = EX 2 (EX) 2 = EX 2 = = 3 2 = 3 5. u 2 f X (u)du u 4 du (c) To find EX 4, we can write 6

7 EX 4 = = 3 2 = 3 7. u 4 f X (u)du u 6 du 7

8 2. Suppose the number of customers arriving at a store obeys a Poisson distribution with an average of λ customers per unit time. That is, if Y is the number of customers arriving in an interval of length t, then Y P oisson(λt). Suppose that the store opens at time t = 0. Let X be the arrival time of the first customer. Show that X Exponential(λ). Solution: We first find P (X > t): P (X > t) = P (No arrival in [0, t]) λt (λt)0 = e 0! = e λt. Thus, the CDF of X for x > 0 is given by F X (x) = 1 P (X > x) = 1 e λx, which is the CDF of Exponential(λ). Note that by the same argument, the time between the first and second customer also has Exponential(λ) distribution. In general, the time between the kth and k + 1th customer is Exponential(λ). 8