ECE353: Probability and Random Processes. Lecture 5 - Cumulative Distribution Function and Expectation

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1 ECE353: Probability and Random Processes Lecture 5 - Cumulative Distribution Function and Expectation Xiao Fu School of Electrical Engineering and Computer Science Oregon State University xiao.fu@oregonstate.edu

2 From PMF to CDF Recall PMF of a discrete RV is P [X = x]. Definition: the Cumulative Distribution Function (CDF): F X (x) := P [X x]. very useful since in many cases we care about P [X x]. it comes very handy in calculating things like P [l X u]. ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 1

3 Example: X with PMF From PMF to CDF, x = , x = 2 0.2, x = 3 0, o.w ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 2

4 From PMF to CDF Example: from PMF to CDF (F X (x) = P X [X x]), x = , x = 2 0.2, x = 3 F X (x) = 0, o.w. 0, x < ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 3

5 From PMF to CDF Example: from PMF to CDF (F X (x) = P X [X x]), x = , x = 2 0.2, x = 3 F X (x) = 0, o.w. 0, x < 1, x = ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 4

6 From PMF to CDF Example: from PMF to CDF (F X (x) = P X [X x]), x = , x = 2 0.2, x = 3 F X (x) = 0, o.w. 0, x < 1, 1 x < ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 5

7 From PMF to CDF Example: from PMF to CDF (F X (x) = P X [X x]), x = , x = 2 0.2, x = 3 F X (x) = 0, o.w. 0, x < 1, 1 x < x = ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 6

8 From PMF to CDF Example: from PMF to CDF (F X (x) = P X [X x]), x = , x = 2 0.2, x = 3 F X (x) = 0, o.w. 0, x < 1, 1 x < x < 3 1 x ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 7

9 Properties of CDF Some important properties of CDF: 1) F X ( ) = 0 and F X (+ ) = 1. 2) F X (x) 0. 3) x x, we have F X (x ) F X (x). 4) F X (x) is a constant between two consecutive values x 1 and x 2. 5) P [α < X β] = F X (β) F X (α). ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 8

10 Sample Mean and Expectation Consider a collection of random samples {X 1,..., X N }. mean: sample mean := 1 N X i N i=1 Compute the sample X i corresponds to the ith sample; or, the outcome of ith trial of a random experiment. The problem with sample mean is that itself is random. To fix this, a solution is to take N, in which case, under certain conditions, it can be shown that the sample mean converges to ensemble mean, or, the expectation of a random variable X. Definition: the expectation of a RV X is definied as E[X] := µ X = x S X xp X (x) ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 9

11 Expectation Example: Let us draw a fair die. We have S X = {1, 2, 3, 4, 5, 6} and { 1 6, x {1, 2, 3, 4, 5, 6} 0, o.w. E[X] = x {1,2,...,6} 1 6 x = 1 6 ( ) = 3.5 Why did we say the sample mean converges to E[X]? ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 10

12 Expectation Consider a collection of samples {X 1,..., X N }. Denote N(x j ) = N 1(X i = x j ), where 1(X = x) = i=1 { 1, X = x 0, o.w. In plain words, N(x j ) is the times of seeing x j in the set of samples. Consequently, we have 1 N N X i = 1 N i=1 x j S X x j N(x j ) = N(x j ) x j N, x j S X where we have lim N N(x j ) N = P X(x j ). ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 11

13 Expectation of Bernoulli RV Example: Bernoulli 0-1 RV: { 0, w.p. 1 p 1, w.p. p. By definition, E[X] = 0 (1 p) + 1 p = p. What if we have { 3, w.p. 1 p 5, w.p. p. ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 12

14 Expectation of Geometric RV Example: Geometric RV: { p(1 p) x 1, x {1, 2, 3,...} 0, o.w. By definition, we have E[X] = = p = p xp(1 p) x 1 x=1 xq x 1 (q = 1 p) x=1 x=1 dq x dq ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 13

15 By definition, we have Expectation of Geometric RV E[X] = xp(1 p) x 1 x=1 = p d ( x=1 qx ) (q = 1 p) dq = p d ( p(1 + p + p ) ) ( ) d p 1 1 q = p dq This is intuitive: Consider the coffee shop example: the number of visits that you need to meet your barista is inversely proportional to the probability that you can meet him/her there each time. dq = 1 p ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 14

16 Example: Poisson RV: Expectation of Poisson RV { α x x! e α, x {0, 1, 2, 3,...} 0, o.w. By definition, we have E[X] = = = x=0 x=1 x αx x! e α x αx α x x! e α = x (x 1)! e α x=1 α y+1 e α = α α y y! y! e α = α y=0 y=0 ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 15

17 Limit of Poisson RV Theorem: The Poisson PMF is limit of the Binomial(n, p) PMF, i.e., n and p 0 = np α. Proof: The Binomial(n, p) PMF is Taking p = α/n, we wish to show ( n k ) (α n ( ) n p k (1 p) n k. k The left hand side (LHS) can be written as n! α k k!(n k)! n k(1 p)n k = αk k! ) k ( 1 α ) n k α k n k! e α [ ] n(n 1)(n 2)... (n k + 1) (1 p) n k n k ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 16

18 Limit of Poisson RV let s continue... α k k! [ ] n(n 1)(n 2)... (n k + 1) (1 p) n k. n n... n We have lim n α k k! [ n(n 1)(n 2)... (n k + 1) n n... n ] (1 p) n k = αk k! = αk k! = αk k! ( lim 1 α ) n k n n ( ) 1 α n lim n ( 1 α n n ) k ( lim 1 α n n ) n By basic Calculus, we have lim n ( 1 1 n) n = e 1. ECE353 Probability and Random Processes X. Fu, School of EECS, Oregon State University 17

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