Probability Theory and Random Processes (MA225) Lecture 05 PDF
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This document contains lecture slides on probability theory and random processes, specifically focusing on random variables and cumulative distribution functions (CDF). Examples and definitions are presented. It appears to be part of a course on probability theory.
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Probability Theory and Random Processes (MA225) Lecture Slides Lecture 05 Random Variables Def: A function X : S → R is called a random variable. Example 1: Tossing a fair coin n times. Assume that the tosses are independent. Let X : S → R be defined by the no. of heads...
Probability Theory and Random Processes (MA225) Lecture Slides Lecture 05 Random Variables Def: A function X : S → R is called a random variable. Example 1: Tossing a fair coin n times. Assume that the tosses are independent. Let X : S → R be defined by the no. of heads. Example 2: Throwing a fair die twice. Assume the throws are independent. Let X : S → R be defined by the sum of the outcomes. Example 3: Suppose we are testing the reliability of a battery. Define X : S → R by X1 (ω) = ω. Now suppose we are mainly interested in whether the battery would last more than 2 years or not. Then X2 = 1(2,∞). Example 4: Take n=2. P(X = 0) = P(X = 2) = 1/4, P(X = 1) = 1/2. Example 5: P(X = 2) = 1/36, P(X = 3) = 2/36, P(X = 4) = 3/36, P(X = 5) = 4/36, P(X = 6) = 5/36, P(X = 7) = 6/36, P(X = 8) = 5/36, P(X = 9) = 4/36, P(X = 10) = 3/36, P(X = 11) = 2/36, P(X = 12) = 1/36. Example 6: P(I ) = I e −t dt, defines a probability on B(0, ∞). R P(X2 = 1) = e −2 , P(X2 = 0) = 1 − e −2. Cumulative Distribution Function Def: The cumulative distribution function (CDF) of a random variable X is a function FX : R → [0, ∞) defined by FX (x) = P(X ≤ x). Example 7: 0 if x < 0, 1/4 if 0 ≤ x < 1, FX (x) = 3/4 if 1 ≤ x < 2, x ≥ 2. 1 if Example 2: 0 if x < 2, 2 ≤ x < 3, 1/36 if 3/36 if 3 ≤ x < 4, 6/36 if 4 ≤ x < 5, 10/36 if 5 ≤ x < 6, 15/36 if 6 ≤ x < 7, FX (x) = 21/36 if 7 ≤ x < 8, 8 ≤ x < 9, 26/36 if 30/36 if 9 ≤ x < 10, 33/36 if 10 ≤ x < 11, 35/36 if 11 ≤ x < 12, 1 if x ≥ 12. Example 3: ( 0 if x < 0, FX1 (x) = 1 − e −x if x ≥ 0. 0 if x < 0, FX2 (x) = 1 − e −2 if 0 ≤ x < 1, 1 if x ≥ 1. Proposition: The CDF of a random variable has the following properties: (1) FX (·) is non-decreasing and hence has only jump discontinuities. (2) lim FX (x) = 1, lim FX (x) = 0. x↑∞ x↓−∞ (3) lim FX (x + h) = FX (x), ∀x ∈ R, thus CDF is right continuous. h↓0 (4) lim FX (x − h) = FX (x) − P(X = x), ∀x ∈ R. h↓0 Theorem: Let F be a function satisfying properties (1)-(3). Then F is a CDF. Remarks I Random variable is just a function and does not depend on the probability. But the distribution of the random variable depends on the probability. So keeping the function same if we change the probability then the random variable will remain the same but its distribution will change. Consider Example 1, but with the probabilities P(HH) = 9/16, P(TT ) = 1/16, P(HT ) = P(TH) = 3/16. What will be the distribution function in this case? I If x ∈ R is such that P(X = x) > 0, then x is said to be an atom of the distribution function of X. Thus if the distribution function of a random variable has no atoms then it is continuous. I P(a < X ≤ b) = FX (b) − FX (a). I P(a ≤ X ≤ b) = FX (b) − FX (a−). I P(a < X < b) = FX (b−) − FX (a). I P(a ≤ X < b) = FX (b−) − FX (a−).