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# ELEG310 Week 3 notes ELEG310

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This 12 page Class Notes was uploaded by Cindy Nguyen on Sunday February 28, 2016. The Class Notes belongs to ELEG310 at University of Delaware taught by Dr. Daniel Weile in Spring 2016. Since its upload, it has received 238 views. For similar materials see Random Signals and Noise in Electrical Engineering at University of Delaware.

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Date Created: 02/28/16

ELEG 310: Random Signals and Noise Lecture 5 2/21/2016 Bernoulli Trial – has 2 outcomes, “success” or “failure.” Success is represented by p Failure is represented by 1 – p. e.g. Flipping a coin, HTTH (heads tails tails heads) p = success of heads “H” p – 1 = failure of heads, “T” 2 2 HTTH = p(p – 1)(p – 1)p = p (1 – p) getting exactly 2 head, regardless of order HHTT and HTHT are the same outcome We are interested in counting the # of successes. To do this, we write: 4 2 2 (2)p (1 – p) “Four choose two” times the probability of getting exactly two heads. Binomial Random Variable n number of trials, k number of successes ???? P[B = k] = ( ???? k n-k = p [1 – p] P[B∈{0, 1, 2, … n}] = 1 Probability in set must be 1 Binomial theorem ???? ???? (???? + ????)???? = ∑ ( )???? ???? ???? ????−???? ???? ????=0 Proof by Example of a simple case: n = 1 (1 trial) According to formula, 1 P[B = 0] = ( ) p (1 – p) = 1 – p 0 1 1 0 P[B = 1] = 1 ) p (1 – p) = p 1 – p + p = 1 (equals 1, formula works) Proof by Induction: When n=1, 1 1 (???? + ????= ∑ ( )???? ????−???? ???? ????=0 1 1 = (0) ???? 1+ (1) ???? 0 = b + a Induction step: Assume it works for n = N, then prove N+1. Therefore, it would work for all N. ???? (a+b)+1= (a+b)(a+b) = (a+b) ∑ ???? ???? ????−???? ????=0 ???? ???? ???? ???? ∑ (????)???? ????+1????????−????+ ∑ (????)????????+1????????−????+1 ????=0 ????=0 Substitution in 1 addend: let j=k+1 ????+1 ???? =∑ ( )???? ???? ????+1−???? ????=1 ????−1 Now rename j to k. ????+1 ???? =∑ ( ???? )???? ????????+1−???? +∑ ( )???? ???? ????+1−???? ????=1 ????−1 ????=0 ???? ???? = ( )???? ????+1+ ∑ ( ???? ) ( ) ] ???? ????????+1−???? + ( )????????+1 ???? ????=1 ????−1 ???? 0 ???? ???? ????! ????! (????−1) + (????)= (????−1 ! ????+1−???? !(???? ! ????−???? ! ???? Multiply by???? ????????! ????−????+1 ????! = ????! ????+1−???? ! ????! ????−???? ! ????!(????+1) (????+1 ! ????+1 = = =( ???? ) by definition ????! ????+1−???? ! ???? ????+1−???? ! N+1 ????+1 ????+1 ???? ????+1 ???? ????+1−???? ????+1 ????+1 (a+b) =(????+1 )???? +∑ ( ???? ) ???? ???? + ( 0 )???? ????=1 ????+1 = ∑ (????+1 )???? ????????+1−???? ????=0 ???? (Proven) ???? k n-k P[B = k] = ???? ) p (1 – p) ∑ ???? P[B = k] = ∑???? ( ) p (1 – p)n-k ????=0 ????=0 ???? = [p+ (1 – p)] = 1 (works) Counting Bernoulli Trials Geometric Distribution Recall and Memorize Geometric Sums: 1 ∑ ∞ i ????=0 r = 1−???? Repeat Bernoulli Trial until we get a success. st p[m=1] = p success on 1 trial p[m=2] = p(1-p) also pq Must fail first to get a success on second trial. p[m=3] = p(1-p) also pq p[m] = pq-1 Checking it: ∞ ∑ ????=1 ???? ???? = 1 Must take somewhere between 1 and ∞ to guarantee getting a success ???? ∑ ????=1 ????????????−1 = p ∑ ????=1 ????????−1 = p ∑????=0 ???? ???? = 1−???? Outcome ꝭ (zeta) X(ꝭ) ꝭ ∈ ???? (ꝭ is a member of S) X is a “mapping.” X : S is the domain ofXthe mapping; S ⊂ ℝ is the range of mapping. The capital letter X denotes a “random variable,” which is not a variable; it is a function that maps the outcome. P[X < x]. e.g. Suppose you toss a coin 3 times, and want to count the heads. ꝭ = TTT TTH THT THH HTT HTH HHT HHH X(ꝭ) = 0 1 1 2 1 2 2 3 SX= {0, 1, 2, 3} range of the random variable Below is the same experiment, but we assign different rules to the outcomes; so we use a different random variable, Y (capital y). SY= {$0, $1, $8} Prize money: $0 for no heads, $1 for getting 2 heads, $8 for getting 3 heads Y (ꝭ) = 0 0 0 $1 0 $1 $1 $1 $8 Discrete Random Variables Probability Mask Function (pmf) A discrete function that assigns probability to each atomic outcome. P Xx) = P(X = x] = P[{ꝭ: X(S) = x}] x ∈ ℝ X is the random variable Axioms of Probability Mask Function (i) PX(x) ≥ 0 (outcome must be positive or 0;⊂ ℝ) (ii) ∑ ????∈ ???? P Xx) = 1 ???? (iii) P[x∈B] = ∑ ????∈ B PX(x) (∀ ???? ∈ SX⊂ SX) Define random variables corresponding to probability laws like geometric, Bernoulli, etc. Example 1: Bernoulli Random Variable (a.k.a success/fail) Define success asꝭ = A. 0 ꝭ ∉ A Indicator function: IA(ꝭ) = {1 ꝭ ∈ A Example 2: Geometric Random Variable outcome: k-1 P Xk) = P[X – k] = (1 – p) p Example 3: Binomial Random Variable ???? k n-k PX(k) ( ???? p (1 – p) What else can we do with these concepts? To fully specify a random variable, we must know the distribution completely – not every value, but we want to observe the center(expected) value of the distribution if we were to repeat the experiment over again. This would be the mean, which “summarizes our gut feelings.” ???? Χ ???? Χ =[ ] ∑ ????∈ ???? x PX(x) ???? Probability All poss. Expectation mask of X outcomes function Assign k as all possible outcomes. ∑ = ???? xkP Xxk) e.g. what is the mean of a Bernoulli Random Variable? MI= (0IP(0) +I(1)P(1) = 0(1 – p) + 1(p) = p Uniform Random Variable 1 PX(x) ???? for x ∈ {0,1,2,…???? − 1} ???? Χ = ∑ ????−1 ???? 1 ????=0 ???? 1 ∑ ????−1 = ???? ????=0 ???? 1 = ???? [0 + 1 + 2… ???? − 3 + ???? − 2 + ???? − 1 ]( ) 1 ???? = [(???? − 1) ???? 2 ????−1 = 2 Geometric Random Variable **(This could be on the test!) ???? = ∑ x PX(x) = ∑ xkPX(xk) ???? ????∈ ???? ???? ∞ k-1 =∑ ????=1 ???? pq = p∑∞ ???? qk-1 this looks like a derivative. ????=1 ???? = p∑????=1 q k use geometric series to simplify sum ???????? ???? ???? = p???????? 1−???? now take derivative ???? = 2 ???? 1 = ???? ELEG 310: Random Signals and Noise Lecture 6 2/23/2016 Expectation E[X] = ∑ ????∈???????? xP Xx) P (k) = ( )???? (1 − ????) ????−???? X ???? ???? ???? ???? ????−???? E[X] ∑ ????(????)???? (1 − ????) definition of Expectation ????=0 ???? ????! =∑ ???? ???? (1 − ????)−???? ????=0 ????! ????−???? ! ???? ????! ???? ????−???? =∑ (????−1)! ????−???? !1 − ????) ????=1 ????−1 = ????∑ (????−1)! ????????+1(1 − ????)−????+1 sub j=k-1 ????=0????! ????−1−???? ! ????−1 =????????∑ ????(????−1)???? (1 − ????)????−????+1 ????=0 ???? =np Purpose: illustrate calculating the Mean, and use of Binomial Theorem. Suppose we define a new Random Variable as a function of another Random Variable. Z = g(X) We now want to compute the mean. E[Z] = E[g(X)] g(X) Expectation Value of a Square of a Random Variable Let X be geometric. 2 ∞ 2 ????−1 E[X ] =∑????=1 ???? ???????? (recall: q = 1-p) ∞ 2 ????−2 = ????????∑ ????=1 ???? ???? = ????????(∑ ∞ k k − 1 ???? ????−2+ ∑ ∞ k????????−2 ) ????=1 ????=1 ∞ ????2 ???? ∞ ????−1 = ???????? ∑ ????????2(???? ) + ????∑ ????=1k???? ????=1 2 = ???????? ???? (∑ ∞ (???? )) + ???? ???? ∑ ∞ ???????? ????????2 ????=1 ???????? ????=1 ????2 ???? ???? ???? = ???????? ????????2 (1−????) + ???????????? 1−????) … simplify 2−???? = ????2 Expectation Properties E[aX] =∑????∈???????? ????xkP Xxk) = ????∑ ????∈????????xkPX(xk) = aE[X] E[X + C] = (???? + ????) PX(xk) ????∈???????? = ∑ ????∈???? xPX(xk) +∑????∈???? ???? PX(xk) = E[X] + ????????∈???? PX(k ) = E[X] + C E[g(X) +h(X)] = E[g(X)] + E[h(X)] Deviance D Xx) = (X – E[x]) Variance ???????? 2= VAR[X] = E[???? ????????)] Standard Deviation 2 STD [X] = √???? ???? = ???????? Purpose of STD when you already have Variance? Standard Deviation has the same units as the variable we are measuring, while variance squares the units. Can simplify computation using: VAR[X] = E[X ] – E[X]2 Examples of computing variance Bernoulli Random Variable, Bernoulli Trial VAR[X] = E[X ] – E[X]2 IA= {1 ????????????ℎ ???????????????????????????????????????????? ???? 0 ????????ℎ???????????????????????? E [A ] = p E [(A ) ] = 0 (1 – p) + 1 (p) = p Plugging in, we get: 2 2 VAR[IA] = EA[I ] – EA[(I )] = p – p = p(1 – p) If p is 0, no variance If p is 1, no variance We maximize variance by p being fair. Variance of Geometric Random Variable VAR[X] = E[X ] – E[X] from previous lecture 2−???? 1 = - ( )2 ????2 ???? =1−???? ????2 Variance of Uniform Random Variable X ∈ {0,1,…???? − 1} 1 ????−1 PX(x) = E[X] = ???? 2 ????−1 E[X]= ∑ (???? )( ) ????=0 ???? 1 ????−1 2 =???? ∑ ????=0(???? ) ???? =∑ (???? ) = an + bn + cn + d ????=0 …Sum at N=0, N=1, N=3, etc. 1 ????−1 2 1 ????−1 = ???? ∑ ????=0 (???? ) =???? 6 (2M -2 +1)(2M -1 + 1) (????−1 (2????−1) = 6 2 2 VAR[X] = E[X ] – E[X] (????−1 (2????−1) (????−1) 2 = - ( ) 6 2 (????−1 (????+1) = 12

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