Probability review. September 11, Stoch. Systems Analysis Introduction 1

Transcription

1 Probability review Alejandro Ribeiro Dept. of Electrical and Systems Engineering University of Pennsylvania September 11, 2015 Stoch. Systems Analysis Introduction 1

2 Conditional probabilities Conditional probabilities Conditional expectation Stoch. Systems Analysis Introduction 2

3 Conditional pmf and cdf for discrete RVs Recall definition of conditional probability for events E and F P(E F ) = P(E F ) P(F ) Change in likelihoods when information is given, renormalization Define the conditional pmf of RV X given Y as (both RVs discrete) p X Y (x y) = P [ X = x Y = y ] = P [X = x, Y = y] P [Y = y] Can rewrite as p X Y (x y) = P [X = x, Y = y] P [Y = y] = p XY (x, y) p Y (y) Pmf for random variable x, given parameter y ( Y not random anymore ) Define conditional cdf as (a range of X conditional on a value of Y ) F X Y (x [ ] y) = P X x Y = y = p X Y (z y) z x Stoch. Systems Analysis Introduction 3

4 Example Example Independent Bernoulli Y and Z, variable X = Y + Z Conditional pmf of X given Y? For X = 0, Y = 0 p X Y (X = 0 Y = 0) = P [X = 0, Y = 0] P [Y = 0] = (1 p)2 1 p = 1 p Or, from joint and marginal pdfs (just a matter of definition) p X Y (X = 0 Y = 0) = p XY (0, 0) p Y (0) = (1 p)2 1 p = 1 p Can compute the rest analogously p X Y (0 0) = (1 p), p X Y (1 0) = p, p X Y (2 0) = 0 p X Y (0 1) = 0, p X Y (1 1) = 1 p, p X Y (2 1) = p Stoch. Systems Analysis Introduction 4

5 Conditional pdf and cdf for continuous RVs Define conditional pdf of RV X given Y as (both RVs continuous) f X Y (x y) = f XY (x, y) f Y (y) For motivation, define intervals x = [x, x+dx] and y = [y, y+dy] Can approximate conditional probability P [ X x Y y ] as P [ X x Y y ] = P [X x, Y y] P [Y y] f XY (x, y)dxdy f Y (y)dy From definition of conditional pdf it follows after simplifying terms P [ X x ] Y y fx Y (x y)dx Which is what we would expect of a density Conditional cdf defined as F X Y (x) = Stoch. Systems Analysis Introduction 5 x f X Y (u y)du

6 Example: Communications channel Random message (RV) Y, transmit signal y (realization of Y ) Received signal is x = y + z (z realization of random noise) Can model communication system as a relation between RVs X = Y + Z Model communication noise as Z N (0, σ 2 ) independent of Y Conditional pdf of X given Y. Use definition: f X Y (x y) = f XY (x, y) f Y (y) =? f Y (y) Problem is we don t know f XY (x, y). Have to calculate Computing conditional probs. typically easier than computing joints Stoch. Systems Analysis Introduction 6

7 Example: Communications channel (continued) If Y = y is given, then Y not random anymore (Dorothy s principle) It still is random in reality, we are thinking of it as given If Y were not random, say Y = y with y given then... Cdf of X, now easily obtained X = y + Z P [X x] = P [y + Z x] = P [Z x y] = But since Z is normal with 0 mean and variance σ 2 P [X x] = 1 2πσ x y e z2 /2σ 2 dz = 1 2πσ x X is normal with mean y and variance σ 2 x y p Z (z) dz e (z y)2 /2σ 2 dz Stoch. Systems Analysis Introduction 7

8 Digital communications channel Conditioning is a common tool to compute probabilities Message 1 (prob. p) Transmit Y = 1 Message 2 (prob. q) Transmit Y = 1 Received signal X = Y + Z Y = ±1 + X Z N (0, σ 2 ) Decoding rule Ŷ = 1 if X 0, Ŷ = 1 if X < 0 ] What is the probability of error, P e := P [Ŷ Y? Red dots to the left and blue dots to the right are errors Ŷ = 1 Ŷ = x Stoch. Systems Analysis Introduction 8

9 Output pdf From communications channel example we know If Y = 1, then X N (1, σ 2 ), conditional pdf is f X Y (x, 1) = 1 2πσ e (x 1)2 /2σ 2 If Y = 1, then X N ( 1, σ 2 ), conditional pdf is f X Y (x, 1) = 1 2πσ e (x+1)2 /2σ 2 f X Y (x) N ( 1, σ 2 ) N (1, σ 2 ) 1 1 x Stoch. Systems Analysis Introduction 9

10 Error probability Write probability of error by conditioning on Y = ±1 (total probability) P e = P { Ŷ Y Y = 1 } P { Y = 1 } + P { Ŷ Y Y = 1 } P { Y = 1 } = P { Ŷ = 1 Y = 1 } p + P { Ŷ = 1 Y = 1 } q But according to the decision rule P e = P { X < 0 Y = 1 } p + P { X 0 Y = 1 } q But X given Y is normally distributed, then P e = p e (x 1)2 /2σ 2 + q 0 e (x+1)2 /2σ 2 = 1 1 2πσ 2πσ 2πσ 0 f X Y (x) e x2 /2σ 2 N ( 1, σ 2 ) N (1, σ 2 ) 1 1 Stoch. Systems Analysis Introduction 10 x

11 Conditional expectation Conditional probabilities Conditional expectation Stoch. Systems Analysis Introduction 11

12 Definition of conditional expectation For continuous RVs X, Y define conditional expectation as E [ X ] y = x f X Y (x y) dx For discrete RVs X, Y conditional expectation is E [ X ] y = x p X Y (x y) Defined for given y E [ X ] y is a value All possible values y of Y random variable E [ X ] Y Y is RV, E [ X y ] value associated with outcome Y = y x Stoch. Systems Analysis Introduction 12

13 Double expectation If E [ X Y ] is a RV, can compute expected value EY [ EX ( X Y )] Subindices are for clarity purposes, innermost expectation is with respect to X, outermost with respect to Y What is E Y [ EX ( X Y )]? Not surprisingly E [X ] = EY [ EX ( X Y )] Show for discrete RVs (write integrals for continuous) [ ( )] ( ) E Y EX X Y = E X X y py (y) = [ ] x p X Y (x y) p Y (y) y y x = [ ] x p X Y (x y)p Y (y) = [ ] x p X,Y (x, y) x y x y = x xp X (x) = E [X ] Yields a method to compute expected values Condition on Y = y X ( y ) Compute expected value over X for given y E X X [ ( y ) ] Compute expected value over all values y of Y E Y EX X Y Stoch. Systems Analysis Introduction 13

14 Example Seniors get A = 4 with prob. 0.5, B = 3 with prob. 0.5 Juniors get B = 3 with prob. 0.6, B = 2 with prob. 0.4 Exchange student s standing: senior (junior) with prob. 0.7 (0.3) Expectation of X = exchange student s grade? Begin conditioning on standing E [ X ] Senior = = 3.5 E [ X ] Junior = = 2.6 Now sum over standing s probability E [X ] = E [ X Senior ] P [Senior] + E [ X Junior ] P [Junior] = = 3.23 Stoch. Systems Analysis Introduction 14

15 Conditioning to compute expectations As with probabilities conditioning is useful to compute expectations. Spreads difficulty into simpler problems Example A baseball player hits X i runs per game Expected number of runs is E [X i ] = E [X ] independently of game Player plays N games in the season. N is random (playoffs, injuries?) Expected value of number of games is E [N] [ N ] What is the expected number of runs in the season? E X i Both, N and X i are random i=1 Stoch. Systems Analysis Introduction 15

16 Sum of random number of random quantities Step 1: Condition on N = n then [( N ) ] X i N = n = i=1 Step 2: Compute expected value with respect to X i [( N ) ] [ n ] E Xi X i N = n = E X i = ne [X ] i=1 Second equality possible because n is a number (not a RV like N) Step 3: Conpute expected value with respect to values n of N [ [( N ) ]] [ ] E N E Xi X i N = E N NE [X ] = E [N] E [X ] i=1 i=1 [ N ] Yielding result E X i = E [N] E [X ] i=1 Stoch. Systems Analysis Introduction 16 n i=1 X i