Phi D 3. Reading 2.4 pg IndicatorExapley. lo if i belongs to the subset. Ear. Let. 2 1 possible non empty subsets and L of

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1 Reading 2.4 pg IndicatorExapley Consider the set of integers 1,2 in A non empty subset of 1,2 n is chosen at radon Let X denote the size of the subset What is ELI Let lo if i belongs to the subset otherwise so fat X Xi Think Hui proble 3 a There are 2 1 possible non empty subsets and L of them contain i Hence Phi D 3 Ear n 3

2 Fact Two RVs X and Y are independent iff ElfhlHkly Ethan E kill for all functions ht and KC The for all in the above fact a an important distinction When h and function k are both the i e hk x ad identity klyl g we say the RVs are uncorrelated The correlation between two random variables X and Y is EEXYJ If EEXY ELITE EY we say X and Y are In general uncorrelated it can be quite difficult to talk about expectations of products so we often use tricks to band them one such trick is the Cauchy Schwarz inequality The inequality for rado variables states that CoalySchi I EExidle VEEXTEEYTT between The two random variables X d Y is covariance xn EEN EEN Y EET

3 When we remove the neon from a radar variable we call it Zero weary or centered A random variable with zero mean ad unit variance B called standard or normalized Fact For uncorrelated RVs Xi Xu k var Xi u Xi Q What value of Cow XY implies X and Y are uncorrelated A coucxy El XY TEENY XETED E Ex E y E XY EEE Y EEXEED E FLEXTECH E XY E FLED where we have used the fact that E X is not radon and hence EEE Y FLEX ELY This shows that X and Y are uncorrelated iff couch't o

4 ConditionalPobability For discrete RVs the conditional probability follows directly from the definition for rado events see Lee 2 notes The conditional PMF is summarized via the equation packing pxiyk.ly pylgj partykj pxkj here pxxlxilyjj PIX x.ly yj Using PMFs the law of total probability becomes pilgj pyixlyjlxilp.hr E Binay symmetric channel Suppose we transmit Xe o B ad receive a flipped bit with probability E Kalled the crossover probability Let Y be the received signal X B Ip Whet s pyly if I E f i Ply o ply ok ojp x o Ply ok 1 PKa Ite l p Ep

5 Conditioning is done to remove r d w ss key so terms easier to deal with One takeaway is that when we condition on something it is no longer radon this leads to the subsfifh.ae aw P glkyl z x xi P gcxi.yj z X xi ExJ Let Z XtY glxy Plz.ge X i PlXtY jlx i PlitY jlx i pfy j ilx i If X and Y are independent this gives us PIZ jlx il.ph ji

6 Conditional Expectation Note that YIX is a RV itself Hence we can compute it expectation YIX Ejyjp.in yjlxi An important tool usig conditional expectation is Expectations We'll consider two variants the woftoh_ LTEVersionl s Let B3 partition the sample space Then FLEX E FLEX1B PCB TIE A biased coin with PCH p is tossed sequentially until 2 consecutive heads appear Let be the number of required tosses Find E EX B T on first toss P B a l p partition BE HT Pl Ba p It p Bz HH PIB p ELI tp Ex ti t p t p EET 2 p 2 This idea be useful for HW1 problem 5 may

7 LTEVersi Let X ad Y be two RVs possibly correlated Then EEN Ey En 4143 EJ JL 1, all outcomes equi probable X to w c we Y to Clearly I x t but let's use LTE to check we 433 we 45,6 I KH o FLEXI WEEK5,63 t.lt Z o FLEX14 1 FLEXI we Iz l t t o Zz Ey Em MY E Iz I I I I I

8 Robley A building has a floors S pp.se M peopleenter at floor 1 where M Poisson lo Each person is equally likely to exit at any fle 21 in Find the expected number of floors on which at least one person exits the elevator Solution We Le e M people ad can envision each as randomly selecting a floor to exit Let X be the number of steps made Then we can again use the LTE version 2 to Efx Em an EXIT see that First fix Man It is easier to talk about the na b of floors where we stop is made so call Y No introduce the indicator RVs 1 i to a stop on floor otherwise n so that Y Ai We have that plait l ut in in n X that number experiments where everyone chooses fro among n 1 floors of n total

9 Therefore EE 1 I IT n l t m and EEXIM on u EEY n n l t M we are now ready to find ELI EEK Em Exam XIN II Exim In 7 peek II In all IT pmh EEE Pff goes u n It It n ne PmH n It Hmmm loll Now use the i po fact that E.to it e to see that t I E Ex n ne exp loll t n l e I

10 indicates N is distributed 17 according to A hen lays N eggs where NEPoisson X Each egg latches with probability p independently at the other eggs Let k be the umber of chicks Find EEKIN EEK ad E NIKI Solution From the problem we have e and Pew kla E ply plate Pula r 1 binomial J Poisson for specific n a Therefore K N pit mean a d for a r do N we have E EKIN p N f bio ial distribution We can find ELEK using LTE version 2 EEK E Eun KINI En pn p I

11 To find El NIKI we c use Bayes rie Polk alle Pkn kla putt if nah In pkwlklmlpi.cm 2 phllp.tk 1Yn e 7 s e a Help'llp 17mi e use and algebra e t psa n K e tp a From this we can compute NIK k n polk alle which goes the final result ktq1 NIK Kt aid