Indicators: Now With Pair - PDF document

1Indicators: Now With Pair-wise Flavo
1Indicators: Now With Pair-wise Flavor!•Recall Iiis indicator variable for event Aiwhen:Let X = # of events that occur:•Now consider pair of events AiAjoccurringIiIj= 1 if both events Aiand Ajoccur, 0 otherwiseNumber of pairs of events that occur is From Event Pairs to Variance•Expected number of pairs of events:•Recall: Var(X) = E[X2] –(E[X])2Let’s Try It With the Binomial•X ~ Bin(n, p)Each trial: Xi~ Ber(p)Let event Ai= trial iis success (i.e., Xi= 1)Computer Cluster Utilization•Computer cluster with N serversRequests independently go to server iwith probability piLet event Ai= server ireceives no requestsX = # of events A1, A2,  Anthat occurY  # servers that receive ≥ 1 request  N –XE[Y] after first nrequests?Since requests independent:Computer Cluster Utilization (cont.)•Computer cluster with N serversRequests independently go to server iwith probability piLet event Ai= server ireceives no requestsX = # of events A1, A2,  Anthat occurY  # servers that receive ≥ 1 request  N –XVar(Y) after first nrequests?Independent requests:( = (-1)2 Var(X) = Var(X) )Computer Cluster = Coupon Collecting•Computer cluster with N serversRequests independently go to server iwith probability piLet event Ai= server ireceives no requestsX = # of events A1, A2,  Anthat occurY  # servers that receive ≥ 1 request  N –X•This is really another Coup

on Collector” problemEach server
on Collector” problemEach server is a coupon type”Request to server = collecting a coupon of that type•Hash table versionEach server is a bucket in tableRequest to server = string gets hashed to that bucket2Product of Expectations•Let X and Y are independent random variables, and g() and h() are real-valued functionsProof:The Dance of the Covariance•Say X and Y are arbitrary random variables•Covariance of X and Y:•Equivalently:X and Y independent, E[XY] = E[X]E[Y] Cov(X,Y) = 0But Cov(X,Y) = 0 does notimply X and Y independent!Dependence and Covariance•X and Y are random variables with PMF:E[X] = 0, E[Y] = 1/3Since XY = 0, E[XY] = 0Cov(X, Y) = E[XY] –E[X]E[Y] = 0 –0 = 0•But, X and Y are clearly dependentXY-101pY(y)01/301/32/3101/301/3pX(x)1/31/31/31Example of Covariance•Consider rolling a 6-sided dieLet indicator variable X = 1 if roll is 1, 2, 3, or 4Let indicator variable Y = 1 if roll is 3, 4, 5, or 6•What is Cov(X, Y)?E[X] = 2/3 and E[Y] = 2/3E[XY]== (0 * 0) + (0 * 1/3) + (0 * 1/3) + (1 * 1/3) = 1/3Cov(X, Y) = E[XY] –E[X]E[Y] = 1/3 –4/9 = -1/9Consider: P(X = 1) = 2/3 and P(X = 1 | Y = 1) = 1/2oObserving Y = 1 makes X = 1 lesslikelyAnother Example of Covariance•Consider the following data:WeightHeightWeight * Height645736487159418953492597676241545551280558502900775542355748273656422352514221427661463668573876E[W] = 62.75E[H] = 52.75E[W*

H]= 3355.83303540455055606540
H]= 3355.833035404550556065404550556065707580HeightWeightCov(W, H) = E[W*H] –E[W]E[H]= 3355.83 –(62.75)(52.75)= 45.77Properties of Covariance•Say X and Y are arbitrary random variables•Covariance of sums of random variablesX1, X2, , Xnand Y1, Y2, , Ymare random variables3Variance of Sum of Variables•Proof:If all Xiand Xjindependent (i j): Note:By symmetry:Hola Compadre: La Distribución Binomial•Let Y ~ Bin(n, p)nindependent trialsLet Xi= 1 if i-th trial is success”, 0 otherwiseXi~ Ber(p)E[Xi] = pVar(Y) = Var(X1) + Var(X2) + ... + Var(Xn)Var(Xi)= E[Xi2] –(E[Xi])2= E[Xi] –(E[Xi])2since Xi2= Xi= p –p2= p(1 –p)Var(Y) = nVar(Xi) = np(1 –p)Variance of Sample Mean•Consider nI.I.D. random variables X1, X2, ... XnXi have distribution Fwith E[Xi] = mand Var(Xi) = s2We call sequence of Xia samplefrom distribution FRecall sample mean: whereWhat is ?Sample Variance•Consider nI.I.D. random variables X1, X2, ... XnXi have distribution Fwith E[Xi] = mand Var(Xi) = s2We call sequence of Xia samplefrom distribution FRecall sample mean: whereSample deviation: for i = 1, 2, ..., nSample variance:What is E[S2]?E[S2] = s2We say S2is unbiased estimate” of s2Proof that E[S2] =s2(just for reference)So, E[S2] = s