Strategic Practice and Homework 8 - Projects at Harvard

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Stat 110 Strategic Practice 8 Solutions, Fall 2011Prof. Joe Blitzstein (Department of Statistics, Harvard University)1 Covariance and Correlation1. Two fair six-sided dice are rolled (one green and one orange), with outcomesX and Y respectively for the green and the orange.(a) Compute the covariance of X + Y and X Y .Cov(X + Y,X Y )=Cov(X, X) Cov(X, Y )+Cov(Y,X) Cov(Y,Y )=0.(b) Are X + Y and X Y independent? Show that they are, or that theyaren’t (whichever is true).They are not independent: information about X + Y may give informationabout X Y , as shown by considering an extreme example. Note that ifX + Y =12,thenX = Y =6,soX Y =0. Therefore,P (X Y =0|X + Y =12)=16= P (X Y =0),whichshowsthatX + Y and X Y arenot independent. Alternatively, note that X + Y and X Y are both even orboth odd, since the di↵erence X + Y (X Y )=2Y is even.2. AchickenlaysaPoisson()numberN of eggs. Each egg, independently,hatches a chick with probability p. Let X be the number which hatch, soX|N ⇠ Bin(N,p).Find the correlation between N (the number of eggs) and X (the number ofeggs which hatch). Simplify; your final answer should work out to a simplefunction of p (the should cancel out).As shown in class, in this story X is independent of Y ,withX ⇠ Pois( p) andY ⇠ Pois( q), for q =1 p. SoCov(N,X) =Cov(X + Y,X) =Cov(X, X)+Cov(Y,X) =Var(X) = p,givingCorr(N,X) =pSD(N)SD(X) =pp p= p p.1
3. Let X and Y be standardized r.v.s (i.e., marginally they each have mean 0 andvariance 1) with correlation ⇢ 2 ( 1, 1). Find a, b, c, d (in terms of ⇢) suchthatZ = aX + bY and W = cX + dY are uncorrelated but still standardized.Let us look for a solution with Z = X, finding c and d to make Z and Wuncorrelated. By bilinearity of covariance,Cov(Z, W) =Cov(X, cX + dY )=Cov(X, cX)+Cov(X, dY )=c + d⇢ =0.Also, Var(W )=c 2 + d 2 +2cd⇢ =1. Solving for c, d givesa =1,b=0,c= ⇢/ p 1 ⇢ 2 ,d=1/ p 1 ⇢ 2 .4. Let (X 1 ,...,X k )beMultinomialwithparametersn and (p 1 ,...,p k ). Use indicatorr.v.s to show that Cov(X i ,X j )= np i p j for i 6= j.First let us find Cov(X 1 ,X 2 ). Consider the story of the Multinomial, where nobjects are being placed into categories 1,...,k.LetI i be the indicator r.v. forobject i being in category 1, and let J j be the indicator r.v. for object j beingin category 2. Then X 1 = P ni=1 I i,X 2 = P nj=1 J j. SoCov(X 1 ,X 2 )=Cov(= X i,jnXI i ,i=1nXJ j )j=1Cov(I i ,J j ).All the terms here with i 6= j are 0 since the ith object is categorized independentlyof the jth object. So this becomessincenXCov(I i ,J i )=nCov(I 1 ,J 1 )= np 1 p 2 ,i=1Cov(I 1 ,J 1 )=E(I 1 J 1 ) (EI 1 )(EJ 1 )= p 1 p 2 .By the same method, we have Cov(X i ,X j )= np i p j for all i 6= j.5. Let X and Y be r.v.s. Is it correct to say “max(X, Y )+min(X, Y )=X + Y ?Is it correct to say “Cov(max(X, Y ), min(X, Y )) = Cov(X, Y )sinceeitherthemax is X and the min is Y or vice versa, and covariance is symmetric”?2
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