Stat 110 Section 9 Covariance, Multinomial and MVN TF: Mingshu Huang ([email protected]) Section Hour: Tue 3-4pm, SC-216 Office Hour: Wed 9-10pm, SC-300H Topics: • Jointly, Marginal, Conditional Distribution: – Joint Distribution: fX,Y (x, y) – Marginal Distribution: (continuous case) fX (x) = – Conditional Distribution: fY |X (y|x) = R∞ −∞ fX,Y (x, y)dy fX|Y (x|y)fY (y) fX,Y (x, y) = R∞ fX (x) −∞ fX,Y (x, y)dy • Covariance: Cov(X, Y ) = E((X − EX)(Y − EY )) = E(XY ) − E(X)E(Y ) – Cov(X, X) = V ar(X) – Cov(X + Y, Z + W ) = Cov(X, Z) + Cov(X, W ) + Cov(Y, Z) + Cov(Y, W ) – V ar(X + Y ) = V ar(X) + V ar(Y ) + 2Cov(X, Y ) Cov(X, Y ) • Correlation: Corr(X, Y ) = p V ar(X)V ar(Y ) – If X and Y are independent, then Cov(X, Y ) = Corr(X, Y ) = 0. – But if Cov(X, Y ) = Corr(X, Y ) = 0, X and Y are NOT necessarily independent. (They are uncorrelated though.) • Multinomial: X ∼ M ultik (n, p) where X = (X1 , ..., Xk ) and p = (p1 , ..., pk ). n! pn1 pn2 pnk n1 !n2 !...nk ! 1 2 k – Marginal: If X ∼ M ultik (n, p), then Xj ∼ Bin(n, pj ). – Joint PMF: P (X1 = n1 , ..., Xk = nk ) = – Lumping: If X ∼ M ultik (n, p), then Xi + Xj ∼ Bin(n, pi + pj ). – Covariance: If X ∼ M ultik (n, p), then for i 6= j, Cov(Xi , Xj ) = −npi pj . • Multivariate Normal (MVN): – Definition: t1 X1 + ... + tk Xk normal for any choice of t1 , ..., tk , then X = (X1 , ..., Xk ) is MVN. – Special case: k = 2 the Bivariate Normal – Joint MGFs: E(eW ) = eE(W )+V ar(W )/2 – Corr=0 means independence! (only holds for MVN!!) 1 • 1D Transformation: fY (y) = fX (x)| dx | where X is continuous R.V. and Y = g(X) is dy differentiable and one-to-one. Problems: 1. (Two Dice) Two fair dice are rolled, with outcome X and Y respectively. (a) Are X+Y and X-Y correlated? Cov(X + Y, X − Y ) = Cov(X, X) + Cov(Y, X) − Cov(X, Y ) − Cov(Y, Y ) = V ar(X) − V ar(Y ) = 0 So they are uncorrelated. (b) Are they independent? No. One example would be if X − Y = 0 then X + Y must be an even number. (c) To generalize, if X and Y are two random variables with Var(X)=Var(Y), what can we say about the correlation of X+Y and X-Y? As shown in 1(a), the correlation of X+Y and X-Y would be 0 if they have the same variance. 2. (Max and Min) Let X, Y be i.i.d. Expo(1). (a) What is the correlation between the minimum L = min(X, Y ) and A = |Y − X|? A can be think as A=max(X,Y)-min(X,Y)=M-L where M=max(X,Y). Recall that if you think of X and Y as the arrival time of two buses, A can be thought as the time between the arrival of the first and the second bus and L the wait time for the first bus. By the memoryless property of Exponential, A and L are independent, thus Corr(L, A)=0. 2 (b) Calculate the correlation between L = min(X, Y ) and the maximum M = max(X, Y ). From 2(b) we know M=A+L, so Cov(L, M ) = Cov(L, A + L) = Cov(L, A) + Cov(L, L) = V ar(L) = 14 (because L is minimum of Exponential, by Poisson process L ∼ Expo(2)) We also know V ar(L) = 41 and V ar(M ) = V ar(A + L) = V ar(A) + V ar(L) = 45 since A ∼ Expo(1). So Corr(L, M ) = √ Cov(L,M ) = √15 . V ar(L)V ar(M ) 3. (Sour Patch Revisited) Suppose you draw n candies from an infinite pool ofPsour patches; each color (red, green, blue, yellow) has pi probability of being drawn, and pi = 1. Let Xi , i = 1, . . . , 4 be the number of red, green, blue and yellow patches of the n patches respectively. (a) What is the joint distribution of X1 , X2 , X3 , X4 ? It’s a M ulti4 (n, p) where p = (p1 , p2 , p3 , p4 ). (b) Are the Xi ’s independent? No, they sum up to n so knowing 3 of them equals knowing all 4 of them. (c) What is the distribution of X1 ? Write the marginal density function of X1 . It’s a Bin(n, p1 ). (d) What is the distribution of X1 + X2 ? Calculate E(X1 + X2 ). It’s a Bin(n, p1 + p2 ) (e) What is the conditional distribution of X2 given X1 = x1 ? It’s equivalent to re-parameterize X1 in n − x1 people, so p1 X2 |X1 = x1 ∼ Bin(n − x1 , 1−p ) 1 3 4. (Bivariate Normal) Let (X, Y ) be Bivariate Normal, with X and Y marginally N (0, 1) and with correlation ρ between X and Y . (a) Show that (X + Y, X − Y ) is also Bivariate Normal. Let T = t1 (X + Y ) + t2 (X − Y ) = (t1 + t2 )X + (t1 − t2 )Y Since (X, Y ) is Bivariate Normal, (t1 + t2 )X + (t1 − t2 )Y is a Normal for any t1 + t2 and t1 − t2 , so T is Normal for any t1 and t2 , therefore (X + Y, X − Y ) is also Bivariate Normal. (b) Find the joint PDF of X + Y and X − Y without using calculus. From problem 1 we know Cov(X + Y, X − Y ) = 0 since X and Y have the same variance. That means X+Y and X-Y are uncorrelated. In the case of Multivariate Normal, uncorrelation leads to independence, thus X+Y and X-Y are independent. 5. (1D Transformation) Determine the PDF of Y in each of the following: (a) Y = Z 2 where Z ∼ N (0, 1). Since Y = Z 2 is not one-to-one, we need to split it into: √ √ x = y when x > 0 and x = − y when x < 0 1 √ In both cases | dx dy | = 2 y By 1D transformation we have dx −y/2 y −1/2 √1 fY (y) = fX (x)| dx dy |I(x > 0) + fX (x)| dy |I(x < 0) = 2φ e (b) U ∼ Unif(0,1) and Y = U 1/α , where α > 0 Find U in terms of Y: u = y α du α−1 for y ∈ (0, 1). dy = αy α−1 . This is a Beta(α, 1) distribution. So fY (y) = fU (u)| du dy | = αy 4
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