F71SM STATISTICAL METHODS
Tutorial on Section 5 MULTIVARIATE DISTRIBUTIONS AND LINEAR COM-
BINATIONS
1. A bag contains 6 tickets numbered 1, 2, 3, 4, 5, and 6. Two tickets are selected at random,
one after the other and without replacement. Let X1 and X2 be the numbers on the first
and second tickets drawn, respectively.
(a) Tabulate the joint probability mass function for (X1, X2) and hence obtain the
marginal distributions of X1 and X2. Noting that X1 and X2 are identically dis-
tributed, show that each has mean 7/2 and variance 35/12.
(b) Find the distribution of X̄ = (X1 +X2)/2 and hence show that X̄ has mean 7/2 and
variance 7/6.
(c) Show that Cov[X1, X2] = −7/12 and find the correlation coefficient between X1 and
X2. [Corr[X1, X2] = -0.2]
2. Show from the definition of covariance that
(a) Cov[aX + b, cY + d] = acCov[X, Y ]
(b) Cov[X, Y + Z] = Cov[X, Y ] + Cov[X,Z]
where X, Y, Z are random variables and a, b, c, d are (non-random) constants.
3. Let (X, Y ) have joint pdf f(x, y) = e−(x+y), x > 0, y > 0 (worked example 5.2 in the
lecture notes; you may use results presented in worked example 5.2).
(a) Find P (X + Y < 1) and the cumulative distribution function F (x, y). [0.2642] (b) State the values of Cov[X, Y ] and Corr[X, Y ] [Cov[X, Y ] = Corr[X, Y ] = 0] (c) Find the mean and variance of S = X+Y and W = X−Y . Are S and W correlated? [E[S] = 2,Var[S] = 2, E[W ] = 0,Var[W ] = 2] 4. Let X and Y be discrete r.v.s with the following joint pmf: Y −2 −1 1 2 −1 1/16 1/8 1/8 1/16 X 0 1/16 1/16 1/16 1/16 1 1/16 1/8 1/8 1/16 Show that X and Y are uncorrelated but not independent. 5. Let X1, X2, . . . , Xn be a set of independent r.v.s and let S = ∑n i=1Xi. Show that, where the expectations are defined and exist, (a) the pgfs satisfy GS(t) = ∏n i=1GXi(t) (b) the mgfs satisfy MS(t) = ∏n i=1MXi(t) 6. Using mgfs, show that, for independent r.v.s X and Y , (a) X ∼ bi(n, p), Y ∼ bi(m, p)⇒ X + Y ∼ bi(n+m, p) 1 (b) X ∼ Poisson(λ1), Y ∼ Poisson(λ2)⇒ X + Y ∼ Poisson(λ1 + λ2) (c) X, Y ∼ exp(λ)⇒ X + Y ∼ gamma(2, λ) (d) X ∼ χ2n, Y ∼ χ2m ⇒ X + Y ∼ χ2n+m (e) X ∼ N(µX , σ2X), Y ∼ N(µY , σ2Y )⇒ X + Y ∼ N(µX + µY , σ2X + σ2Y ) 7. Cans of food have a mean weight of 274.5g and a standard deviation of 11.6g. The cans alone have a mean weight of 25.3g and a standard deviation of 2.3g. The labels on the cans state ‘net weight 240g’. You may assume that the weights of the cans alone and of their contents are independently and normally distributed. Find (a) the mean and standard deviation of the weights of the contents of the cans [Mean = 249.2g, SD = 11.37g]; (b) the percentage of cans whose contents are underweight [20.9%]; (c) the probability that the total weight of the contents of four cans exceeds 960g. [0.9474] 8. Claim sizes on policies of type A are normally distributed about a mean of £1100 and with standard deviation £200. Claim sizes on policies of type B are normally distributed about a mean of £1300 and with standard deviation £300. Assuming independence among all claim sizes, find the probability that the total size of four type A claims exceeds the total size of three type B claims. [0.7764] Hint: Find the distribution of (X1 +X2 +X3 +X4)− (Y1 +Y2 +Y3) where Xi is a type A claim size and Yj is a type B claim size. 2