Statistics STAT:5100 (22S:193), Fall Sample Final Exam B

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1 Statistics STAT:5 (22S:93), Fall 25 Sample Final Exam B Please write your answers in the exam books provided.. Let X, Y, and Y 2 be independent random variables with X N(µ X, σ 2 X ) and Y i N(µ Y, σ 2 Y ). Let Z i X + Y i for i, 2. (a) Find the means and variances of Z and Z 2. (b) Find the covariance and correlation of Z and Z The conditional density of Y given X x is xe xy for y >. The marginal density of X is e x for x >. (a) Find the marginal density of Y. (b) Find the conditional density of X given Y y; if this is a standard distribution identify it by name. 3. The random variables U and V are independent Uniform[, ] random variables. Find the density of X V/U. 4. Let X,..., X n be a random sample from an Exponential(λ) population with density λe λx for x f(x λ) for some λ >. Recall that E[X i ] /λ and Var(X i ) /λ 2. (a) Show that X n AN(/λ, /(nλ 2 )) as n. (b) Show that n(x n /λ)/x n AN(, ) as n (c) Show that /X n AN(λ, λ 2 /n) as n. Recall that Y n AN(a n, b 2 n) as n means that (Y n a n )/b n converges in distribution to a standard normal random variable. 5. Suppose n different letters and envelopes are typed, but the letters accidentally are assigned randomly to the envelopes. Let X be the number of letters assigned to the correct envelopes. Compute the mean and variance of X. Hint: Let if the i-th letter is assigned to the correct envelope Y i, and consider how X is related to Y,..., Y n.

2 Statistics STAT:5 (22S:93), Fall 25 Some Distributions Bernoulli(p) pmf P (X x p) p x ( p) x ; x, ; p mean, variance E[X] p, Var(X) p( p) mgf M X (t) ( p) + pe t Binomial(n, p) pmf P (X x n, p) ( n x) p x ( p) n x ; x,,..., n; p mean, variance E[X] np, Var(X) np( p) mgf M X (t) (( p) + pe t ) n Poisson(λ) pmf P (X x λ) λx e λ ; x,,...; λ < x! mean, variance E[X] λ, Var(X) λ mgf M X (t) e λ(et ) Geometric(p) pmf P (X x p) p( p) x ; x, 2,...; < p mean, variance E[X] p, Var(X) p p 2 pe mgf M X (t) t ( p)e t Negative Binomial(r.p) pmf P (X x r, p) ( ) r+x x p r ( p) x ; x,,...; < p mean, variance E[X] r( p), Var(X) r( p) ( p p 2 mgf M X (t) Beta(α, β) pdf ) r p ( p)e t f(x α, β) Γ(α+β) Γ(α)Γ(β) xα ( x) β ; < x < mean, variance E[X] α, Var(X) αβ α+β (α+β) 2 (α+β+) Cauchy(θ, σ) pdf f(x θ, σ) πσ +( x θ σ ) 2 ; < x < ; < θ < ; σ > mean, variance do not exist mgf does not exist Gamma(α, β) pdf f(x α, β) x α e x/β ; < x < ; α, β > Γ(α)β α mean, variance E[X] αβ, Var(X) αβ 2 ( mgf M X (t) βt ) α, t < β Normal(µ, σ 2 ) pdf f(x µ, σ 2 ) 2πσ exp (x µ) 2 }; σ 2 > 2σ 2 mean, variance E[X] µ, Var(X) σ 2 mgf M X (t) expµt + 2 t2 σ 2 } 2

3 Statistics STAT:5 (22S:93), Fall 25 Solutions. (a) The means are E[Z i ] E[X + Y i ] E[X] + E[Y i ] µ X + µ Y and the variances are Var[Z i ] Var[X + Y i ] Var[X] + Var[Y i ] σx 2 + σy 2. since X and Y i are independent. (b) The covariance of Z and Z 2 is Cov(Z, Z 2 ) Cov(X + Y, X + Y 2 ) Cov(X, X + Y 2 ) + Cov(Y, X + Y 2 ) Cov(X, X) + Cov(X, Y 2 ) + Cov(Y, X + Y 2 ) Var(X) σx 2 since X, Y, and Y 2 are independent. The correlation is therefore ρ Y,Y 2 2. (a) The marginal density of Y is Cov(Z, Z 2 ) Var(Z )Var(Z 2 ) σ 2 X σ 2 X + σ2 Y. + σy 2 /σ2 X for y >. f Y (y) ( + y) 2 ( + y) 2 xe xy e x dx xe (+y)x dx ue u du (b) The conditional density of X given Y y for y > is f(x, y) f X Y (x y) f Y (y) ( + y) 2 xe (+y)x for x >. This is a Gamma(2, ( + y) ) density. defined for y <. The conditional density is not 3

4 Statistics STAT:5 (22S:93), Fall Let Y U. The inverse transformation is u y v xy. and the image of A [, ] [, ] under the transformation is B (x, y) : y, xy }. All x values with x are possible, and for a given x value the corresponding y value must satisfy both y and xy, or y /x. So y values must satisfy y min(, /x), and B can be written as B (x, y) : x, y min(, /x)}. The Jacobian determinant of the inverse transformation is ( ) det y y x The joint density of U, V is f U,V (u, v) The joint density of X, Y is thus f X,Y (x, y) f U,V (y, xy)y for u and v. y for x and y min(, /x) and the marginal density of X is min(,/x) ydy for x f X (x) y2 min(,/x) for x 2 2 min(, /x2 ) for x for x > 2x 2 for x 2 for x <. 4

5 Statistics STAT:5 (22S:93), Fall (a) The mean and variance of X i are E[X i ] /λ and Var(X i ) /λ 2. By the central limit theorem, X n AN(/λ, /(nλ 2 )), or n(xn /λ)λ D Z N(, ). (b) By the law of large numbers, X P n /λ, and by the continuous mapping theorem /(λx n ) P. Slutsky s theorem then implies that n(xn /λ)/x n n(x n /λ)λ λx n (c) By the delta method with f(x) /x and f (x) /x 2 f(x n ) /X n AN(f(/λ), f (/λ) 2 /(nλ 2 )) AN(λ, λ 4 /(nλ 2 )) AN(λ, λ 2 /n) D Z N(, ). In this problem parts (b) and (c) are algebraically equivalent, so either of the two proofs shown here works in both cases. 5. The number of correctly assigned letters is n X Y i. i The Y i are Bernoulli random variables, and by symmetry E[Y i ] for all i. So n [ n ] n E[X] E Y i E[Y i ] n n and i Var(Y i ) n i ( ) n n n 2 The Y i are not independent, so their covariances need to be considered in computing the variance of X. By symmetry, for all i j E[Y i Y j ] E[Y Y 2 ] P (letters and 2 both in correct envelopes) P (letter correct)p (letter 2 correct letter correct) n n. So the covariance of Y i and Y j for i j is Cov(Y i, Y j ) E[Y i Y j ] E[Y i ]E[Y j ] ( n(n ) n 5 ) 2 n ( n ) n n 2 (n )

6 Statistics STAT:5 (22S:93), Fall 25 and therefore ( n ) Var(X) Var Y i i i nvar(y ) + n(n )Cov(Y, Y 2 ) n n + n(n ) n 2 n 2 (n ) n + n n. Var(Y i ) + Cov(Y i, Y j ) i j 6