18.443 Problem Set 1 Solutions

18.443 Problem Set 1 Spring 2014 Statistics for Applications Due Date: 2/13/2015 prior to 3:00pm Problems from John A. Rice, Third Edition. [Chapter.Section.P roblem] 1. Problem 6.4.1 Z ∼ N (0, 1) and U ∼ χ2n and Z and U are independent. T = Z/ U/n a Student’s t random variable with n degrees of freedom. Find the density function of T. Solution: • The density of Z is fZ (z) = −1 2 √1 e 2 x , 2π −∞ < x < +∞. • The density of U is fU (u) = 1 u(n/2)−1 e−u/2 Γ(n/2)2n/2 0 , if , if u>0 u≤0 • Because Z and U are independent their joint density is fZ,U (z, u) = fZ (z)fU (u) • Consider transforming (Z, U ) to (T, V ), where T = Z/ U/n and V = U, computing the joint density of (T, V ) and then integrating out V to obtain the marginal density of T . – Determine the functions g(T, V ) = Z and h(T, V ) = U g(T, V ) = V /nT h(T, V ) = V Then the joint density of (T, V ) is given by fT,V (t, v) = fZ,U (g(t, v), h(t, v)) × J where J is the Jacobian of the transformation from (Z, U ) to (Z, U ). Compute J: ∂g(t, v) ∂g(t, v) T 1 V /n ( √ ) V −1/2 ∂t ∂v = V /n J = ∂h(t, v) ∂h(t, v) = n 2 0 1 ∂t ∂v 1 The joint density of (T, V ) is thus fT,V (t, v) = fZ (g(t, v)) × fU (h(t, v)) × J √ p 1 1 1 2 (n/2)−1 −v/2 = √ e− 2 ( v/nt) × v e × v/n Γ(n/2)2n/2 2π 2 v t 1 1 = √ v 2 (n+1)−1 e− 2 (1+ n ) n/2 2πnΓ(n/2)2 Integrate over v to obtain the marginal density of T : a∞ fT (t) = J0 fT,V (t, v)dv ∞ v t2 1 1 √ = v 2 (n+1)−1 e− 2 (1+ n ) dv 2πnΓ(n/2)2Jn/2 0 ∞ 1 v t2 1 = √ v 2 (n+1)−1 e− 2 (1+ n ) dv 2πnΓ(n/2)2n/2 0 The integral factor can be evaluated by recognizing that it is identical to integrating a Gamma(α, λ) density function apart from the normalization constant, with α = (n + 1)/2 2 and λ = 12 (1 + tn ), that is a ∞ 1 α α−1 −λv 1 = 0 Γ(α) λ v e dv a ∞ α−1 −λv −α So Γ(α)λ = 0 v e dv which gives J ∞ v t2 1 v 2 (n+1)−1 e− 2 (1+ n ) dv = Γ(α) × λ−α 0 = Γ((n + 1)/2) × [ 12 (1 + t2 −(n+1)/2 n )] Finally we can a ∞write fT (t) = 0 fT,V (t, v)dv J ∞ 1 v t2 1 = √ v 2 (n+1)−1 e− 2 (1+ n ) dv 2πnΓ(n/2)2n/2 0 1 1 t2 = √ Γ((n + 1)/2) × [ (1 + )]−(n+1)/2 2 n 2πnΓ(n/2)2n/2 −(n+1)/2 2 Γ((n + 1)/2) t = √ × 1+ n πnΓ(n/2) 2. Suppose the random variable X has a t distribution with n degrees of freedom. (a). For what values of n is the variance finite/infinite. (b). Derive a formula for the variance of X (when it is finite). Solution: (a). For the variance of the t distribution to be finite it must have a finite second moment: 2 E[T 2 ] = a t2 fT (t)dt < ∞. The integrand of this second moment calculation is proportional to t→∞ t2 (n+1)/2 × t2−(n+1) ∝ t1−n (n+1)/2 −−−−−→ n 2 t 1+ n The integral of this integrand thus converges if and only if t2 fT (t) ∝  (1 − n) < (−1), which is equivalent to n > 2. (b). If n > 2, then the variance of T is finite.pFor such n, the mean of T exists and is zero, so writing T = Z/ U/n for independent Z ∼ N (0, 1) and U ∼ χ2n p 2 /U ] V ar(T ) = E[T 2 ] = E[[Z/ U/n]2 ] = E[nZ E 1 2 = n × E[Z ] × E U 1 = 1 × n × (n−2) n = n−2 1 The expectation E[ U ] = 1/(n − 2) can be computed directly for n > 2. (Note that the formula is undefined for n = 2 and gives negative values for n < 2) 3. 6.4.4. Also, add part (c) answer the question if the random variable T follows a standard normal distribution N (0, 1). Comment on the differences and why that should be. Solution: We are given that T follows a t7 distribution. The problem is solved by finding an expression for t0 in terms of the cumulative distribution function of T. (a). To find the t0 such that P (|T | < t0 ) = .9 this is equivalent to P (T < .95), which is solved in R using the function qt() – the quantile function for the t distribution > args(qt) function (p, df, ncp, lower.tail = TRUE, log.p = FALSE) >qt(.95,df=7) [1] 1.894579 3 So, t0 = 1.894579. (b). P (T > t0 ) = .05 is equivalent to P (T ≤ t0 ) = 1 − .05 = .95. This is the same t0 found in (a). > args(qt) function (p, df, ncp, lower.tail = TRUE, log.p = FALSE) > qt(p=.95, df=7) [1] 1.894579 # Which is equivalent to > qt(p=.05, df=7, lower.tail=FALSE) [1] 1.894579 (c). For the standard normal distribution we use qnorm() – the quan­ tile function for the N ormal(0, 1) distribution > args(qnorm) function (p, mean = 0, sd = 1, lower.tail = TRUE, log.p = FALSE) NULL > qnorm(.95) [1] 1.644854 > qnorm(p=.05, lower.tail=FALSE) [1] 1.644854 So for both parts (a) and (b) t0 = 1.644854 for the N (0, 1) r.v. versus t0 = 1.894579 for the t distribution with 7 degrees of freedom. The t0 values are larger for the t distribution indicating that the t distribution has heavier tail areas than the N ormal(0, 1) distribution. This makes sense because the t distribution equals a N ormal(0, 1) random variable divided by a random variable with expectation equal to 1 but positive variance. The possibility of the denominator of the t ratio being less than 1 increases the probability of larger values. 4. Problem 8.10.10. Use the normal approximation of the Poisson distribution to sketch the approximate sampling distribution of λ̂ of Example A of Section 8.4. According to this approximation, what is 4 P (|λ0 − λ̂| > δ) for δ = −.5, 1, 1.5, 2, 2.5 where λ0 is the true value of λ. Solution: In the example, the estimate λ̂ = X = n 1 Xi 1 n = 24.9, with n = 23. The X1 , . . . , Xn are assumed to be i.i.d. (independent and identically distributed) P oisson(λ0 ) random variables with E[Xi ] = λ0 and V ar[Xi ] = λ0 By the Central Limit Theorem √ (X−λ0 ) n→∞ n √λ −−−−→ N (0, 1). 0 The approximate sampling distribution of λ̂ is thus a Normal distri­ bution centered λ0 with p p standard deviation equal to λ0 /n ≈ nλ̂ = 24.9/23 = 1.040485. For the probability computations: P (|λ0 − λ̂| > δ) = P (|λ0 − X| > δ) √ n|λ0 − X| √ δ √ = P( > n√ ) λ0 √ λ0 ≈ P (|N (0, 1)| > n √δ ) √ λ̂ 23 = P (|N (0, 1)| > √24.9 δ) p = 2 × (1 − Φ( 23/24.9 × δ)) Using R and the function pnorm we can compute the desired values: > 2 * 1-pnorm( sqrt(23/24.9) * c(.5,1.,1.5,2.,2.5)) [1] 1.315420 1.168253 1.074703 1.027292 1.008137 5 5. Problem 8.10.13. In Example D of Secton 8.4, the method of moments estimate was found to be α̂ = 3X. In this example, consider the sampling distribu­ tion of α̂. (a). Show that E(α̂) = α, that is, that the estimate is unbiased. (b). Show that V ar(α̂) = (3 − α2 )/n. (c). Use the central limit theorem to deduce a normal approximation to the sampling distribution of α̂. According to this approximation, if n = 25 and α = 0, what is the P (|α̂| > .5). Solution The sample of values X1 , . . . , Xn giving X are i.i.d. with density func­ tion f (x | α) = 1+αx 2 , for −1 ≤ x ≤ +1, with parameter α : −1 ≤ α ≤ 1. (The values are such that xi = cos(θi ), where θi is the angle at which electrons are emitted in muon decay.) (a). Since the Xi are i.i.d. a1 E[X] = E[Xi ] = −1 x × ( 1+αx 2 )dx = α/3 It follows that E[α̂] = E[3X] = 3E[X] = 3(α/3) = α (b). Since the Xi are i.i.d. V ar[Xi ]/n = ( n1 ) × (E[X 2 ] − E[X]2 ) a1 2 ( n1 ) × ([ −1 x2 × ( 1+αx 2 )dx] − (α/3) ) a 2 1 ( n1 ) × ([ −1 x2 dx] − (α/3)2 ) ( n1 ) × ([1/3] − (α/3)2 ) 3 − α2 = 9n It follows that: 3 − α2 V ar[α̂] = V ar[3X] = 9 × V ar[X] = . n (c) By the central limit theorm, for true parameter α = α0 , it follows that 3 − α02 n→∞ α̂ −−−−−→ N (α0 , ) n V ar[X] = = = = 6