# Statistiek (WISB361)

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1 Statistiek (WISB361) Final exam June 29, 2015 Schrijf uw naam op elk in te leveren vel. Schrijf ook uw studentnummer op blad 1. The maximum number of points is 100. Points distribution: Consider the sample X = X i n of i.i.d. random variables with probability density function: 2 2 f(x ) = x 3 for x, 0 otherwise for > 0. (a) [7pt] Determine the maximum likelihood estimator ˆ MLE and 2 MLE of and 2 respectively. The likelihood for a realization x of the sample can be written as: n n lik() = The likelihood is increasing in, hence: 2 2 x 3 1 (,+ ) (x i ) = 2n 2n n i x3 i MLE = X (1) where X (1) := min i X i. By the invariance principle: 2 MLE = ( MLE ) 2 = X 2 (1) 1 (0,xi)() = 2n 2n n x3 i 1 (0,x(1) )() (b) [4pt] Determine the distribution of ˆ MLE. The probability distribution function (PDF) of MLE is the PDF of X (1). Therefore with for x >. Therefore F X(1) (x) = 1 P(X (1) > x) = 1 F X1 (x) = and the probability density function (pdf) is: x n P(X i > x) = 1 (1 F X1 (x)) n, F X(1) (x) = (1 2n f X(1) (x) = 2 2 /y 3 dy = 1 2 x 2 x 2n (2n 2n x 2n+1 ) 1 (, ) (x) ) 1 (, ) (x) (c) [3pt] Is ˆ MLE a biased estimator? Is ˆ MLE asymptotically unbiased? We have: so that MLE is biased. However, E( MLE ) = E(X (1) ) = 2n so that MLE is asymptotically unbiased. lim E(X (1)) = n x 2n dx = 2n x2n+1 2n 1 (1)

2 (d) [3pt] Find the method of moment estimator ˆ MoM of. Since E(X (1) ) = E(X 1 ), by (1) with n = 1, we have that E(X 1 ) = 2. Thus: where we denote with X n the sample mean. (e) [3pt] Find the variance of ˆ MoM. Is it finite? Var(ˆ MoM ) = Var(X1) 4n = +, because E(X 2 1 ) = ˆ MoM = X n 2 x 2 2 dx = + x3 (f) [3pt] Compare the mean squared errors (MSE) of ˆ MLE and ˆ MoM. Which estimator is the most efficient? By point (d) it follows that ˆ MoM is unbiased. Hence MSE(ˆ MoM ) = Var(ˆ MoM ) = +. As regards the MSE( MLE ), we have: Since when n 2, we get: ( ) 2 MLE = Var( MLE ) + (E( MLE ) ) 2 = Var(X (1) ) + 2n 1 E(X 2 (1) ) = MSE( MLE ) = x 2 2n 2n x 2n+1 x = n d n (n 1)(2n 1) Therefore if the sample size is larger than 1, MLE is more efficient than MoM. 2. Coffee abuse is often considered related to an increase of heart rate (i.e. number of poundings of the heart per unit of time). In order to support this claim, an experiment was planned on a sample of 10 subjects. The hearth rate was measured twice for each subject: the first time at rest, the second after having drank a cup of coffee. The measurements at rest are: and after a cup of coffe: x = x 1,..., x 10 = 74, 68, 67, 71, 69, 65, 70, 70, 66, 67 y = y 1,..., y 10 = 71, 72, 69, 66, 73, 77, 69, 68, 71, 78 We can assume that the random variables Z i = Y i X i, denoting the difference between the heart rate after the cup of coffee and the heart rate at rest, are i.i.d. normal random variables. (a) [12pt] Test the hypothesis that the coffee increase the heart rate at α = 0.05 level of significance. We have paired normal observations and we will perform a paired t test. Since Z i := Y i X i N(µ, σ 2 ), with µ := E(Y i X i ), and unknown variance σ 2, we want to test the hypotheses: H0 : µ = 0, H 1 : µ > 0. at α = 0.05 level of significance. Given the test statistics T : T = Z S/ n 2

3 with S 2 = 1 n 1 (Z i Z) 2, under H 0 T t(n 1). With the data of the problem, we have the following realization t of the test statistic: t = z s/ n = = We perform a one sided t test, so that the rejection region B is B = [t 9,0.05, + ) = [1.833, + ). Since / B, we do not reject H 0 at 0.05 level of significance. (b) [5pt] Calculate the p-value of the test. p value = P(T H 0 ). From the tables, we can just say that 0.05 < p value < (c) [3pt] In case the normality assumption does not hold, how you could test the hypothesis of point (a)? (It is enough to explain which test is the most appropriate, without performing the analysis). Signed rank test. 3. Let X 1 and X 2 be i.i.d. random variables such that X i Unif[, + 1], for i = 1, 2, where Unif[, + 1] denotes the uniform distribution in the interval [, + 1]. In order to test: H0 : = 0, H 1 : > 0 at α = 0.05 level of significance. we have two competing tests, with the following rejection regions: with C R. (a) [2pt] Find the significance level α of TEST1. TEST1 : Reject H 0 if X 1 > 0.95, TEST2 : Reject H 0 if X 1 + X 2 > C, α 1 = P(X 1 > 0.95 = 0) = 0.05 (b) [4pt] Find the value of C so that TEST2 has the same significance level α of TEST1. Since P(X 1 + X 2 > C = 0) is the area of region inside the unit square (with vertices (0, 0), (0, 1), (1, 0) and (1, 1)) above the line x 2 = C x 1, we have: 1 C 2 /2 : if 0 C 1, α 2 (C) = P(X 1 + X 2 > C = 0) = (2 C) 2 /2 if 1 < C 2, 0 otherwise We see that the equation α 2 (C) = α 1 has solution only for 1 < C 2. Therefore, the solution is (c) [4pt] Calculate the power function of each test. For the first test: C = 2 2α 1 = π 1 () = P(X 1 > 0.95 ) = 0 if 0.5, if 0.5 < 0.95, 1 otherwise 3

4 For the second test, we notice that π 2 () = P(X 1 + X 2 > C ) is the area of the unit square with vertices (, ), (, + 1), ( + 1, ) and ( + 1, + 1), over the line x 2 = C x 1. Hence, 0 if C/2 1, (2 + 2 C) π 2 () = 2 /2 if (C/2) 1 < (C 1)/2, 1 (C 2) 2 /2 if (C 1)/2 < C/2 1 if > C/2. (d) [6pt] Is TEST2 more powerful than TEST1? From point (c) it follows that TEST1 is more powerful for near 0, but TEST2 is more powerful for larger. Hence, TEST2 is not uniformly more powerful than TEST1. (e) [4pt] Show how to get a test that has the same significance level but more powerful than TEST2. If either X 1 1 or X 2 1 we should reject H 0, because if = 0 then P(X i < 1) = 1. Thus, if we consider the rejection region: B := (x 1, x 2 ) : x 1 + x 2 > C (x 1, x 2 ) : x 1 > 1 (x 1, x 2 ) : x 2 > 1 The first set is the rejection region for TEST2. The test with this rejection region B has the same significance as TEST2 because the last two sets both have probability 0 if = 0. But for 0 < < C 1 the power of this test is strictly larger than π 2 (). 4. Let the independent normal random variables Y 1, Y 2,..., Y n be such that Y i N(µ, α 2 x 2 i ), for i = 1,..., n, where the given constants x i are not all equal and no one of which is zero. (a) [13pt] Derive the least squares estimators of µ and α 2, after you have properly rescaled the random variables Y i. The linear model for Y i is: Y i = µ + ɛ i, i = 1, 2,..., n, where ɛ i are independent normal RV with zero mean and different variances: Var(ɛ i ) = α 2 x 2 i. Let Therefore, we have the linear model for Z i : Z i = Y i /x i Z i = µ 1 x i + ɛ i i = 1, 2,..., n, where ɛ i are i.i.d. RV with zero mean and common variance: Var( ɛ i ) = α 2. Minimizing the sum of the squares S(µ) = (Z i µ x i ) 2, we obtain the LS estimator ˆµ LSE of µ: ˆµ LSE = Z i/x i 1/x2 i = Y i/x 2 i 1/x2 i and, from the residual sum of the squares (RSS), we get the estimator of the variance: α 2 = 1 n 1 n (Z i ˆµ LSE ) 2 = 1 n (Y i /x i ˆµ LSE ) 2 n 1 (b) [4pt] Which is the distribution of (n 1)ˆα 2 /α 2? It follows that (n 1)ˆα 2 /α 2 has a χ 2 n 1 distribution. 4

5 (c) [3pt] Discuss the test of hypotheses: H0 : α = 1, H 1 : α 1. By point (b) we can perform a two sided χ 2 n 1 test, using the test statistics (n 1)ˆα Consider the sample X = X i n of i.i.d. random variables such that X i N(, σ 2 ) with σ 2 known and Ω, where the parameter space is Ω = 2, 0, 1. (a) [2pt] Show that X = 1/n X i is a sufficient statistics for and that the likelihood lik() can be factorized in lik() = h(x)g ( x), where x is a realization of X, x is a realization of X and h(x) = (2πσ 2 ) n/2 exp 1 2σ 2 n (x i x) 2, g ( x) = exp n ( x )2 2σ2 lik() = (2πσ 2 ) n/2 exp 1 n 2σ 2 (x i ) 2 = (2πσ 2 ) n/2 exp 1 n 2σ 2 (x i x + x ) 2 = (2πσ 2 ) n/2 exp 1 n 2σ 2 (x i x) 2 exp n ( x )2 2σ2 In the figure above the functions g (y) are plotted for the three possible values of the parameter. (b) [8pt] Find a maximum likelihood estimator ˆ MLE of. From the Figure follows that: 2 if x < 1, ˆ MLE = 0 if 1 x if x > 0.5 (c) [4pt] Find the probability mass function of ˆ MLE. 5

6 If we denote with 0 the true value of the parameter, we have: Φ(( 1 0 ) n/σ) if t = 2, P(ˆ MLE = t 0 ) = Φ((0.5 0 ) n/σ) Φ(( 1 0 ) n/σ) if t = 0, 1 Φ((0.5 0 ) n/σ) if t = 1, for 0 2, 0, 1 and where Φ( ) denote the CDF of the standard normal distribution. (d) [3pt] Is ˆ MLE a biased estimator? ˆ MLE is a biased estimator. In fact, in case 0 = 2, we have: E(ˆ MLE 0 = 2) = 2Φ( n/σ)+1 Φ(2.5 n/σ) = 2Φ( n/σ)+φ( 2.5 n/σ) > 2Φ( n/σ) > 2 because 0 < Φ( ) < 1. 6

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