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1 Solutions to 2014 MAS exam : Theory 1. a) 2 P (gray | litter 2) = . 5 b) P (brown) = P (brown | litter 1)P (litter 1) + P (brown | litter 2)P (litter 2) 2 1 3 1 19 = ( )( ) + ( )( ) = . 3 2 5 2 30 c) P (brown | litter 1)P (litter 1) P (brown) 2 1 ( )( ) 10 = 3 19 2 = . 19 30 P (litter 1 | brown) = 2. A negative binomial random variable has mean r p and variance r(1−p) . p2 a) By the weak law of large numbers (Mean of a sequence of independent random variables, each with mean µ and variance σ 2 , converges in probability to the mean µ), we have X̄ = Pn i=1 n Xi converges in probability to pr . That means, for any given small value , the probability that X̄ is beyond the given distance of r p goes to zero as n increases to ∞. b) Central limit theorem tells “For a sequence of independent and identically distributed random variables, each with mean µ and variance σ 2 , X̄ √ σ/ n √ converges in distribution to N (0, 1).” By the Central limit theorem, we have √n(X̄−r/p)2 converges in distribution to r(1−p)/p N (0, 1). 2 c) For a negative binomial random variable with n=30, r=10, p=.5, mean is r/p = 20 and variance is r(1 − p)/p2 = 20. √ 30(X̄ − 20) √ P (X̄ ≤ 11) = P ( ≤ 20 √ 30(19 − 20) √ ) ≈ P (Z ≤ −1.225) = 0.1103 20 3. a) The marginal density of X is fX (x) = = Z 1 0 Z 1 0 fX,Y (x, y)dy 1 (x + 2y)dy 4 1 (xy + y 2 )|y=1 = y=0 4 1 = (x + 1), 0 ≤ x < 2. 4 b) The conditional density of Y given X is fX,Y (x, y) fX (x) 1 (x + 2y) = 41 (x + 1) 4 x + 2y = , 0 ≤ x < 2, 0 ≤ y < 1. x+1 fY |X (y|x) = X and Y are not independent because the conditional density of Y given X depends on X. c) Z = 9 (X+1) ⇒X= 9 Z − 1. So, 3 < z ≤ 9 and J = dx dz = − z92 . Therefore, the density of Z is 1 9 81 fZ (z) = ( − 1 + 1)|J| = z −3 , 3 < z ≤ 9. 4 z 4 3 4. a) By looking at the density function, it is actually a beta distribution with parameters a = θ and b = 1. Therefore, mean of the distribution is θ . θ+1 Solving θ̂ θ̂+1 = X̄, we have the method of moments estimator of θ is X̄ . 1 − X̄ θ̂ = b) The log likelihood function of θ is n Y l(θ) = log( f (xi |θ)) i=1 n Y = log( θxθ−1 ) i i=1 = log(θn ( n Y xi )θ−1 i=1 = n log θ + (θ − 1) n X log xi . i=1 Then ∂l(θ) ∂θ = n θ + Pn i=1 log xi . Solving ∂l(θ) ∂θ θ̂ = = 0, we have n − Pn i=1 log xi . Because the second derivative of l(θ) is − θn2 , which is always negative, the solution above is a maximizer of l(θ). Therefore θ̂ = − Pn n i=1 log xi is the MLE of θ. c) By the large sample theory, we know the asymptotic variance of the MLE is the inverse of the fisher information, which is i−1 ∂2 l(θ) ∂θ2 n −1 = − E(− 2 ) θ n −1 θ2 = ( 2) = θ n I(θ)−1 = −E h 4 5. a) The joint probability mass function of the six random variables, X1 , . . . , X6 , is given by 6 P f (x1 , x2 , x3 , x4 , x5 , x6 |θ) = e−6θ · θi=1 6 Q xi . xi ! i=1 The prior distribution of θ is given by Gamma(2, 1.5), Γ2 2−1 −1.5θ θ e . 1.52 π(θ) = Thus for the posterior distribution we have the following: π(θ|x) ∝ f (x|θ) · π(θ) 6 P ∝ e−6θ · θi=1 xi · θ2−1 e−1.5θ 6 P ∝ θi=1 P 6 which is the kernel of the Gamma P 6 distribution of θ is Gamma xi +2−1 e−7.5θ , xi + 2 , 7.5 distribution. Therefore, the posterior i=1 xi + 2 , 7.5 . i=1 b) A Bayes estimator is the mean of the posterior distribution. Therefore, a Bayes estimator of θ can be 6 P θ̂ = Xi + 2 i=1 7.5 c) A 90% credible interval for θ can be given by the 5th and the 95th percentile of the P 6 Gamma i=1 xi + 2 , 7.5 distribution. 5 6. a) The common probability density function of X1 , X2 , · · · , X5 is given by f (x) = θe−θx , x ≥ 0. To derive the distribution of Y , we calculate P (Y > y) = P (X1 > y, X2 > y, · · · , X5 > y) = [P (Xi > y)]5 " = R∞ #5 θe−θx dx = e−5θy y Hence fY (y) = 5θe−5θy . b) The power function of the test is given as follows: π(θ) = Pθ (rejecting H0 ) = Pθ (Y > 18) = e−5θ·18 = e−90θ Note that θ1 > θ2 ⇒ −90θ1 < −90θ2 ⇒ e−90θ1 < e−90θ2 Therefor the power function is a decreasing function of θ. c)The size (the significance level) of the test is: sup π(θ) = sup e−90θ = e−90/20 = 0.011109 θ∈Ω0 1 θ≥ 20