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# quiz4BS-f07 - a Find the method of moment estimator of...

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ECON 506-M Quiz4-B, November 29, 2007 NAME: 1. y i are observations from a normal distribution with mean = 10 and variance = 4. The following is an estimator of the population mean: T = (y 2 + y 3 ) /2, Find the bias and mean square error of the estimator. Solution: E (T) = E [y 2 + y 3 ) /2 ] = 10, unbiased. V (T2) = (1/4) * 8 = 2, MSE = 2 2. Write the log likelihood function for a sample of m = 2 observations, Y 1 & Y 2 from the following Gamma distribution: f(y) = 2 β β y ye - for 0 < y < ∞ Solution: L = f (Y 1 =y 1 , Y 2 =y 1 ) = f (Y 1 =y 1 ) f (Y 2 =y 2 ) = = 2 1 1 β β y e y - * 2 2 2 β β y e y - = 4 ) ( 2 1 2 1 β β y y e y y + - ln L = β β Ln y y Lny Lny 4 ) ( 2 1 2 1 - + - + 3. A log likelihood function for a certain distribution is the following. Find the maximum likelihood estimator of the parameter θ . Ln L = 4 ln θ - 5 θ - 6.0 Solution: d (ln L) / d θ = 4 θ -1 - 5 = 0 θ = 4/5 4. Let n Y Y Y , , , 2 1 denote a random sample from the probability density function F(y) = 1 - θ θ y 0<y<1

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Unformatted text preview: a. Find the method of moment estimator of . Solution: 1 ) ( 1 + = = âˆ« dy y Y E . Therefore: Y Y Y-= â‡’ + = 1 Ë† 1 b. Show that Y is a consistent estimator of 1 + Î¸ . Solution: 1 ) ( ) ( + = = Y E Y E âˆž â†’ = â‡’ = n Y V n Y V Y V as ) ( lim ) ( ) ( 5. Y i are independent, identically distributed from a normal distribution. In our sample we find n = 19; 10 = Y , S 2 = 49 a. Construct a 95% confidence interval for the mean. Solution: S = 7. 95% CI is: n S t Y 18 025 . Â± 10 Â± 2.101 (7/âˆš18) 10 Â± 3.47 [6.53, 13.47] b. Construct a 90% confidence interval for Ïƒ 2 . Solution: 39046 . 9 , 8693 . 28 2 95 . 2 05 . = = Ï‡ 90% CI is [ ] 39046 . 9 ) 49 )( 18 ( , 8693 . 28 ) 49 )( 18 ( [30.55, 93.93] 2...
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