Econometric take home APPS_Part_43

Econometric take home APPS_Part_43 - (a From the previous...

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(a) From the previous problem, M x ( t ) = exp[ λ ( e t - 1)]. Suppose y is distributed as Poisson with parameter μ . Then, M y ( t )=exp[ μ ( e t -1)]. The product of these two moment generating functions is M x ( t ) M y ( t )= exp[ λ ( e t - 1)]exp[ μ ( e t - 1)] = exp[( λ + μ )( e t - 1)], which is the moment generating function of the Poisson distribution with parameter λ + μ . Therefore, on the basis of the theorem given in the problem, it follows that x + y has a Poisson distribution with parameter λ + μ . (b) The density of the Chi-squared distribution with n degrees of freedom is [from (C-39)] fx n ex x n x n () (/) ,. / => −− 12 2 0 2 1 22 1 Γ Let the constant term be k for the present. The moment generating function is M ( t ) = k ee x d x tx x n /( / ) 1 0 = k . d x xt n (1/ ) ( / ) 1 0 This is a gamma integral which reduces to M ( t ) = k (1/2 - t ) - n /2 Γ ( n /2). Now, reinserting the constant k and simplifying produces the moment generating function M ( t ) = (1 - 2 t ) - n /2 . Suppose that x i is distributed as chi-squared with n i degrees of freedom. The moment generating function of Σ i x i is Π i M i ( t ) = / 2 t n i i which is the MGF of a chi-squared variable with n = Σ i n i degrees of freedom. (c) We let y = σ z + μ . Then, M y ( t ) = E [exp( ty )] = [ ] [ ][ Ee e Ee tz t t z t t z ( ) σμ μ μ + == σσ ] = [ ] t t tt μσ σ =− / exp ( ) / 2 2 μ 2 Using the same approach as in part b., it follows that the moment generating function for a sum of random variables with means μ i and standard deviations σ i is Mt x i i i i i i ∑∑ exp 1 2 . ± 171
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Appendix C Estimation and Inference 1. The following sample is drawn from a normal distribution with mean μ and standard deviation σ : x = 1.3, 2.1, .4, 1.3, .5, .2, 1.8, 2.5, 1.9, 3.2. Compute the mean, median, variance, and standard deviation of the sample. x x n i i n = = 1 = 1.52, s 2 = ( ) xx n i i n = 2 1 1 = .9418, s = .97 m e d i a n = 1.55, midway between 1.3 and 1.8. 2. Using the data in the previous exercise, test the following hypotheses: (a) μ > 2. (b) μ < .7. (c) σ 2 = .5. (d) Using a likelihood ratio test, test the following hypothesis μ = 1.8, σ 2 = .8. (a) We would reject the hypothesis if 1.52 is too small relative to the hypothesized value of 2. Since the data are sampled from a normal distribution, we may use a t test to test the hypothesis. The t ratio is t [9] = (1.52 - 2) / [.97/ 10 ]= -1.472. The 95% critical value from the t distribution for a one tailed test is -1.833. Therefore, we would not reject the hypothesis at a significance level of 95%. (b) We would reject the hypothesis if 1.52 is excessively large relative to the hypothesized mean of .7. The t ratio is t [9] = (1.52 - .7) / [.97/ 10 ]= 2.673. Using the same critical value as in the previous problem, we would reject this hypothesis.
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Econometric take home APPS_Part_43 - (a From the previous...

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