answerkey5_5

The conditional variance of y i given x i is then

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Unformatted text preview: The conditional variance of Y i given X i is then obtained from the product of P ( Y i = 1 | X i ) and P ( Y i = 0 | X i ): V ar ( Y i | X i ) = ( β + β 1 X i )(1- β- β 1 X i ). (If Z is a random variable which is one with probability p and zero with probability q = 1- p (i.e., Z is a Bernoulli random variable), then E ( Z ) = p and V ar ( Z ) = p · q . Likewise, the variance of Y i given X i is p · q where p is P ( Y i = 1 | X i ) and q is P ( Y i = 0 | X i )). 3. (a) (Ans) The probability mass function can be written by P ( Y i = y i ) = p · 1 { y i =1 } + q · 1 { y i =2 } + (1- p- q ) · 1 { y i =3 } 2 where 1 { y i = c } for c = 1 , 2 , 3 is an indecator fuctions defined as 1 { y i = c } = 1 for y i = c otherwise . Let n j denote the number of data point at which y i = j for j = 1 , 2. I.e., n 1 is the number of data points for Y i = 1 while n 2 is the number of data points when Y i = 2. The number of data points for Y i = 3 automatically becomes n- n 1- n 2 as Y i only takes three values 1 , 2 or 3 with positive probabilities. Given the probability mass function of Y i , the likelihood function is L n = p n 1 · q n 2 · (1- p- q ) n- n 1- n 2 and the log likelihood function is ln L n = n 1 ln p + n 2 ln q + ( n- n 1- n 2 )ln(1- p- q ). (b) (Ans) Solving the following two equations simultaneously ∂ ln L n ∂p = n 1 p- n- n 1- n 2 (1- p- q ) = 0 ∂ ln L n ∂q = n 1 q- n- n 1- n 2 (1- p- q ) = 0, we obtain the MLE of p and q , ˆ p = n 1 n ˆ q = n 2 n . 3...
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The conditional variance of Y i given X i is then obtained...

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