Question
Please explain both questions. Show work.
7. Suppose X and Y are the random variables with joint PMF given by: 10.16 0.24 2 0.2 0.3 3 0.04 0.06 Are X and Y independent? (a) Yes Answer: Ta (b) No 8. Let X and Y be two random variables such that where Z is independent of X and Z N(0, ?2). A random sample of n 25 pairs (Xi, Yi),..., (Xn,Yn) resulted in the following statistics: X=-0.14; Y-5.04 x-x)2-220.11 (xi-X)Y, 100.23 25 i-1 25 Suppose ?2-419. Which of the following corre- sponds to an approximate 68% confidence interval for the true value of ? (considering Xi, , xn fixed) (a) -7.486 +0.204 (b) 0.455 ±0.138 (d) 7.882±0.203 (c) 2.772 ±0.136 (e) 5.262±0.856 Answer: 8b
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Answer #1

Question 7

Two discrete random variables Z1 and Z2 are independent if and only if P(Z_1 = z_1, Z_2 = z_2) = P(Z_1 = z_1) \ P(Z_2 = z_2) , for all possible pairs (z1, z2).

Here,

\\ P(X = 1) = 0.16 + 0.24 = 0.4 \\ P(X = 2) = 0.2 + 0.3 = 0.5 \\ P(X = 3) = 0.04 + 0.06 = 0.1 \\ P(Y = 1) = 0.16 + 0.2 + 0.04 = 0.4 \\ P(Y = 2) = 0.24 + 0.3 + 0.06 = 0.6 \\

Now,

\\ P(X = 1, Y = 1) = 0.16 = 0.4 * 0.4 = P(X = 1) \ P(Y = 1) \\ P(X = 1, Y = 2) = 0.24 = 0.4 * 0.6 = P(X = 1) \ P(Y = 2) \\ P(X = 2, Y = 1) = 0.2 = 0.5 * 0.4 = P(X = 2) \ P(Y = 1) \\ P(X = 2, Y = 2) = 0.3 = 0.5 * 0.6 = P(X = 2) \ P(Y = 2) \\ P(X = 3, Y = 1) = 0.04 = 0.1 * 0.4 = P(X = 3) \ P(Y = 1) \\ P(X = 3, Y = 2) = 0.06 = 0.1 * 0.6 = P(X = 3) \ P(Y = 2) \\

Hence, X and Y are mutually independent.

Question 8

Let b be the least square estimator of \beta .

Then, b = \frac {\frac {1} {n} \sum _ 1 ^ n (x _ i - \bar{x}) (y _ i - \bar{y})} {\frac {1} {n} \sum _ 1 ^ n (x _ i - \bar{x}) ^ 2} = \frac {\sum _ 1 ^ n (x _ i - \bar{x}) y _ i} {\sum _ 1 ^ n (x_i - \bar{x}) ^ 2} \sim N(\beta, \frac {\sigma ^ 2} {\sum _ 1 ^ n (x _ i - \bar{x}) ^ 2})

Hence, a 68% confidence indeval for \beta will be given by \\ P(\mid \frac {b - \beta} {\sqrt{\frac {\sigma ^2} {\sum _ 1 ^ n (x _ i - \bar{x}) ^ 2}}} \mid \leq \tau _ {\frac {\alpha} {2}}) = 1 - \alpha, \ where \ \tau _ {\alpha} \ is \ the \ upper \ \alpha \ point \ of \ \mathbb{N}(0,1) \ distribution \\ P((b - d, b + d) \ni \beta) = 0.68, \ where \ d = \tau _ {0.16} \ \frac {\sigma} {\sqrt{\sum _ 1 ^ n (x _ i - \bar{x}) ^ 2}} \\

Here, b = 100.23 / 220.11 = 0.455 (approximately)

Since we do not know the value \sigma ^ 2 , we will have to use its estimate to get the value of d. But we should use the cut-off based on t23 distribution, as \hat{\sigma ^ 2} \sim \chi ^ 2 _ {n - 2} distribution. But as an approximation, one can use the normal cutoff also.

Hence, d = 0.994 * sqrt(4.19 / 220.11) = 0.137 (approximately)

Thus, the 68% confidence interval is given by (0.455 - 0.137, 0.455 + 0.137)

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