Statistics, Paper I

To minimize $S$ with respect to $a$ and $b$ , we take partial derivatives and set them to zero: 1. Derivative with respect to $a$ : $\frac{\partial S}{\partial a} = -2 \sum (Y_i - a - bX_i) = 0 \implies \sum Y = na + b \sum X$ 2. Derivative with respect to $b$ : $\frac{\partial S}{\partial b} = -2 \sum X_i(Y_i - a - bX_i) = 0 \implies \sum XY = a \sum X + b \sum X^2$ These two expressions constitute the Normal Equations . Sum of Squares of Residuals The sum of squares of residuals ( $SSE$ ) is given by S = ∑ ( Y i − a − b X i ) Y i − a ∑ ( Y i − a − b X i ) − b ∑ X i ( Y i − a − b X i ) S = \sum (Y_i - a - bX_i)Y_i - a\sum(Y_i - a - bX_i) - b\sum X_i(Y_i - a - bX_i) S = ∑ ( Y i ​ − a − b X i ​ ) Y i ​ − a ∑ ( Y i ​ − a − b X i ​ ) − b ∑ X i ​ ( Y i ​ − a − b X i ​ ) . Since the normal equations imply $\sum(Y_i - a - bX_i) = 0$ and $\sum X_i(Y_i - a - bX_i) = 0$ , the last two terms vanish, leaving: S S E = $SSE = \sum Y_i(Y_i - a - bX_i) = \sum Y_i^2 - a\sum Y_i - b\sum X_i Y_i$ Substituting the least squares estimates $a$ and $b$ , the expression is simplified to the required identity: ∑ Y 2 − a ∑ Y − b ∑ X Y \boxed{\sum Y^2 - a\sum Y - b\sum XY} ∑ Y 2 − a ∑ Y − b ∑ X Y ​ 8

When two fair dice are thrown, the total number of possible outcomes is 6 $6 \times 6 = 36$ . Each outcome is equally likely. (i) Probability of getting a sum between 4 and 8 inclusive. The possible sums range from 2 (1+1) to 12 (6+6). We are interested in sums S such that 4 $4 \le S \le 8$ . The outcomes that yield these sums are: Sum = 4: (1,3), (2,2), (3,1) - 3 outcomes Sum = 5: (1,4), (2,3), (3,2), (4,1) - 4 outcomes Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) - 5 outcomes Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - 6 outcomes Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) - 5 outcomes The total number of favourable outcomes is 3 + 4 + 5 + 6 + 5 = 23 3 + 4 + 5 + 6 + 5 = 23 3 + 4 + 5 + 6 + 5 = 23 . The probability is given by: $P(\text{Sum between 4 and 8}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{23}{36}$ (ii) Probability of getting a product between 4 and 8 inclusive. We are interested in products P such that 4 $4 \le P \le 8$ . The outcomes that yield these products are: Product = 4: (1,4), (2,2), (4,1) - 3 outcomes Product = 5: (1,5), (5,1) - 2 outcomes Product = 6: (1,6), (2,3), (3,2), (6,1) - 4 outcomes Product = 7: No outcomes Product = 8: (2,4), (4,2) - 2 outcomes The total number of favourable outcomes is 3 + 2 + 4 + 0 + 2 = 11 3 + 2 + 4 + 0 + 2 = 11 3 + 2 + 4 + 0 + 2 = 11 . The probability is given by: $P(\text{Product between 4 and 8}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{11}{36}$ 3

Find conditional expectation of bivariate normal distribution.