Financial Risk Manager Part 1

Cook's Distance Calculation

Cook's distance measures the influence of each observation on the regression coefficients. For the 5th observation, we need to calculate:

$$D_i = \frac{(\hat{Y}_i - \hat{Y}_{i(i)})^2}{p \cdot MSE} \cdot \frac{h_{ii}}{(1 - h_{ii})^2}$$

Where:

• $\hat{Y}_i$ is the predicted value using all observations
• $\hat{Y}_{i(i)}$ is the predicted value when observation i is excluded
• p is the number of parameters (2 for simple linear regression)
• MSE is the mean squared error
• $h_{ii}$ is the leverage

Step 1: Calculate predicted values

Using full dataset (5 observations): $\hat{Y} = 1.4110 + 0.1512X$ For observation 5 (X = 2.45): $\hat{Y}_5 = 1.4110 + 0.1512 \times 2.45 = 1.4110 + 0.37044 = 1.78144$

Using first 4 observations: $\hat{Y} = 0.3169 + 1.3667X$ For observation 5 (X = 2.45): $\hat{Y}_{5(5)} = 0.3169 + 1.3667 \times 2.45 = 0.3169 + 3.348415 = 3.665315$

Step 2: Calculate leverage $h_{55}$

Leverage for simple linear regression: $h_{ii} = \frac{1}{n} + \frac{(X_i - \bar{X})^2}{\sum(X_j - \bar{X})^2}$

From the full dataset:

• n = 5
• Mean of X: $\bar{X} = \frac{1.85 + 0.65 + 0.63 + 1.24 + 2.45}{5} = \frac{6.82}{5} = 1.364$
• $X_5 - \bar{X} = 2.45 - 1.364 = 1.086$
• $\sum(X_j - \bar{X})^2 = (1.85-1.364)^2 + (0.65-1.364)^2 + (0.63-1.364)^2 + (1.24-1.364)^2 + (2.45-1.364)^2$ $= 0.236 + 0.510 + 0.538 + 0.015 + 1.179 = 2.478$

$h_{55} = \frac{1}{5} + \frac{(1.086)^2}{2.478} = 0.2 + \frac{1.179}{2.478} = 0.2 + 0.476 = 0.676$

Step 3: Calculate MSE

Using full dataset:

• Residuals: $e_i = Y_i - \hat{Y}_i$
• $e_1 = 3.67 - (1.4110 + 0.1512\times1.85) = 3.67 - 1.69072 = 1.97928$
• $e_2 = 1.88 - (1.4110 + 0.1512\times0.65) = 1.88 - 1.50928 = 0.37072$
• $e_3 = 1.35 - (1.4110 + 0.1512\times0.63) = 1.35 - 1.506256 = -0.156256$
• $e_4 = 0.34 - (1.4110 + 0.1512\times1.24) = 0.34 - 1.598488 = -1.258488$
• $e_5 = 0.89 - (1.4110 + 0.1512\times2.45) = 0.89 - 1.78144 = -0.89144$

$SSE = \sum e_i^2 = 3.917 + 0.137 + 0.024 + 1.584 + 0.795 = 6.457$ $MSE = \frac{SSE}{n-p} = \frac{6.457}{5-2} = \frac{6.457}{3} = 2.1523$

Step 4: Calculate Cook's distance

$D_5 = \frac{(\hat{Y}_5 - \hat{Y}_{5(5)})^2}{p \cdot MSE} \cdot \frac{h_{55}}{(1 - h_{55})^2}$ $D_5 = \frac{(1.78144 - 3.665315)^2}{2 \times 2.1523} \cdot \frac{0.676}{(1 - 0.676)^2}$ $D_5 = \frac{(-1.883875)^2}{4.3046} \cdot \frac{0.676}{(0.324)^2}$ $D_5 = \frac{3.549}{4.3046} \cdot \frac{0.676}{0.105}$ $D_5 = 0.824 \times 6.438 = 5.305$

However, this calculation seems to give a different result than the options. Let me recalculate using the standard formula:

$D_i = \frac{e_i^2}{p \cdot MSE} \cdot \frac{h_{ii}}{(1 - h_{ii})^2}$

$D_5 = \frac{(-0.89144)^2}{2 \times 2.1523} \cdot \frac{0.676}{(1 - 0.676)^2}$ $D_5 = \frac{0.795}{4.3046} \cdot \frac{0.676}{0.105}$ $D_5 = 0.1847 \times 6.438 = 1.189$

This is still not matching the options. Let me use the alternative formula:

$D_i = \frac{(\hat{\beta} - \hat{\beta}_{(i)})'X'X(\hat{\beta} - \hat{\beta}_{(i)})}{p \cdot MSE}$

Given the regression coefficients:

• Full dataset: $\hat{\beta}_0 = 1.4110$, $\hat{\beta}_1 = 0.1512$
• Without observation 5: $\hat{\beta}_0 = 0.3169$, $\hat{\beta}_1 = 1.3667$

Difference: $\Delta\beta_0 = 1.4110 - 0.3169 = 1.0941$, $\Delta\beta_1 = 0.1512 - 1.3667 = -1.2155$

$X'X = \begin{bmatrix} n & \sum X_i \\ \sum X_i & \sum X_i^2 \end{bmatrix} = \begin{bmatrix} 5 & 6.82 \\ 6.82 & 11.7364 \end{bmatrix}$

$(\hat{\beta} - \hat{\beta}_{(i)})'X'X(\hat{\beta} - \hat{\beta}_{(i)}) = [1.0941 \quad -1.2155] \begin{bmatrix} 5 & 6.82 \\ 6.82 & 11.7364 \end{bmatrix} \begin{bmatrix} 1.0941 \\ -1.2155 \end{bmatrix}$

First compute: $X'X(\hat{\beta} - \hat{\beta}_{(i)}) = \begin{bmatrix} 5\times1.0941 + 6.82\times(-1.2155) \\ 6.82\times1.0941 + 11.7364\times(-1.2155) \end{bmatrix} = \begin{bmatrix} 5.4705 - 8.2897 \\ 7.4618 - 14.265 \end{bmatrix} = \begin{bmatrix} -2.8192 \\ -6.8032 \end{bmatrix}$

Then: $[1.0941 \quad -1.2155] \begin{bmatrix} -2.8192 \\ -6.8032 \end{bmatrix} = 1.0941\times(-2.8192) + (-1.2155)\times(-6.8032) = -3.084 + 8.268 = 5.184$

$D_5 = \frac{5.184}{2 \times 2.1523} = \frac{5.184}{4.3046} = 1.204$

This is still not matching exactly. Given the options and the calculations, the closest match is 1.6268, which is option B.