Question

Consider the regression model $y_{i}=\beta_{0}+i\beta_{1}+\epsilon_{i} (i=1,2,...,n>2)$ where $\beta_{0}$ and $\beta_{1}$ are unknown parameters and $\epsilon_{i}$ are random errors. Let $y_{i}$ be the observed value of $Y_{i}(i=1,2,...,n)$. Using the method of ordinary least squares, the estimate of $\beta_{1}$ is

• A. $\frac{1}{n^{2}-1}[\frac{12}{n}\sum_{i=1}^{n}iy_{i}-6(n+1)\overline{y}]$
• B. $\frac{1}{(n^{2}-1)n}[12\sum_{i=1}^{n}iy_{i}-6(n+1)\overline{y}]$
• C. $\frac{1}{n(n^{2}-1)}[12\sum_{i=1}^{n}iy_{i}-6n\overline{y}]$
• D. $\frac{1}{n(n^{2}-1)}[12\sum_{i=1}^{n}iy_{i}-6\overline{y}]$

Hint:

For regression where $x$ is just an index $1 \dots n$, the variance of $x$ is always $\frac{n^2-1}{12}$. Using this standard result makes calculating least-squares parameters for time-series or indexed data much faster.

Answer 1

The Correct Option is A

In a simple linear regression $y = \beta_0 + \beta_1 x + \epsilon$, the Ordinary Least Squares (OLS) estimator for the slope is $\hat{\beta}_1 = \frac{\sum (x_i - \overline{x})(y_i - \overline{y})}{\sum (x_i - \overline{x})^2}$.

Step 1: \color{redIdentify the Independent Variable
Here, the independent variable $x_i$ is simply the index $i$.
So, $x_i \in \{1, 2, \dots, n\}$.
The mean of $x$ is $\overline{x} = \frac{\sum_{i=1}^n i}{n} = \frac{n(n+1)/2}{n} = \frac{n+1}{2}$.

Step 2: \color{redCalculate the Denominator (Sum of Squares of x)
$\sum (x_i - \overline{x})^2 = \sum i^2 - n\overline{x}^2$.
$\sum i^2 = \frac{n(n+1)(2n+1)}{6}$.
$n\overline{x}^2 = n \left(\frac{n+1}{2}\right)^2 = \frac{n(n+1)^2}{4}$.
$\sum (x_i - \overline{x})^2 = \frac{n(n+1)}{12} [2(2n+1) - 3(n+1)] = \frac{n(n+1)(n-1)}{12} = \frac{n(n^2-1)}{12}$.

Step 3: \color{redCalculate the Numerator
$\sum (i - \overline{x})(y_i - \overline{y}) = \sum iy_i - n\overline{x}\overline{y}$.
Substitute $\overline{x} = \frac{n+1}{2}$:
Numerator $= \sum iy_i - n \frac{n+1}{2} \overline{y} = \sum iy_i - \frac{n(n+1)}{2} \overline{y}$.

Step 4: \color{redCombine to find \hat{\beta}_1
$\hat{\beta}_1 = \frac{\sum iy_i - \frac{n(n+1)}{2} \overline{y}}{\frac{n(n^2-1)}{12}} = \frac{12}{n(n^2-1)} \left[ \sum iy_i - \frac{n(n+1)}{2} \overline{y} \right]$.
Distributing $\frac{12}{n}$ inside the bracket:
$\hat{\beta}_1 = \frac{1}{n^2-1} \left[ \frac{12}{n} \sum iy_i - \frac{12}{n} \cdot \frac{n(n+1)}{2} \overline{y} \right]$.
$\hat{\beta}_1 = \frac{1}{n^2-1} \left[ \frac{12}{n} \sum iy_i - 6(n+1) \overline{y} \right]$. [cite: 4926, 4941]
This matches Option (1).