Winston Todd

2022-10-15

Given this regression model: ${y}_{i}={\beta }_{0}+{\beta }_{1}{x}_{i}+{E}_{i}$.
All the assumptions are valid except that now: ${E}_{i}\sim N\left(0,{x}_{i}{\sigma }^{2}\right)$
Find Maximum likelihood parameters for ${\beta }_{0}$, ${\beta }_{1}$

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espava8b

Expert

Let us define the following quantities:
${\overline{y}}_{i}={x}_{i}^{-1/2}{y}_{i},\phantom{\rule{1em}{0ex}}{\overline{x}}_{i-}={x}_{i}^{-1/2},\phantom{\rule{1em}{0ex}}{\overline{x}}_{i+}={x}_{i}^{+1/2},\phantom{\rule{1em}{0ex}}{\overline{E}}_{i}={x}_{i}^{-1/2}{E}_{i}$
Then the equation becomes
${\overline{y}}_{i}={\beta }_{0}{\overline{x}}_{i-}+{\beta }_{1}{\overline{x}}_{i+}+{\overline{E}}_{i},\phantom{\rule{1em}{0ex}}{\overline{E}}_{i}\in \mathcal{N}\left(0,{\sigma }^{2}\right)$
What we have now is a standard, unweighted linear regression in two variables.
To solve this we define two matrices
$X=\left[\begin{array}{cc}{\overline{x}}_{1-}& {\overline{x}}_{1+}\\ ⋮& ⋮\\ {\overline{x}}_{n-}& {\overline{x}}_{n+}\end{array}\right],\phantom{\rule{1em}{0ex}}Y=\left[\begin{array}{c}{\overline{y}}_{1}\\ ⋮\\ {\overline{y}}_{n}\end{array}\right]$
What happens if we cannot assume ${x}_{i}>0$? Well, in that case the original relation reduces to ${y}_{i}={\beta }_{0}$ and ${E}_{i}=0$. So that means that ${\beta }_{0}$ is forced to be equal to ${y}_{i}$ in that case. If there are multiple values of ${x}_{i}>0$, then we have two scenarios:
- All of the corresponding values of ${y}_{i}$ are identical, in which case ${\beta }_{0}={y}_{i}$.
- The corresponding values of ${y}_{i}$ are not identical, in which case the model is infeasible and cannot be solved.
If the model is feasible, then we fix ${\beta }_{0}$ and eliminate those $\left({x}_{i},{y}_{i}\right)$ pairs from the model, leaving us with a single-variable regression.
${\overline{z}}_{i}={\beta }_{1}{\overline{x}}_{i+}+{\overline{E}}_{i},\phantom{\rule{1em}{0ex}}{\overline{z}}_{i}\triangleq {\overline{y}}_{i}-{\beta }_{0}{\overline{x}}_{i-}\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}{\beta }_{1}=\frac{\left(\sum _{i}{y}_{i}\right)-n{\beta }_{0}}{\sum _{i}{x}_{i}}.$

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