Winston Todd

Answered

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: ${E}_{i}\sim N(0,{x}_{i}{\sigma}^{2})$

Find Maximum likelihood parameters for ${\beta}_{0}$, ${\beta}_{1}$

Answer & Explanation

espava8b

Expert

2022-10-16Added 12 answers

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

$${\overline{y}}_{i}={\beta}_{0}{\overline{x}}_{i-}+{\beta}_{1}{\overline{x}}_{i+}+{\overline{E}}_{i},\phantom{\rule{1em}{0ex}}{\overline{E}}_{i}\in \mathcal{N}(0,{\sigma}^{2})$$

What we have now is a standard, unweighted linear regression in two variables.

To solve this we define two matrices

$$X=\left[\begin{array}{cc}{\overline{x}}_{1-}& {\overline{x}}_{1+}\\ \vdots & \vdots \\ {\overline{x}}_{n-}& {\overline{x}}_{n+}\end{array}\right],\phantom{\rule{1em}{0ex}}Y=\left[\begin{array}{c}{\overline{y}}_{1}\\ \vdots \\ {\overline{y}}_{n}\end{array}\right]$$

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 $({x}_{i},{y}_{i})$ pairs from the model, leaving us with a single-variable regression.

$${\overline{z}}_{i}={\beta}_{1}{\overline{x}}_{i+}+{\overline{E}}_{i},\phantom{\rule{1em}{0ex}}{\overline{z}}_{i}\triangleq {\overline{y}}_{i}-{\beta}_{0}{\overline{x}}_{i-}\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}{\beta}_{1}=\frac{(\sum _{i}{y}_{i})-n{\beta}_{0}}{\sum _{i}{x}_{i}}.$$

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