2
$\begingroup$

Im working on some school questions regarding regression models. Here I have to show that:

$$ Y_i = \tilde{\alpha} + \beta X_i + \tilde{\epsilon_i}$$ where $$ \tilde{\epsilon_i} \overset{}{\sim} N(\mu_\tilde{\epsilon},\sigma^2_\tilde{\epsilon})$$ and $$\mu_\tilde{\epsilon} \neq 0 $$

Is equivalent to:

$$ Y_i = \alpha + \beta X_i + \epsilon_i$$ where $$ \epsilon_i \overset{}{\sim} N(0,\sigma^2_\epsilon)$$

My problem is that I cant image these equations being equal if the error is greater than 0 which violates the full ideal conditions. Appreciate any hints!

$\endgroup$
  • 1
    $\begingroup$ $\tilde \alpha + μ_{\tilde \epsilon} = \alpha$ and $σ_{\tilde \epsilon} ^2 = σ_{\epsilon}^2$ then two models are the same. $\endgroup$ – user158565 May 27 '17 at 20:26
  • $\begingroup$ Hi @a_statistician, thank you for the answer. However, I do not really follow. Could you please explain particularly the first part? $\endgroup$ – Googme May 28 '17 at 7:05
  • $\begingroup$ On the face of it, Googme, you appear to be asking why $(\tilde\alpha+\mu_{\tilde\epsilon})+0$ (the intercept in the second model) is equal to $\tilde\alpha+\mu_{\tilde\epsilon}$ (the intercept in the first model). Since that is so obvious, could you be a little more specific about what you're not following? $\endgroup$ – whuber May 28 '17 at 15:04
2
$\begingroup$

Let $ Y_i = \tilde{\alpha} + \beta X_i + \tilde{\epsilon_i}$ where $ \tilde{\epsilon_i} \overset{}{\sim} N(\mu_\tilde{\epsilon},\sigma^2_\tilde{\epsilon})$ and $\mu_\tilde{\epsilon} \neq 0 $ be model 1, and

$ Y_i = \alpha + \beta X_i + \epsilon_i$ where $ \epsilon_i \overset{}{\sim} N(0,\sigma^2_\epsilon)$ be model 2.

From model 1, we have $\operatorname{E}(Y_i) = \tilde{\alpha} + \beta X_i +\mu_\tilde{\epsilon}$

From model 2. we have $\operatorname{E}(Y_i) = {\alpha} + \beta X_i$

Obviously, $\tilde{\alpha} + \mu_\tilde{\epsilon} = {\alpha}$

From model 1, we have $\operatorname{Var}(Y_i|X_i) = \operatorname{Var}(\tilde{\epsilon}) = \sigma^2_\tilde{\epsilon}$

From model 2, we have $\operatorname{Var}(Y_i|X_i) = \operatorname{Var}({\epsilon}) = \sigma^2_{\epsilon}$

So, $\sigma^2_\tilde{\epsilon} =\sigma^2_{\epsilon}$.

Therefore, the two models are the same. In fact, $\tilde{\alpha}$ and $ \mu_\tilde{\epsilon}$ in model 1 are two intercepts and have infinite number of solutions.

$\endgroup$
  • 1
    $\begingroup$ There are some subtleties that probably should be addressed. They can be appreciated by thinking about what additional justification would be needed if this were a generalized linear model. Although everything in your argument would apply to a GLM, its conclusion might be incorrect. That shows that the argument is incomplete. $\endgroup$ – whuber May 28 '17 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.