# How to proof equivalence of two regression models with different errors?

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!

• $\tilde \alpha + μ_{\tilde \epsilon} = \alpha$ and $σ_{\tilde \epsilon} ^2 = σ_{\epsilon}^2$ then two models are the same. – user158565 May 27 '17 at 20:26
• Hi @a_statistician, thank you for the answer. However, I do not really follow. Could you please explain particularly the first part? – Googme May 28 '17 at 7:05
• 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? – whuber May 28 '17 at 15:04

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.

• 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. – whuber May 28 '17 at 15:19