# Constant conditional variance does not imply constant conditional mean?

by Syd Amerikaner   Last Updated September 22, 2019 21:19 PM

Let $$y = X\beta + u$$ be a regression model. If we assume $$\mathbb V[u|X] = \sigma^2$$ then does this imply $$\mathbb E[u|X] = c$$? Clearly, $$\mathbb V[u|X] = \mathbb E[uu'|X] - \mathbb E[u|X]\mathbb E[u|X]'$$, so $$\mathbb E[uu'|X] = \sigma^2 + \mathbb E[u|X]\mathbb E[u|X]'$$. Since both, $$\mathbb E[uu'|X]$$ and $$\mathbb E[u|X]\mathbb E[u|X]'$$ can vary, $$\mathbb E[u|X]$$ need not be constant. But what would be a counter example?

If we assume iid observations and only a single regressor, it suffices to regard a single $$u_i$$. So, $$E[u_i^2|x_i] = \sigma^2 + \mathbb E[u_i|x_i]^2$$. If we take the derivative wrt to $$x_i$$, we find $$D\mathbb E[u_i^2|x_i] = 2D\mathbb E[u_i|x_i]$$ so I thought about assuming $$\mathbb E[u_i|x_i] = 0.5x_i$$ and $$\mathbb E[u^2_i|x_i] = x_i$$. Is this reasoning correct or do I miss a point?

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#### Answers 1

$$\mathbb E[u|X]$$ can be whatever you want. The variance assumption alone does not imply it must be anything. Most commonly, $$\mathbb E[u|X]$$ is assumed to be $$0$$. If this was not done, and instead you assumed it was equal to some parameter to be estimated, like $$c$$, then all of the parameters would probably not be identifiable, and they would be less interpretable.

If you assume $$\mathbb E[u_i|x_i] = 0.5x_i$$ then you can write your model as $$y_i = \beta_0 + \beta_1 x_i + (.5x_i + \sigma z_i)$$ where $$z_i$$ is a standard normal variate. The parameters $$(\beta_0, \beta_1, .5)$$ yield the same likelihood as $$(\beta_0, \beta_1 + .5, 0)$$. You are not estimating $$.5$$, so technically your model is still identifiable; but I don't see any advantage to be gained from not following the convention of assuming your noise is mean zero.

Taylor
September 22, 2019 20:49 PM