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We know that a crucial assumption of employing OLS is that the independent variable and the error terms are uncorrelated. That is the "textbook" definition. I've seen in many (1, 2) online sources that if the independent variables and the error terms are correlated, violating strict exogeneity, this also means that the dependent variable $X$ is dependent on the independent variable, $Y$. What I don't understand is how correlation between $X$ and $\mu$ translates into a depdendence of $X$ on $Y$. For example,

Consider the model: $$ Y = \beta Y_{t-1} + \mu_t $$

For this model the independent variables are correlated with the terror term: $$ E[y_t \epsilon_t]=E[(\beta y_{t-1}+ \epsilon_t)\epsilon_t] \qquad (by \ \ \ y_t=\beta y_{t-1}+ \epsilon_t) $$ $$ \quad \qquad =\beta E[y_{t-1} \epsilon_t]+E[\epsilon_t^2] $$ $$ \quad \qquad =E[\epsilon_t^2] \qquad \qquad \qquad \quad (by \ \ \ E[y_{t-1} \epsilon_t]=0) $$ $$ \quad \qquad =\sigma^2 \qquad \qquad \qquad \quad \quad (by \quad \epsilon_t \sim N(0, \sigma^2)) $$

That is, for $ y_{t+1} $, $y_t$ is an independent variable but is correlated with an error term.

My question is: This doesn't mean in anyway that $y_t$ is dependent on $y_{t+1}$. Is that statement wrong or am I missing something?

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  • $\begingroup$ On the contrary, this is an explicit dependence between $y_t$ and $y_{t+1}.$ Are you perhaps confusing dependent (in its probability-related definition) with "causally related"? $\endgroup$
    – whuber ♦
    Commented Feb 2, 2021 at 16:27
  • $\begingroup$ But in that case, wouldn't the dependent and independent variable be dependent even if the errors are not correlated? (Since the correlation between X & Y is what lets us model Y as a function of X?) $\endgroup$
    – WorldGov
    Commented Feb 2, 2021 at 18:13
  • $\begingroup$ Dependence and correlation differ. To check independence, you need to apply the definition; but one way to demonstrate lack of independence is to show the correlation is nonzero. Generally, when $Y$ is any non-constant random variable, and $\mu$ is an independent random variable, and $Z=f(Y,\mu)$ is some function of those two variables, you would not expect $Z$ to be independent of either $Y$ or $\mu.$ In the case $f(Y,\mu)=\beta Y+\mu$ with $\beta\ne 0$ it is straightforward to show non-independence because the covariance of $Y$ and $Z$ is nonzero. $\endgroup$
    – whuber ♦
    Commented Feb 2, 2021 at 18:19

1 Answer 1

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Hi: When dealing with lagged dependent variables in time series models, things can become clearer if you write the model out this way: ( at first you used $u$ and then later you used $\epsilon$. I will use $\epsilon$ ).

First, use the lag operator:

$ Y_t(1- \beta L) = \epsilon_t$

Then, divide both sides by $(1-\beta L)$:

$ Y_t = \sum_{i=0}^{\infty} \beta^{i} \epsilon_i$

So, $Y_t$ is dependent on the current $\epsilon$ and all past $\epsilon$ but $\epsilon$ is the only thing on the RHS so the OLS orthogonality condition doesn't really apply because there aren't any regressors for $\epsilon$ to be correlated with. Does that make sense ? $\epsilon$ is the only regressor in the model really.

Note: This AR(1) model still cannot be estimated by OLS because the $\beta$ estimate will be biased but its not due to the condition you mentioned. The biased-ness arises in a complicated way and I can't remember how at the moment. There was another thread where someone explained how it arises but I forget the name of the thread. maybe search for "Why is AR(1) OLS estimate biased ?".

UPDATE: Below is not what I was looking for but still quite useful-interesting. It explains the biasedness using your argument but the dependence is on PAST $\epsilon$ rather than the current $\epsilon$. That dependence also violates the OLS condition. So, if you want to think of the reason as stemming from the argument below, that might be more suitable for you since it's closer to yours. Still, there's another argument for how the biasedness arises that focuses on the formula for $\hat{\beta}$ in OLS but I failed to find it. Anyway, sorry for confusion and maybe below is the best way to think about it since I failed to find the other argument !!!!!

What's wrong if I fit the auto-regression with OLS?

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  • $\begingroup$ Another way to think of above is the following. In an AR(1), $y_t$ is meant to be dependent on $y_{t-1}$. But it is shown above that $y_{t-1}$ is a function of it's current and past error terms. But $y_t$ is dependent on $y_{t-1}$ the latter of which is a function of it's own current and past error terms. This makes the $y_t$ dependent on past error terms and this violates the OLS assumption. $\endgroup$
    – mlofton
    Commented Jul 8 at 3:50

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