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What is the Stambaugh bias? Why is it important for predictability regressions?

Can anyone explain it in simple terms?

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  • $\begingroup$ Thanks, I was wrong... Will be more careful in the future. $\endgroup$
    – Alex C
    Mar 15, 2018 at 15:58
  • $\begingroup$ No worries, you're right way more often than wrong ;) $\endgroup$
    – Bob Jansen
    Mar 15, 2018 at 16:18

1 Answer 1

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The bias comes from the paper Stambaugh (1999) and has nothing to do with small sample bias. It has to do with point (1) below.

The argument goes as follows:

  1. Typical lagged explanatory variables for stock-return regressions are correlated with contemporaneous stock returns
  2. This contemporaneous correlation biases forecasting regressions

First review OLS bias of AR(1):

\begin{equation} x_t = \alpha + \rho x_{t-1} + v_t \end{equation}

\begin{equation} \hat{\rho} = \frac{\hat{Cov} (x_t, x_{t-1})}{\hat{Var} (x_{t-1})} \end{equation}

\begin{equation} \hat{\rho} = \rho + \frac{\hat{Cov} (v_t, x_{t-1})}{\hat{Var} (x_{t-1})} \end{equation}

Stambaugh shows that there is no analytical formula but as an approximation the bias is given by:

\begin{equation} E_t(\hat{\rho}) - \rho = - \frac{1+3\rho}{T} \end{equation}

Now assume that the predictor of stock returns follows the process $x_t$ above. If returns $r_t$ follows:

\begin{equation} r_t = \alpha + \beta x_{t-1} + u_t \end{equation}

Then you can see the bias of $\beta$:

\begin{equation} E(\hat{\beta}) - \beta = \frac{Cov(u_t, v_t)}{Var(v_t)}[E{(\hat{\rho})}-\rho] \end{equation}

Depending on the sign of $Cov(u_t, v_t)$ you get the sign of the bias.

I strongly recomment reading the reference above.

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  • $\begingroup$ Better than the one I linked, upvoted. $\endgroup$
    – Bob Jansen
    Mar 15, 2018 at 16:19

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