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

Can anyone explain it in simple terms?

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

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.

  • $\begingroup$ Better than the one I linked, upvoted. $\endgroup$ – Bob Jansen Mar 15 '18 at 16:19

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.