2
$\begingroup$

I've been reading Cochrane's 2006 paper "The Dog that did not bark: A Defense of Return Predictability", but i am still struggling to understand what the dog was, and why it wasn't barking?

If anyone could shed some brief intuition that would be appreciated. Perhaps my knowledge of existing literature, in order to identify the 'dog' is lacking.

$\endgroup$
  • $\begingroup$ Google "the Adventure of Silver Blaze" for the Sherlock Holmes mystery. A prize racehorse goes missing. The absence of the bark suggests the thief was known... "Is there any point to which you would wish to draw my attention?' 'To the curious incident of the dog in the night-time.' 'The dog did nothing in the night-time.' 'That was the curious incident,' remarked Sherlock Holmes.” $\endgroup$ – demully Jan 8 at 2:43
6
$\begingroup$

I'm not sure how deep of a question you are asking. The dog that did not bark is from a Sherlock Holmes murder mystery. The dog at the house did not bark at the intruder, so Holmes believed the dog knew the intruder. Therefore, the lack of evidence like barking, was itself the evidence. In the Cochrane paper, the introduction mentions that the lack of evidence of returns predictability is itself evidence. In this case, it is evidence of dividend growth being predictable. Clearly the author is just trying to come up with a literary twist to make his paper more interesting so I would not take it very seriously. I can't defend Cochrane's actual use of this evidence because, although I only skimmed it, I did not find the paper very convincing.

$\endgroup$
2
$\begingroup$

As it was pointed above the phrase is taken from Sherlock Holme's novel. It describes the case when the dog should have bark, but didn't. Now if we come to the Cochrane paper. He introduces the system of equations ($r_{t+1}$ - returns, $\Delta d_{t+1}$ - dividend growth and $d_t - p_t$ - dividend-price ratio): $$ r_{t+1} = a_r + \beta_r(d_t - p_t) + \epsilon^r_{t+1}, \\ \Delta d_{t+1} = a_d + \beta_d(d_t - p_t) + \epsilon^d_{t+1}, \\ d_{t+1} - p_{t+1} = a_{dp} + \phi(d_t - p_t) + \epsilon^{dp}_{t+1}. $$

He argues that if you just test $H_0: \beta_r = 0$, basically it tests the predictability of returns and you won't find significance. However, if you jointly test $H_0: \beta_r = 0\land \beta_d = \rho\phi - 1$ (I'll explain later from where it comes from). This new null gives your more "power" to reject the null (although Cochrane is wrong in the definition of power since $power = \mathbb{P}[ reject\text{ } H_0 | H_A]$, but now it's not that important). To construct this null he uses Campbell-Shiller 1988 log-linearization for the returns to obtain:

$$ r_{t+1} \approx \kappa + \rho(p_{t+1} - d_{t+1}) + \Delta d_{t+1} - (p_t - d_t), $$ where $\kappa$ - cosntant and $\rho$ - point of log-linearization. From this equation and previous system we can form the following identities:

$$ \beta_r = 1 + \beta_d - \rho\phi, \\ a_r = \kappa + a_d - \rho a_{dp}, \\ \epsilon^r_{t+1} = \epsilon^d_{t+1} - \rho\epsilon^{dp}_{t+1}. $$

And now comes the most important part. In order to have the $\beta_r = 0$ we have to have $\beta_d = \rho\phi - 1 \approx -0.1$, but this is supported much less by the data and we adress the absence of this coefficient in the data. And here $\hat{\beta}_d = 0$ (estimated in the data) represents the dog that didn't bark!

$\endgroup$

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.