4
$\begingroup$

Hansen and Jagannathan distance, or HJ-distance for time-series regression of excess test assets return on excess factor return reads:

$HJ = \sqrt{\alpha'(E[RR']^{-1})\alpha}$

However, I am little bit confused whether the alpha here is the intercept term or the residual? Second, how do I calculate the middle term because it is involved with the expectation ?


More detail: For the equation above, you can refer to the document (equation 4, page 18) https://goo.gl/ZBh1Tu

Besides, you can read equation 12 on page 7 of this paper: https://goo.gl/VnHZTQ

@phdstudent: When you mean the covariance matrix $\Sigma = E[RR']$. Is this correct for the explicit form:

$E[RR'] = \frac{(R-\bar{R})*(R-\bar{R})'}{T-1}$ where $\bar{R}$ is average return

Btw, should $R$ in this case is return or excess return ?

$\endgroup$
4
$\begingroup$

It would be easier to answer if you tell us where that equation came from (there are many ways of deriving the HJ distance) - in any case the numerator of your equation should be the expected return on the efficient portfolio and the denominator the expected variance/covariance.

Let me give you a the same equation using simpler notation (and derive it!). From the law of one price:

\begin{equation} 1 = E [R_{i,t+1} m^\star_{t+1}] \end{equation}

Therefore:

\begin{equation} 1 = E(R_{i,t+1}) E(m^\star_{t+1}) + Corr(R_{i,t+1}, m^\star_{t+1}) Std(R_{i,t+1}) Std(m^\star_{t+1}) \end{equation}

Rewrite the equation above using $R_{f,t+1} = 1/E(m^\star_{t+1})$ to get:

\begin{equation} \frac{E(R_{i,t+1}) - R_{f,t+1}}{Std(R_{i,t+1}) } \leq \frac{Std(m^\star_{t+1})}{E(m^\star_{t+1})} \end{equation}

The left hand side equation is the equivalent to your equation (the maximum attainable sharpe ratio). And the equation on the right gives you the bound so: The max Sharpe ratio in the economy is then bounded by the minimum variance SDF volatility over mean!

How do we use these?

  • Take $N$ assets. Compute excess returns.
  • Estimate variance covariance matrix of returns $\Sigma = E[R R']$ and average payoffs $E(R_{t+1})$. Usually the first one we estimate by taking a large sample and computing covariance matrix and the latter just by averaging returns.
  • Plot the above locus and compare with your candidate SDF;

The locus should deliver something like this:

enter image description here

Edit: After some clarifications above.

I now see where your derivation comes from (see equation 10 from Hodrick and Zhang), which is not about the bound itself but the distance. Basically your equation comes form solving the following problem:

\begin{equation} \min_m E[(y_{t+1}-m_{t+1})^2] + 2\lambda (E[m_{t+1}R_{t+1}]-1) \end{equation}

where the first term is the JG distance and the second term the constraint.

Take f.o.c. with respect to $m$ for every asset and you get: \begin{equation} E[(y_{t+1}-m^\star_{t+1})^2] = [1-E(R_{t+1}y_{t+1})]'[E(R_{t+1}R'_{t+1}][1-E(R_{t+1}y_{t+1})] \end{equation}

So indeed your $\alpha$ are the time-series intercepts of the time-series regression and the denominator is just a matrix with expected values of products of returns for all assets (just like a variance-covariance matrix when means are close to zero).

$\endgroup$
  • $\begingroup$ Can you say a bit more about the diagram. What does the shaded gray area represent, what is the equation of the curve that bounds that area, what are the black dots $\endgroup$ – Alex C May 2 '18 at 15:51
  • 1
    $\begingroup$ The shaded grey area are admissible portfolios. So the lower bound of that area gives the maximum return for a given volatility (you can ignore the part with negative returns). So it is the LHS of my third equation. The black boxes are the RHS of my third equation for the case of power utility as you vary $\gamma$. Only for high values of $\gamma$ you enter the admissible region. $\endgroup$ – phdstudent May 2 '18 at 15:58
  • $\begingroup$ Could you please explain for me a little more, what is the dimension of the denominator ($E[RR']$), is it TxT matrix if R is TxN matrix ? And if the means are not close to zero, $E[RR'] = \Sigma - E[R](E[R])'$ is the way to calculate $E[RR']$ ? $\endgroup$ – BlueFx May 2 '18 at 17:47

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.