following Cochrane (2005) and other insights, we know that under Arbitrage Pricing Theory (Ross, 1976), if investors believe returns follow a linear multifactor structure of the form


where $x^i$ is a the asset $i$ return, $r^f$ is the return on a risk-free asset, $\beta_{ij}$ is the factor loading of asset $i$ with respect to factor $j$ and $f_j$ is the $j$-th factor; then we have that the pricing kernel or stochastic discount factor $m$ which prices assets by defintion according to $p(x^i)=E[mx^i]$, also satisfies


for some arbitrary $a$ and $b$ i.e. it satisfies a linear multifactor structure. Is anyone able to derive the second equation starting from the first one using the classical assumptions in the APT or at least give me some hints about it?

  • $\begingroup$ I think you have your APT assumptions slightly wrong. We assume returns follow the top model; we also assume that we can diversify away the effect of $\epsilon_i$; and, we assume markets do not allow the persistence of arbitrage. Limits to arbitrage is trouble for the last assumption and the possibility of misspecification is a threat to the second assumption. Finally, if those assumptions do hold, we assume excess returns $X^i=x^i-r^f$ hold to the (hand-wavey) $X^i \approx m = \sum_{j=1}^M b_j f_j$. Equality with $m$ is not assumed, $m$ has no $a$, and that there's troubling "$\approx$". $\endgroup$ – kurtosis Jul 31 '20 at 19:25
  • $\begingroup$ See chapter 6 in Cochrane's AP book. $\endgroup$ – fesman Aug 1 '20 at 8:18
  • $\begingroup$ @kurtosis Why would the SDF be approximately equal to excess return? $\endgroup$ – fesman Aug 1 '20 at 8:19
  • $\begingroup$ The SDF is for the excess return; risk factors do not describe the risk-free part of returns. You can get this from combining equations (1)-(3) in Ross (1976). The approximation is there too, in all its undefined sloppy glory. Kind of a dodge, IMHO. $\endgroup$ – kurtosis Aug 1 '20 at 10:18
  • $\begingroup$ @kurtosis See the proof in Cochrane's book. The relationship between excess return and SDF is $E(mR^e)=0$. You claim $m\approx R^e$. This doesn't make sense to me. It would mean roughly $E(m^2)=E(R^{e2})\approx0$. $\endgroup$ – fesman Aug 1 '20 at 10:44

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