Deriving a One-Factor Pricing Relation from the SDF

# deriving asset pricing equation


# deriving asset pricing equation












I am reading Deep Learning in Asset Pricing by Chen, Pelger and Zhu (2019) and in the first section it says:

> The fundamental no-arbitrage assumption is equivalent to the existence of a strictly positive stochastic discount factor(SDF) $M_{t+1}$ such that for any return in excess of the risk-free rate $R^e_{t+1,i} = R_{t+1,i} - R^f_{t+1}$, it holds $$\mathbb{E}_t [M_{t+1} R^e_{t+1,i}] = 0 \Leftrightarrow \underbrace{\mathbb{E}_t[R_{t+1,i}^e] = \Big( -\frac{Cov_t(R_{t+1,i}^e, M_{t+1})}{Var_t(M_{t+1})}\Big)}_{\beta_{t,i}} \cdot \underbrace{\Big(\frac{Var_t(M_{t+1})}{\mathbb{E}_t [M_{t+1}]} \Big)}_{\lambda_t},$$ where $\beta_{t,i}$ is the exposure to systematic risk and $\lambda_t$ is the price fo risk. The SDF is an affine transformation of the tangency portfolio. Without loss of generality we consider the SDF formulation $$M_{t+1} = 1 - \sum_{i=1}^N w_{t,i} R_{t+1,i}^e = 1 - w_t^T R_{t+1}^e.$$ The fundamental pricing equation $\mathbb{E}_t[R_{t+1}^e M_{t+1}] = 0$ implies the SDF weights $$w_t =\mathbb{E}_t [R_{t+1}^e (R_{t+1}^e)^T]^{-1} \mathbb{E}_t [R_{t+1}^e],$$ which are the portfolio weights of the conditional mean variance efficient portfolio. We define the tangency portfolio as $F_{t+1} = w_t^TR_{t+1}^e$ and will refer to this traded factor as the SDF. The asset pricing equation can now be formulated as $$\mathbb{E}_t[R^e_{t+1,i}] = \frac{Cov_t(R_{t+1,i}^e, F_{t+1})}{Var_t({F_{t+1})}}\cdot \mathbb{E}_t(F_{t+1}) = \beta_{t,i} \mathbb{E}_t[F_{t+1}].$$ Hence no arbitrage implies a one factor model $$R_{t+1,i}^e = \beta_{t,i} F_{t+1} + \epsilon_{t+1,i}$$

I am trying to derive the last part: \begin{align*} 0 &= \mathbb{E}_t(M_{t+1}R_{t+1,i}^e) = \mathbb{E}[(1-F_{t+1})R_{t+1,i}^e] \\ &= \mathbb{E}[R^e_{t+1,i}] - \mathbb{E} [F_{t+1} R_{t+1,i}^e] \\ &= \mathbb{E}[R^e_{t+1,i}] - Cov(F_{t+1}, R_{t+1,i}^e) - \mathbb{E}[F_{t+1}] \mathbb{E}[R^e_{t+1,i}]\\ \mathbb{E}[R^e_{t+1,i}](1-\mathbb{E}(F_{t+1})) &= Cov(F_{t+1}, R^e_{t+1, i})\\ \mathbb{E}[R^e_{t+1,i}] &= \frac{Cov(F_{t+1}, R^e_{t+1,i})}{1-\mathbb{E}(F_{t+1})} \\ &= \frac{Cov(F_{t+1}, R^e_{t+1,i})}{\mathbb{E}[F_{t+1}] - \mathbb{E}(F_{t+1})^2} \cdot \mathbb{E}[F_{t+1}] \end{align*} This is where I'm stuck at. Any help would be much appreciated.

## Answer by ysimsek (score 2, accepted)

https://quant.stackexchange.com/a/81251

Suppose that F, tangency portfolio, is tradable. Then it must satisfy no arbitrage condition as well. That means:

$\mathbb{E}_t[F_{t+1}M_{t+1}] = 0$

and

$\mathbb{E}_t[F_{t+1}(1-F_{t+1})] = 0$

Using this, you can easily derive the formula in the paper.