Dynamic risk, accounting-based valuation and firm fundamentals

Abstract

This study extends the accounting-based valuation framework of Ohlson (Contemp Acc Res 11(2):661–687, 1995) and Feltham and Ohlson (Acc Rev 74(2):165–183, 1999) to incorporate dynamic expectations about the level of systematic risk in the economy. Our model explains recent empirical findings documenting a strong negative association between changes in economy-wide risk and future stock returns. Importantly, the model also generates costs of capital that are solely a linear function of accounting variables and other firm fundamentals, including the book-to-market ratio, the earnings-to-price ratio, the forward earnings-to-price ratio, size and the dividend yield. This result provides a theoretical rationale for the inclusion of these popular variables in cost of capital (expected return) computations by the accounting and finance literatures and obviates the need to estimate costs of capital from unobservable (future) covariances. The model also generates an accounting return decomposition in the spirit of Vuolteenaho (J Finance 57(1):233–264, 2002). Empirically, we find that costs of capital generated by our model are significantly associated with future returns both in and out of sample in contrast to standard benchmark models. We further obtain significantly lower valuation errors in out-of-sample tests than traditional models that ignore dynamic risk expectations.

Appendices

Appendix 1

1.1 Derivation of the discount factor

The stochastic discount factor m _t,t+1 is defined as the marginal rate substitution of aggregate consumption between two periods (see Cochrane 2001):

$$ m_{t,t+1} =\frac{M_{t+1}}{M_{t}}, $$

(19)

$$ M_{t} =\frac{\partial U(C_{t})}{\partial C_{t}}, $$

(20)

where $U(\cdot)$ is a representative agent’s utility function, and C _t is aggregate consumption. Assuming a standard power utility economy with a risk aversion parameter α, then

$$ U(C_{t}) =\frac{C_{t}^{1-\alpha }}{1-\alpha } $$

(21)

$$ \Rightarrow M_{t} =C_{t}^{-\alpha }. $$

(22)

If consumption grows at the rate g _t+1, then next period consumption is $C_{t+1}=C_{t}\exp (g_{t+1})$ so that:

$$ m_{t,t+1} =\exp (-\alpha g_{t+1}) $$

(23)

$$ \approx a-bg_{t+1}, $$

(24)

where a and b > 0 are constants. This result accords both with intuition and general linear asset pricing models. Firms whose (abnormal) earnings are positively correlated with aggregate consumption are discounted more because of risk aversion—note the negative sign in Eq. (24)—consistent with the implications of the consumption CAPM.

The expectation of m _t,t+1 must be the gross risk-free discount rate, E _t [m _t,t+1] = R ⁻¹ _f , which implies in turn that

$$ m_{t,t+1} =E_{t}[m_{t,t+1}]+(m_{t,t+1}-E_{t}[m_{t,t+1}]) $$

(25)

$$ =R_{f}^{-1}(1-bR_{f}(g_{t+1}-E_{t}[g_{t+1}]). $$

(26)

Assuming that growth in aggregate consumption takes the dynamics g _t+1 = g + σ _g,t e _t+1 and that σ _g,t+1 = σ _g,t  + ξ _t+1 (where e _t+1 and ξ _t+1 are mean zero random variables as in the text), then

$$ m_{t,t+1}=R_{f}^{-1}(1-bR_{f}\sigma_{g,t}e_{t+1}). $$

(27)

Calling bR _f σ _g,t ≡ σ _m,t yields the same form of the dynamic discount factor used in this study:

$$ m_{t,t+1}=R_{f}^{-1}(1-\sigma_{m,t}e_{t+1}). $$

This discount factor readily relates to the CAPM. Assume that consumption can be proxied by the wealth in the market portfolio, then Eq. (24) can be replaced with the discount factor that generates the CAPM:

$$ m_{t,t+1}=a-br_{m,t+1}, $$

(28)

where r _m,t+1 is the return on the market portfolio.

Appendix 2

2.1 Proof of Proposition 1

The underlying dynamics are given by:

$$ \begin{aligned} x_{t+1}^{a} &=\omega x_{t}^{a}+(1-\omega) x_{L}^{a}+v_{t}+\epsilon_{t+1},\\ v_{t+1} &=\gamma v_{t}+u_{t+1}, \\ m_{t,t+1} &=m_{t+1}=R_{f}^{-1}(1-\sigma_{m,t}e_{t+1}), \\ m_{t,t+2} &=m_{t+1}m_{t+2}=R_{f}^{-1}(1-\sigma_{m,t}e_{t+1})R_{f}^{-1}(1-\sigma_{m,t+1}e_{t+2}) \\ &=R_{f}^{-2}(1-\sigma_{m,t}e_{t+1})(1-\sigma_{m,t+1}e_{t+2}), \\ \sigma_{m,t+1} &=\sigma_{m,t}+\xi_{t+1}. \end{aligned} $$

All error terms are assumed to be i.i.d and are orthogonal to one another except for $\epsilon_{t+1}$ and e _t+1, which have a correlation coefficient of ρ. The expectation operator $E_{t}[\cdot ]$ represents a conditional expectation given information at time t.

From Eq. (1) in the text, the price of the equity is given by:

$$ \begin{aligned} S_{t} &=B_{t}+\sum_{i=1}^{\infty }R_{f,t,t+i}^{-1}E_{t}[x_{t+i}^{a}]+\sum_{i=1}^{\infty }cov_{t}(m_{t,t+i},x_{t+i}^{a})\\ &=B_{t}+\sum_{i=1}^{\infty }E_{t}[m_{t,t+i}x_{t+i}^{a}]. \end{aligned} $$

Let z _t+1 = x ^a _t+1 m _t,t+1, then

$$ \begin{aligned} E_{t}[z_{t+1}] &=E_{t}[x_{t+1}^{a}m_{t+1}] \\ &=E_{t}[m_{t+1}]E_{t}[x_{t+1}^{a}]+E_{t}[(m_{t+1}-E_{t}[m_{t+1}])(x_{t+1}^{a}-E_{t}[x_{t+1}^{a}])\\ &=\frac{1}{R_{f}}(\omega E_{t}[x_{t}^{a}]+E_{t}[v_{t}]+(1-\omega) x_{L}^{a})-\frac{1}{R_{f}}E_{t}(\sigma_{m,t}e_{t+1}\epsilon_{t+1})\\ &=\frac{1}{R_{f}}(\omega E_{t}[x_{t}^{a}]+E_{t}[v_{t}]+(1-\omega) x_{L}^{a}-\sigma_{m,t}\rho \sigma_{x}). \end{aligned} $$

By the law of iterated expectations:

$$ \begin{aligned} E_{t}[z_{t+2}] &=E_{t}[E_{t+1}[z_{t+2} ] ] \\ &=E_{t}[\frac{1}{R_{f}^{2}}(1-E_{t+1}[\sigma_{m,t}e_{t+1}])(\omega E_{t+1}[x_{t+1}^{a}] \\ &\quad +E_{t+1}[v_{t+1}]+(1-\omega) x_{L}^{a}-E_{t+1}[\sigma_{m,t+1}]\rho \sigma_{x})] \\ &=E_{t}[\frac{1}{R_{f}}m_{t+1}(\omega x_{t+1}^{a}+v_{t+1}+(1-\omega) x_{L}^{a}-\sigma_{m,t+1}\rho \sigma_{x})] \\ &=\frac{1}{R_{f}}E_{t}[\omega z_{t+1}+m_{t+1}(v_{t+1}+(1-\omega) x_{L}^{a}-\sigma_{m,t+1}\rho \sigma_{x})] \\ &=\frac{1}{R_{f}}\omega E_{t}[z_{t+1}]+\frac{1}{R_{f}^{2}} (E_{t}[v_{t+1}]+(1-\omega) x_{L}^{a}-\sigma_{m,t}\rho \sigma_{x})\\ &=\frac{1}{R_{f}}\omega E_{t}[z_{t+1}]+\frac{1}{R_{f}^{2}}(\gamma v_{t}+(1-\omega) x_{L}^{a}-\sigma_{m,t}\rho \sigma_{x}) \\ &=\frac{1}{R_{f}^{2}}(\omega (\omega E_{t}[x_{t}^{a}]+E_{t}[v_{t}]+(1-\omega) x_{L}^{a}-\sigma_{m,t}\rho \sigma_{x})+\gamma v_{t}+(1-\omega) x_{L}^{a}-\sigma_{m,t}\rho \sigma_{x}) \\ &=\frac{1}{R_{f}^{2}}(\omega^{2}x_{t}^{a}+(\omega +\gamma) v_{t}+(\omega +1)(1-\omega) x_{L}^{a}-(\omega +1)\sigma_{m,t}\rho \sigma_{x}). \end{aligned} $$

Again, using the law of iterated expectations, we can write:

$$ E_{t}[z_{t+3}]=E_{t}[E_{t+1}[E_{t+2}[z_{t+3}]]]. $$

We solve the expectation by moving from right to left:

$$ \begin{aligned} E_{t+2}[z_{t+3}] &=E_{t+2}[\frac{1}{R_{f}}m_{t+1}m_{t+2}(\omega x_{t+2}^{a}+v_{t+2}+(1-\omega) x_{L}^{a}-\sigma_{m,t+2}\rho \sigma_{x})] \\ &=\frac{1}{R_{f}}E_{t+2}[m_{t+1}m_{t+2}\omega x_{t+2}^{a}]\\ &\quad +\frac{1}{R_{f}}E_{t+2}[m_{t+1}m_{t+2}(v_{t+2}+(1-\omega) x_{L}^{a}-\sigma_{m,t+2}\rho \sigma_{x})]. \end{aligned} $$

Similarly,

$$ \begin{aligned} E_{t+1}[E_{t+2}[z_{t+3} ] ] &=\frac{1}{R_{f}}E_{t+1}[m_{t+1}m_{t+2}\omega x_{t+2}^{a}]\\ &\quad +\frac{1}{R_{f}^{2}}E_{t+1}[m_{t+1}(\gamma v_{t+1}+(1-\omega) x_{L}^{a}-\sigma_{m,t+1}\rho \sigma_{x})]\\ &=\frac{1}{R_{f}^{2}}E_{t+1}[m_{t+1}\omega (\omega x_{t+1}^{a}+v_{t+1}+(1-\omega) x_{L}^{a}-\sigma_{m,t+1}\rho \sigma_{x})]\\ &\quad +\frac{1}{R_{f}^{2}}E_{t+1}[m_{t+1}(\gamma v_{t+1}+(1-\omega) x_{L}^{a}-\sigma_{m,t+1}\rho \sigma_{x})]. \end{aligned} $$

Finally,

$$ \begin{aligned} E_{t}[E_{t+1}[E_{t+2}[z_{t+3} ] ] ] &=\frac{1}{R_{f}^{2}}\omega^{2}E_{t}[m_{t+1}x_{t+1}^{a}]+\frac{1}{R_{f}^{2}}\omega E_{t}[m_{t+1}(v_{t+1}+(1-\omega) x_{L}^{a}-\sigma_{m,t+1}\rho \sigma_{x})] \\ &\quad +\frac{1}{R_{f}^{2}}E_{t}[m_{t+1}(\gamma v_{t+1}+(1-\omega) x_{L}^{a}-\sigma_{m,t+1}\rho \sigma_{x})]\\ &=\frac{1}{R_{f}^{2}}\omega^{2}E_{t}[m_{t+1}x_{t+1}^{a}]+\frac{1}{R_{f}^{3} }\omega E_{t}[v_{t+1}+(1-\omega) x_{L}^{a}-\sigma_{m,t+1}\rho \sigma_{x}] \\ &\quad +\frac{1}{R_{f}^{3}}E_{t}[\gamma v_{t+1}+(1-\omega) x_{L}^{a}-\sigma_{m,t+1}\rho \sigma_{x}]\\ &=\frac{1}{R_{f}^{3}}\omega^{2}(\omega E_{t}[x_{t}^{a}]+E_{t}[v_{t}]+(1-\omega) x_{L}^{a}-\sigma_{m,t}\rho \sigma_{x}) \\ &\quad +\frac{1}{R_{f}^{3}}\omega (\gamma v_{t}+(1-\omega) x_{L}^{a}-\sigma_{m,t}\rho \sigma_{x})\\ &\quad +\frac{1}{R_{f}^{3}}(\gamma^{2}v_{t}+(1-\omega) x_{L}^{a}-\sigma_{m,t}\rho \sigma_{x}). \end{aligned} $$

Thus,

$$ \begin{aligned} E_{t}[z_{t+3}] &=\frac{1}{R_{f}^{3}}[(\omega^{3}x_{t}^{a}+\omega^{2}v_{t}+\omega^{2}(1-\omega) x_{L}^{a}-\omega^{2}\sigma_{m,t}\rho \sigma_{x})\\ &\quad +\omega \gamma v_{t}+\omega (1-\omega) x_{L}^{a}-\omega \sigma_{m,t}\rho \sigma_{x} \\ &\quad +\gamma^{2}v_{t}+(1-\omega) x_{L}^{a}-\sigma_{m,t}\rho \sigma_{x}] \\ &=\frac{1}{R_{f}^{3}}[\omega^{3}x_{t}^{a}+(\omega^{2}+\omega \gamma +\gamma^{2})v_{t}+(\omega^{2}+\omega +1)(1-\omega) x_{L}^{a} \\ &\quad -(\omega^{2}+\omega +1)\sigma_{m,t}\rho \sigma_{x}] \\ &=\frac{1}{R_{f}^{3}}\left[ \omega^{3}x_{t}^{a}+\frac{\omega^{3}-\gamma^{3}}{\omega -\gamma }v_{t}+\frac{\omega^{3}-1^{3}}{\omega -1}(1-\omega) x_{L}^{a}-\frac{\omega^{3}-1^{3}}{\omega -1}\sigma_{m,t}\rho \sigma_{x} \right] . \end{aligned} $$

The pattern emerges, and we see that

$$ E_{t}[z_{t+i}]=\frac{1}{R_{f}^{i}}\left[ \omega^{i}x_{t}^{a}+\frac{\omega^{i}-\gamma^{i}}{\omega -\gamma }v_{t}+\frac{\omega^{i}-1}{\omega -1} (1-\omega )x_{L}^{a}-\frac{\omega^{i}-1}{\omega -1}\sigma_{m,t}\rho \sigma_{x}\right] . $$

Now evaluating the summation yields:

$$ \sum_{i=1}^{\infty }E_{t}[z_{t+i}]=\frac{R_{f}(1-\omega )}{(R_{f}-\omega) (R_{f}-1)}x_{L}^{a}+\frac{\omega }{R_{f}-\omega }x_{t}^{a}+\frac{R_{f}}{ (R_{f}-\omega) (R_{f}-\gamma )}v_{t}-\frac{R_{f}\sigma_{x}\rho }{ (R_{f}-\omega )(R_{f}-1)}\sigma_{m,t}. $$

Thus the form of the stock price is:

$$ \begin{aligned} S_{t} &=B_{t}+\alpha_{1}x_{L}^{a}+\alpha_{2}x_{t}^{a}+\alpha_{3}v_{t}-\lambda_{1}\sigma_{m,t},\\ \alpha_{1} &=\frac{R_{f}(1-\omega) }{(R_{f}-\omega) (R_{f}-1)}, \\ \alpha_{2} &=\frac{\omega }{R_{f}-\omega }, \\ \alpha_{3} &=\frac{R_{f}}{(R_{f}-\omega) (R_{f}-\gamma) }, \\ \lambda_{1} &=\frac{R_{f}\sigma_{x}\rho }{(R_{f}-\omega) (R_{f}-1)}, \end{aligned} $$

which is Eq. (6) of Proposition 1.

To prove Eq. (7) of Proposition 1, substitute out "other information" as follows:

$$ \begin{aligned} x_{t+1}^{a} &=\omega x_{t}^{a}+(1-\omega) x_{L}^{a}+v_{t}+\epsilon_{t+1}, \\ E_{t}[x_{t+1}^{a}] &=\omega x_{t}^{a}+(1-\omega) x_{L}^{a}+v_{t}. \end{aligned} $$

This implies that

$$ v_{t}=E_{t}[x_{t+1}^{a}]-\omega x_{t}^{a}-(1-\omega) x_{L}^{a}. $$

Substituting the latter into the price equation, we can now rewrite price in terms of accounting data and expected next period abnormal earnings. Specifically,

$$ \begin{aligned} S_{t} &=B_{t}+[\alpha_{1}-\alpha_{3}(1-\omega) ]x_{L}^{a}+(\alpha_{2}-\omega \alpha_{3})x_{t}^{a}+\alpha_{3}E_{t}[x_{t+1}^{a}]-\lambda_{1}\sigma_{m,t} \\ &=B_{t}+R_{f}(1-\omega) \left(\frac{1}{(R_{f}-\omega) (R_{f}-1)}-\frac{1}{ (R_{f}-\omega) (R_{f}-\gamma) }\right) x_{L}^{a}\\ &\quad +\omega \left(\frac{1}{R_{f}-\omega }-\frac{R_{f}}{(R_{f}-\omega) (R_{f}-\gamma) }\right) x_{t}^{a}+\alpha_{3}E_{t}[x_{t+1}^{a}]-\lambda_{1}\sigma_{m,t} \\ &=B_{t}+\left(\frac{R_{f}(1-\omega) (1-\gamma) }{(R_{f}-\omega) (R_{f}-\gamma) (R_{f}-1)}\right) x_{L}^{a}\\ & \quad -\left(\frac{\gamma \omega }{(R_{f}-\omega) (R_{f}-\gamma) }\right) x_{t}^{a}+\alpha_{3}E_{t}[x_{t+1}^{a}]-\lambda_{1}\sigma_{m,t}. \end{aligned} $$

Noting that

$$ E_{t}[x_{t+1}^{a}]=E_{t}[x_{t+1}-(R_{f}-1)B_{t}], $$

and substituting, yields Eq. (7):

$$ \begin{aligned} S_{t} &=B_{t}+\gamma_{1}x_{L}^{a}-(R_{f}-1)\alpha_{3}B_{t}-\left(\frac{ \gamma \omega }{(R_{f}-\omega) (R_{f}-\gamma) }\right) x_{t}^{a}+\alpha_{3}E_{t}[x_{t+1}]-\lambda_{1}\sigma_{m,t}. \\ &=\gamma_{1}x_{L}^{a}+\left(\frac{R_{f}(1-\omega -\gamma) +\omega \gamma }{(R_{f}-\omega) (R_{f}-\gamma) }\right) B_{t}-\left(\frac{\gamma \omega }{ (R_{f}-\omega) (R_{f}-\gamma) }\right) x_{t}^{a}+\alpha_{3}E_{t}[x_{t+1}]-\lambda_{1}\sigma_{m,t}. \\ &=\gamma_{1}x_{L}^{a}+g_{2}B_{t}-\left(\frac{\gamma \omega }{ (R_{f}-\omega) (R_{f}-\gamma) }\right) (x_{t}-(R_{f}-1)B_{t-1})+\alpha_{3}E_{t}[x_{t+1}]-\lambda_{1}\sigma_{m,t}.\\ &=\gamma_{1}x_{L}^{a}+g_{2}B_{t}+g_{3}x_{t}-(R_{f}-1)g_{3}B_{t-1}+\alpha_{3}E_{t}[x_{t+1}]-\lambda_{1}\sigma_{m,t}. \\ &=\gamma_{1}x_{L}^{a}+g_{2}B_{t}+g_{3}x_{t}-(R_{f}-1)g_{3}(B_{t}-x_{t}+D_{t})+\alpha_{3}E_{t}[x_{t+1}]-\lambda_{1}\sigma_{m,t}. \\ &=\gamma_{1}x_{L}^{a}+[g_{2}-(R_{f}-1)g_{3}]B_{t}+[g_{3}+(R_{f}-1)g_{3}]x_{t}-(R_{f}-1)g_{3}D_{t}+\alpha_{3}E_{t}[x_{t+1}]-\lambda_{1}\sigma_{m,t}. \\ &=\gamma_{1}x_{L}^{a}+\frac{R_{f}(1-\omega) (1-\gamma) }{(R_{f}-\omega) (R_{f}-\gamma) }B_{t}-\frac{R_{f}\gamma \omega }{(R_{f}-\omega) (R_{f}-\gamma) }x_{t}+\frac{\omega \gamma (R_{f}-1)}{(R_{f}-\omega) (R_{f}-\gamma) }D_{t}+\alpha_{3}E_{t}[x_{t+1}]-\lambda_{1}\sigma_{m,t}.\\ &=\gamma_{1}x_{L}^{a}+\gamma_{2}B_{t}+\gamma_{3}x_{t}+\gamma_{4}D_{t}+\gamma_{5}E_{t}[x_{t+1}]-\lambda_{1}\sigma_{m,t}. \end{aligned} $$

$$ \begin{aligned} \gamma_{1} &=\frac{R_{f}(1-\omega) (1-\gamma) }{(R_{f}-\omega) (R_{f}-\gamma) (R_{f}-1)}\geq 0, \\ \gamma_{2} &=\frac{R_{f}(1-\omega) (1-\gamma) }{(R_{f}-\omega) (R_{f}-\gamma) }\geq 0, \\ \gamma_{3} &=-\frac{R_{f}\gamma \omega }{(R_{f}-\omega) (R_{f}-\gamma) } \leq 0, \\ \gamma_{4} &=\frac{\omega \gamma (R_{f}-1)}{(R_{f}-\omega) (R_{f}-\gamma) } \geq 0,\\ \gamma_{5} &=\frac{R_{f}}{(R_{f}-\omega) (R_{f}-\gamma) }>0. \end{aligned} $$

$\square$

2.2 Proof of proposition 2

$$ \begin{aligned} S_{t+1} &=B_{t+1}+\alpha_{1}x_{L}^{a}+\alpha_{2}x_{t+1}^{a}+\alpha_{3}v_{t+1}-\lambda_{1}\sigma_{m,t+1} \\ &=B_{t}+x_{t+1}-D_{t+1}+\alpha_{1}x_{L}^{a}+\alpha_{2}x_{t+1}^{a}+\alpha_{3}v_{t+1}-\lambda_{1}\sigma_{m,t+1} \\ &=B_{t}R_{f}+x_{t+1}^{a}+\alpha_{1}x_{L}^{a}+\alpha_{2}x_{t+1}^{a}+\alpha_{3}v_{t+1}-\lambda_{1}\sigma_{m,t+1}-D_{t+1} \\ &=R_{f}B_{t}+\alpha_{1}x_{L}^{a}+(1+\alpha_{2})x_{t+1}^{a}+\alpha_{3}v_{t+1}-\lambda_{1}\sigma_{m,t+1}-D_{t+1}. \end{aligned} $$

Therefore,

$$ S_{t+1}+D_{t+1}=R_{f}B_{t}+\alpha_{1}x_{L}^{a}+(1+\alpha_{2})x_{t+1}^{a}+\alpha_{3}v_{t+1}-\lambda_{1}\sigma_{m,t+1}. $$

Now subtracting R _f S _t from the latter equation gives:

$$ \begin{aligned} S_{t+1}+D_{t+1}-R_{f}S_{t} &=\alpha_{1}x_{L}^{a}+(1+\alpha_{2})x_{t+1}^{a}+\alpha_{3}v_{t+1}-\lambda_{1}\sigma_{m,t+1} \\ &\quad - R_{f}(\alpha_{1}x_{L}^{a}+\alpha_{2}x_{t}^{a}+\alpha_{3}v_{t}-\lambda_{1}\sigma_{m,t}) \\ &=\alpha_{1}(1-R_{f})x_{L}^{a}+(1+\alpha_{2})x_{t+1}^{a}+\alpha_{3}v_{t+1}-\lambda_{1}\sigma_{m,t+1}\\ &\quad -R_{f}(\alpha_{2}x_{t}^{a}+\alpha_{3}v_{t}-\lambda_{1}\sigma_{m,t}). \end{aligned} $$

and

$$ \begin{aligned} x_{t+1}^{a} &=\omega x_{t}^{a}+(1-\omega) x_{L}^{a}+v_{t}+\epsilon_{t+1}, \\ (1+\alpha_{2})x_{t+1}^{a} &=\omega (1+\alpha_{2})x_{t}^{a}+(1+\alpha_{2})(1-\omega) x_{L}^{a}+(1+\alpha_{2})v_{t}+(1+\alpha_{2})\epsilon_{t+1}, \\ \alpha_{3}v_{t+1} &=\alpha_{3}\gamma v_{t}+\alpha_{3}u_{t+1}. \end{aligned} $$

Collecting like-terms yields:

$$ \begin{aligned} S_{t+1}+D_{t+1}-R_{f}S_{t} &=\alpha_{1}(1-R_{f})x_{L}^{a}+x_{t}^{a}((1+\alpha_{2})\omega -R_{f}\alpha_{2})\\ &\quad+[\alpha_{3}(\gamma -R_{f})+(1+\alpha_{2})]v_{t}-\lambda_{1}(1-R_{f})\sigma_{m,t} \\ &\quad+(1+\alpha_{2})(1-\omega) x_{L}^{a}+(1+\alpha_{2})\epsilon_{t+1}+\alpha_{3}u_{t+1}-\lambda_{1}\xi_{t+1}. \end{aligned} $$

$$ \begin{aligned} (1+\alpha_{2})\omega -R_{f}\alpha_{2} &=\frac{R_{f}}{R_{f}-\omega }\omega -R_{f}\frac{\omega }{R_{f}-\omega }=0,\\ \alpha_{3}(\gamma -R_{f})+(1+\alpha_{2}) &=\frac{-R_{f}}{R_{f}-\omega }+ \frac{R_{f}}{R_{f}-\omega }=0, \\ \alpha_{1}(1-R_{f})x_{L}^{a} &=-\frac{R_{f}(1-\omega) }{(R_{f}-\omega) } x_{L}^{a}=(1+\alpha_{2})(1-\omega) x_{L}^{a}. \end{aligned} $$

So,

$$ \begin{aligned} S_{t+1}+D_{t+1}-R_{f}S_{t} &=-\lambda_{1}(1-R_{f})\sigma_{m,t} \\ &+(1+\alpha_{2})\epsilon_{t+1}+\alpha_{3}u_{t+1}-\lambda_{1}\xi_{t+1}. \end{aligned} $$

Thus,

$$ \begin{aligned} S_{t+1}+D_{t+1}-R_{f}S_{t} &=\lambda_{1}(R_{f}-1)\sigma_{m,t} \\ &\quad+(1+\alpha_{2})\epsilon_{t+1}+\alpha_{3}u_{t+1}-\lambda_{1}\Updelta \sigma_{m,t}. \end{aligned} $$

The gross stock return is then given by:

$$ \frac{S_{t+1}+D_{t+1}}{S_{t}}=R_{f}+(R_{f}-1)\lambda_{1}\frac{\sigma_{m,t} }{S_{t}}+(1+\alpha_{2})\frac{\epsilon_{t+1}}{S_{t}}+\alpha_{3}\frac{ u_{t+1}}{S_{t}}-\lambda_{1}\frac{\Updelta \sigma_{m,t}}{S_{t}}. $$

Taking conditional expectations provides the expected stock return:

$$ E_{t}[\frac{S_{t+1}+D_{t+1}}{S_{t}}]=R_{f}+(R_{f}-1)\lambda_{1}\frac{\sigma_{m,t}}{S_{t}}. $$

$\square$

2.3 Proof of proposition 3

From the preceding proof, we have

$$ \begin{aligned} \mu_{t+1} &=R_{f}+(R_{f}-1)\lambda_{1}\frac{\sigma_{m,t}}{S_{t}},\\ \sigma_{m,t}\lambda_{1} &=\frac{1}{R_{f}-1}(\mu_{t+1}-R_{f})S_{t}. \end{aligned} $$

Substitute the latter into the value equation yields:

$$ \begin{aligned} S_{t} &=B_{t}+\alpha_{1}+\alpha_{2}x_{t}^{a}+\alpha_{3}v_{t}-\frac{1}{ R_{f}-1}(\mu_{t+1}-R_{f})S_{t},\\ S_{t}(1+\frac{1}{R_{f}-1}(\mu_{t+1}-R_{f})) &=B_{t}+\alpha_{1}+\alpha_{2}x_{t}^{a}+\alpha_{3}v_{t}, \\ \mu_{t+1} &=(R_{f}-1)(\frac{B_{t}}{S_{t}}+\frac{\alpha_{1}}{S_{t}}+\alpha_{2}\frac{x_{t}^{a}}{S_{t}}+\alpha_{2}\frac{v_{t}}{S_{t}}-1)+R_{f},\\ \mu_{t+1} &=R_{f}+(R_{f}-1)(\frac{B_{t}}{S_{t}}+\frac{\alpha_{1}}{S_{t}} +\alpha_{2}\frac{x_{t}^{a}}{S_{t}}+\alpha_{2}\frac{v_{t}}{S_{t}}-1) \\ &=1+(R_{f}-1)(\frac{B_{t}}{S_{t}}+\frac{\alpha_{1}}{S_{t}}+\alpha_{2} \frac{x_{t}^{a}}{S_{t}}+\alpha_{2}\frac{v_{t}}{S_{t}}). \end{aligned} $$

To prove Eq. (11), we use (7) of Proposition 1. That is, we substitute "other information," v _t , with next period expected earnings as outlined in the proof of Proposition 1. Combining this with the results in Proposition 2 yields the (net) return dynamics:

$$ R_{t+1}-1=(R_{f}-1)\left(\gamma_{1}\frac{x_{L}^{a}}{S_{t}}+\gamma_{2} \frac{B_{t}}{S_{t}}+\gamma_{3}\frac{x_{t}}{S_{t}}+\gamma_{4}\frac{D_{t}}{ S_{t}}+\gamma_{5}\frac{E_{t}[x_{t+1}]}{S_{t}}\right) +(1+\alpha_{2})\frac{ \epsilon_{t+1}}{S_{t}}+\alpha_{3}\frac{u_{t+1}}{S_{t}}-\lambda_{1}\frac{ \Updelta \sigma_{m,t}}{S_{t}}. $$

Taking conditional expectations of both sides yields the desired result.

$\square$