Brealey & Myers provide a certainty-equivalent version of the present value rule, using CAPM, as follows:

$$PV_0=\frac{C_1 - \lambda_m *cov(C_1, r_m)}{1 + r_f}$$

$PV_0$ - Present Value of cash flow 1 at time 0.

$\lambda_m$ - Market price of risk = $\frac{r_m-r_f}{\sigma_m^2}$

$cov(C_1, r_m)$ - Covariance of the cash flow at time 1 with the return on the market.

I want to create an n-factor version of this same model. However, using Fama French 3-factor model as an example, the following doesn't seem to work on a toy example I've set up:

$$PV_0=\frac{C_1 - \lambda_m *cov(C_1, r_m)- \lambda_{smb} *cov(C_1, r_{smb})- \lambda_{hml} *cov(C_1, r_{hml})}{1 + r_f}$$

$\lambda_m$ = $\frac{r_m-r_f}{\sigma_m^2}$

$\lambda_{smb}$ = $\frac{r_s-r_b}{\sigma_{smb}^2}$

$\lambda_{hml}$ = $\frac{r_h-r_l}{\sigma_{hml}^2}$

Question: what am I doing wrong? Is there some way I need to adjust for the covariance amongst the factors?


In checking my toy example again, I realized that I might in fact have the right formula above. So points/checkmarks to anyone who can either prove the above right or wrong or provide a citation to the more general form:

$$PV_0=\frac{C_1 - \displaystyle\sum_{i=1}^n\lambda_i *cov(C_1, r_i)}{1 + r_f}$$

for orthogonal risk factors $i_1,i_2,\dotsc,i_n$.


1 Answer 1


Alright, here's the proof (I think):

Statement of APT:

$$E(r_a)=r_f + \displaystyle\sum_{i=1}^n\lambda_i * cov(r_a, r_i)$$

Expand $E(r_a)$:

$$\frac{E(C_1)}{PV_0} - 1 =r_f + \displaystyle\sum_{i=1}^n\lambda_i * cov(\frac{C_1}{PV_0} - 1, r_i)$$

Since $PV_0$ doesn't have any covariance with $r_i$, we can reduce the above to the following:

$$\frac{E(C_1)}{PV_0} - 1 =r_f + \displaystyle\sum_{i=1}^n\frac{\lambda_i * cov(C_1, r_i)}{PV_0}$$


$$\frac{E(C_1) -\displaystyle\sum_{i=1}^n\lambda_i * cov(C_1, r_i)}{PV_0} = 1 + r_f $$

And finally:

$$\frac{E(C_1) -\displaystyle\sum_{i=1}^n\lambda_i * cov(C_1, r_i)}{1 + r_f} = PV_0 $$

QED (until somebody points out a dumb error I've made).

  • $\begingroup$ I'm not qualified to judge this work, which is why I personally couldn't help on the question anyway. But since I want to award the bounty, and since you've taken the time to answer your own question, I've award the bounty to you. Hopefully someone more qualified can assess what's going on. $\endgroup$ Mar 11, 2012 at 18:04
  • $\begingroup$ Appreciate it. I'd be happy to give it back since I was too lazy to figure out it before I posted the original question. $\endgroup$
    – MikeRand
    Mar 11, 2012 at 22:07

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