0
$\begingroup$

Good evening,

I am studying the CAPM and I have a doubt regarding the variance $σ_i^2$ of the expected return of an asset $i$.

In particular, how can I derive the following formula?

$$σ_i^2 = β_i^2 σ_M^2 + var(\epsilon_i)$$

In my book, I read that the formula derives from the correlation of

$$r_i = r_f + \beta_i(r_M - r_f) + \epsilon_i$$ with $r_M$, using also the definition

$$\beta_i = \frac{cov(i,m)}{\sigma_i \sigma_M}.$$

Can you help me in deriving the $\sigma_i^2$ formula? Thanks

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ Please have a look at the difference between the CAPM and a single index model... There is no $\varepsilon_i$ in the CAPM $\endgroup$
    – Kevin
    Nov 22, 2020 at 17:53
  • 1
    $\begingroup$ The formula $\sigma_i^2=\beta_i^2\sigma_M^2+\mathbb{V}\text{ar}[\varepsilon_i]$ doesn't make any sense in the CAPM framework. The CAPM makes no predictions about the returns themselves nor their variances. It merely concerns itself with the expected (excess) returns of an asset. You can use the single index model to empirically test the CAPM but that needs to be differentiated from the underlying economic theory of the CAPM. $\endgroup$
    – Kevin
    Nov 22, 2020 at 18:21
  • 1
    $\begingroup$ If you study the CAPM, then $\sigma_i^2=\beta_i^2\sigma_M^2+\mathbb{V}\text{ar}[\varepsilon_i]$ is not true. It just doesn't make any sense at all. Look at the derivation of the CAPM. There's no $\varepsilon_i$. You simply can't compute the variance of returns in the CAPM. You're studying a different model which is called single index model. $\endgroup$
    – Kevin
    Nov 22, 2020 at 18:26
  • 1
    $\begingroup$ Kevin, I don't reason by sandboxes. I am studying risky assets: int his context, the CAPM is useful. Also, it is useful to know how the asset risk can be composed. Bringing these two topics one after the other doesn't mean the the latter is part of the first. Then, if this topic disposition seems unreasonable to you, I suggest writing to David G. Luenberger, the author of the book I am studying from. Consider I am a poor student that reads from page 1 onwards. Nevertheless, I found your contribute really precious as it shed a light on a valuable argument $\endgroup$
    – Bernheart
    Nov 22, 2020 at 18:38
  • 2
    $\begingroup$ In terms of math, this is largely semantics. Given zero alpha, your equation is equivalent to the CAPM equation. You can always split a variable to expectation and independent error. $\endgroup$
    – fes
    Nov 22, 2020 at 19:01

2 Answers 2

Reset to default
1
$\begingroup$

We have:

\begin{align} Var(r_i){} & =Var\left[r_f+\beta_i(r_m-r_f)+\epsilon_i\right] \\ & =Var\left[r_f+\beta_i r_m-\beta_i r_f +\epsilon_i\right] \\ & \stackrel{\dagger}{=} Var\left[ \beta_i r_m +\epsilon_i\right] \\ & \stackrel{\ddagger}{=} Var\left[ \beta_i r_m \right] +Var\left[ \epsilon_i\right] \\ & \stackrel{\star}{=} \beta_i^2Var\left[ r_m \right] +Var\left[ \epsilon_i\right] \\ \end{align}

In $\dagger$ we have used the property $Var(\alpha +X)= Var(X)$ for $\alpha \in \mathbb{R}$ and $X$ a r.v

In $\ddagger$ we have used the property $Var(X+Y)= Var(X)+Var(Y)$ for $X,Y$ independent r.v

In $\star$ we have used the property $Var(\alpha X)= \alpha ^2 Var(X)$ for $\alpha \in \mathbb{R}$ and $X$ a r.v

Improve this answer
$\endgroup$
1
$\begingroup$

The derivation of stock variance based on the single index model is \begin{align} \sigma_i^2 &= Var ( r_i) = \mathbb{E}(r_i - \bar{r}_i)^2\\ &\stackrel{\dagger}{=} \mathbb{E}\left[\left( \alpha_i + \beta_i r_M + \epsilon_i \right) - \left( \alpha_i + \beta_i \bar{r}_M + \epsilon_i \right)\right]^2\\ &= \mathbb{E}\left[ \beta_i \left( r_M - \bar{r}_M \right) + \epsilon_i \right]^2\\ &\stackrel{\ddagger}{=} \beta_i^2 \mathbb{E}(r_M - \bar{r}_M)^2 + 2 \beta_i^2 \mathbb{E}\left[(r_M - \bar{r}_M)\epsilon_i\right] + \mathbb{E}(\epsilon_i)^2\\ &= \beta_i^2 \mathbb{E}\left(r_M - \bar{r}_M\right)^2 + \mathbb{E}(\epsilon_i)^2\\ &= \underbrace{\beta_i^2 \sigma_M^2}_{\text{systematic risk}} + \underbrace{\sigma_i \epsilon_i^2}_{\text{non-systematic risk}} \end{align}

Given that $$\mathbb{E}(\epsilon_i) = 0 $$ $$\mathbb{E}\left[\epsilon_i(r_M - \bar{r}_M)\right] = 0 $$

and \begin{align} r_i &= \alpha_i + \beta_i r_M + \epsilon_i\\ \mathbb{E}(r_i) &= \mathbb{E}(\alpha_i + \beta_i r_M + \epsilon_i)\\ &= \mathbb{E}(\alpha_i) + \mathbb{E}(\beta_i r_M) + \mathbb{E}(\epsilon_i)\\ &= \alpha_i + \mathbb{E}(\beta_i r_M) \end{align}

Note: In $\dagger$ we have used the property $r_i = \alpha_i + \beta_i r_M + \epsilon_i$, while for $\ddagger$, $\mathbb{E}(\epsilon_i) = 0$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Hi develarist, maybe it's E[(\epsilon_i) ^ 2]? $\endgroup$
    – Bernheart
    Nov 22, 2020 at 18:31
  • $\begingroup$ only your question title, but not the question itself, makes mention of non-systematic risk, so yeah non-systematic risk is $\sigma_i \epsilon_i^2 = \mathbb{E}(\epsilon_i)^2$. An equivalent expression of the decomposition of stock variance is $$\sigma_i^2 = \rho_{i,M}^2 \sigma_i^2 + (1-\rho_{i,M}^2) \sigma_i^2$$ where $(1-\rho_{i,M}^2) \sigma_i^2$ is non-systematic risk $\endgroup$
    – develarist
    Nov 22, 2020 at 18:39

Not the answer you're looking for? Browse other questions tagged or ask your own question.