Deriving Asset Variance in the CAPM Single-Index Model
Summary
The document derives the single-index variance decomposition: an asset’s return variance consists of the market-driven component, scaled by beta squared, plus residual variance. It starts with the return model, removes the risk-free constant because constants do not affect variance, and expands the variance of the market and residual terms. The cross term vanishes when the residual is uncorrelated with market returns; equivalently, the variance sum requires zero covariance, with independence being a stronger sufficient assumption.
The answers identify the market component as systematic risk and the residual component as nonsystematic risk. The derivation is a model-based identity under those assumptions, not an empirical finding. One answer contains notation and algebra inconsistencies in its expansion and labels the residual variance imprecisely, so the result should be read using the stated zero-mean, zero-covariance assumptions rather than relying on every displayed intermediate line.
Key ideas
- A constant risk-free rate does not contribute to return variance.
- The market contribution to asset variance is beta squared times market variance.
- The residual contributes its own variance when it has zero covariance with the market.
- Independence is sufficient for the variance sum, but zero covariance is the key condition.
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Full text
# Nonsystematic risk in a random rate of return
# Nonsystematic risk in a random rate of return
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
## Answer by alexbougias (score 1, accepted)
https://quant.stackexchange.com/a/59527
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
## Answer by develarist (score 1)
https://quant.stackexchange.com/a/59526
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 $\ddagger$, $\mathbb{E}(\epsilon_i) = 0$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.