Definitions
- $\mu_t$: Expected return of tangent portfolio
- $\omega_i$ be the weight of asset $i$ in the tangent portfolio
- $\mu_i$: Expected return of asset i
- $\mu_m$: Expected return of market
- $r_f$: Risk free rate (assumed to be constant)
- $\sigma_{i,j}$: Covariance between asset i and j
- $\sigma_{i,M}$: Covariance between asset i and the market
Why is the Market Portfolio along the Capital Allocation Line?
The CAPM model assumes all market participants only care about mean and variance and that they completely rational, aka they follow markovitz portfolio theory. So
participant $n$'s portfolio can be decomposed into the risk free rate and and the tangent portfolio as so: $\alpha_n \mu_t + \beta_n \mu_f$. Note these are dollar amounts with no restrictions, they can be positive or negative and dont have to equal 1.
The return of the market is defined as the pnl devided by the total dollar amount
$$\mu_m = \frac{\sum_n \alpha_n \mu_t + \beta_n \mu_f}{\sum_n \alpha_n + \beta_n}$$
let $\alpha'_n = \frac{\alpha_n}{\sum_n \alpha_n + \beta_n}$ and $\beta'_n = \frac{\beta_n}{\sum_n \alpha_n + \beta_n}$, so $\sum_n \alpha'_n + \beta'_n = 1$.
$$\mu_m = \sum_n \alpha'_n \mu_t + \beta'_n \mu_f$$
Then $1-\sum_n\alpha'_n = \sum_n\beta'_n$
$$\mu_m = \sum_n \alpha'_n \mu_t + (1-\sum_n\alpha'_n) \mu_f$$
Let $A = \sum_n \alpha'_n$ be the total weight in the tangent portfolio across the entire market.
$$\mu_m = A r_t + (1 - A)r_f$$
Let $\omega_i$ be weight in asset i, so $A = \sum_i \omega_i$
$$\mu_m = \sum_i \omega_i \mu_i + \left(1 - \sum_i \omega_i\right)r_f$$
Derivation of CAPM
Derivatives to be used later
- $\mu'_m = \frac{\partial \mu_m}{\partial \omega_i}$
- $\sigma'_m = \frac{\partial \sigma_m}{\partial \omega_i}$
- $\sigma'_m = \frac{\sigma_{i,M}}{\sigma_m}$ See appendix.
The market portfolio is along the capital allocation line, so it has maximum sharpe ratio. Thus the derivative of sharpe w.r.t to all $\omega_i$ is 0.
$$\forall i,\quad
\frac{\partial}{\partial \omega_i} \left( \frac{\mu_m - r_f}{\sigma_m} \right) = 0 \rightarrow
\frac{\mu'_m\sigma_m-\sigma'_m(\mu_m-r_f)}{\sigma_m^2} = 0
$$
Using our derivatives above gives us:
$$\forall i,\quad
\frac{(\mu_i-r_f)\sigma_m-\left(\frac{\sigma_{i,M}}{\sigma_m}\right)(\mu_m-r_f)}{\sigma_m^2} = 0 \rightarrow
\frac{\mu_i-r_f}{\sigma_m} = \frac{\sigma_{i,M}(\mu_m-r_f)}{\sigma_m^3} \rightarrow
\mu_i = \frac{\sigma_{i,M}}{\sigma_m^2} (\mu_m-r_f)
$$
We can define $\beta_i = \frac{\sigma_{i,M}}{\sigma_m^2}$, leading us to the final, familiar CAPM equation:
$$\forall i,\quad
\mu_i - r_f = \beta_i(\mu_m - r_f)$$
Appendix for $\sigma'_m = \frac{\partial \sigma_m}{\partial \omega_i}$
Note that $\sigma_m = (\sigma_m^2)^{1/2}$, and $\sigma_m^2 = \sum_{i,j} \omega_i \omega_j \sigma_{ij}$, so
$$\sigma'_m =
\frac{1}{2(\sigma_m^2)^{1/2}} \left(2\sum_j \omega_j \sigma_{ij} \right)$$
Using the fact that $\sigma_{i,M} = \text{Cov}[r_i,r_m] = \text{Cov}\left[r_i, \sum_i \omega_i r_i\right] = \sum_j \omega_i \sigma_{ij}$, we get:
$$\sigma'_m =
\frac{\sigma_{i,M}}{\sigma_m}$$