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I know how to derive at the CAPM from a microeconomic foundation. In a recent University course I stumbled over a slide that derived the CAPM solely from the Sharpe ratio:

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I cant come up with that steps by myself and it seems I am missing something down the line.

Do you guys here either have a source / video / etc. that derives the CAPM similarly in a step by step fashion? Or is someone here who could break the involved derivation steps into parts?

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  • $\begingroup$ which course slides were the images taken from? $\endgroup$
    – develarist
    Commented Oct 26, 2020 at 6:22

2 Answers 2

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I imagine that after 3 years you don't need to maximize the shape ratio anymore, but I actually stumbled across the same question when studying for the exam and thus ended up at your post here, so here's my approach to deriving the CAPM equation. Maybe it'll help some other of Professor Lawrence's students in the future! https://imgur.com/oVsQPME

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  • $\begingroup$ In case the link breaks, you should add your derivation to the post using MathJax. $\endgroup$
    – amdopt
    Commented Jul 10, 2023 at 11:22
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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

  • $\mu'_m = \frac{\partial \mu_m}{\partial \omega_i}$
    • $\mu'_m = \mu_i - r_f$
  • $\sigma'_m = \frac{\partial \sigma_m}{\partial \omega_i}$
    • $\sigma'_m = \frac{\sigma_{i,M}}{\sigma_m}$ 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:

$$\sigma'_m = \frac{\sigma_{i,M}}{\sigma_m}$$

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