# CAPM Model, is this exercise done correctly?

Hey i need to know if the task is done correctly, please help :) Standard deviation of the rate of return on the market portfolio is equal to $$\sigma_{MP}=1,5\%=\frac{15}{1000}$$. I have portoflio with 3 assets with weights $$w=(\frac{3}{10},\frac{2}{10},\frac{5}{10})$$ and Beta of this assets is equal $$\beta_1=-\frac{5}{4}, \beta_2=\frac{85}{100}. \beta_3=-\frac{2}{10}$$. I have to estimate the value of the standard deviation of the rate of return on the investor's portfolio. Skip the risk unsystematic.

My solution:

$$\sigma_w^2=\beta^2\sigma_{MP}^2 \Rightarrow \sigma_w=|\beta|\sigma_{MP}$$. I calculate

$$\beta=\frac{cov(K_w,K_{MP})}{\sigma_{MP}^2}$$ I calculate:

$$cov(K_1,K_{MP})=\beta_1\cdot\sigma_m^2=-\frac{5}{4}\cdot \frac{15^2}{1000^2}=-\frac{9}{32000}$$

$$cov(K_2,K_{MP})=\beta_2\cdot\sigma_m^2=\frac{85}{100}\cdot \frac{15^2}{1000^2}=\frac{153}{800000}$$

$$cov(K_3,K_{MP})=\beta_3\cdot\sigma_m^2=-\frac{2}{10}\cdot \frac{15^2}{1000^2}=-\frac{9}{200000}$$

Now I can calculate:

$$cov(K_w,K_{MP})=cov(w_1\cdot K_1+w_2\cdot K_2+ w_3\cdot K_3, K_{MP})=w_1 cov(K_1,K_{MP})+ w_2cov(K_2,K_{MP})+ w_3cov(K_1,K_{MP})=\frac{3}{10}\cdot \frac{-9}{32000}+\frac{2}{10}\cdot \frac{153}{800000}+\frac{5}{10}\cdot\frac{-9}{200000}=-\frac{549}{8000000}$$

So i Can calculate:

$$\beta_w=\frac{cov(K_w,K_{MP})}{\sigma_{MP}^2 }=-\frac{549}{8000000}\cdot\frac{1000000}{225}=-\frac{61}{200}$$

So $$\sigma_{w}=\beta_w\cdot\sigma_{MP}=-\frac{61}{200}\cdot \frac{1000}{15}=-\frac{183}{40000}\approx 0,4575\%$$

please tell me if my solution is correct or somewhere is it wrong?

• can anyone confirm if my solution is correct or has an error crept along the way? Apr 30 '20 at 19:48

That looks correct to me apart from the calculations at the end which aren't in line with the formula you posted above. You could however immediately use the fact that the beta of the portfolio is just the weighted average of the beta of the stocks and save yourself some notation. Just insert $$\beta_i * \sigma_w^2$$ instead of the covariance terms in your calculations and you'll see that it simplifies nicely with the $$\sigma_w^2$$ term cancelling out. So $$\beta_w = -(5/4)*(3/10)+(85/100)*(2/10)-(2/10)*(5/10) = -(61/200)$$ with $$\sigma_w$$ following as |(-61/200)|*(15/1000)= 0.004575