When a Sharpe-Style Rule Supports Adding an Asset to a Portfolio

# How to derive this mathematical equation from the perspective of the mean-variance portfolio optimization?


# How to derive this mathematical equation from the perspective of the mean-variance portfolio optimization?












## Question

- I found a simplified inequation to decide whether the new asset A should be added to my current portfolio B. If the following inequation is satisfied, the new asset A should be added to my portfolio. (source: Mackenzie Investment's research report Correlation vs. Beta: What is the difference)

$$\frac{E(R_{a})}{\sigma_{a}} > \frac{E(R_{b})}{\sigma_{b}} \times corr(R_{a}, R_{b})$$

- One colleague of mine suggested to me that the inequation shown above seems to be derived from the mathematical equation written below, when the condition $W_{a}> 0$ is satisfied.

- $W_{a}$: how much percentage of my total wealth is invested in the asset A $E(R_{a})$: the expected return of the asset A $\sigma_{a}$: the standard deviation of the returns of the asset A $r_{f}$: risk-free return such as the US government bonds I assume that $R_{A}$ is the same thing as $R_{a}$, which means the return of the asset A.

- Is there anyone who can show me how the equation written at the bottom can be simplified to the inequation written at the top, when the condition $W_{a}> 0$ is satisfied?

## Answer by Pleb (score 3, accepted)

https://quant.stackexchange.com/a/67901

Assume that the denominator is positive (ie. $>0$) and let $r_f=0$. For ease of notation, we will rewrite $W_A$ as:

\begin{align*} W_A = \frac{\mathbb{E}\left[R_A\right] \sigma_B^2 - \mathbb{E}\left[R_B\right]\sigma_A\sigma_B \mathbb{C}orr(R_A, R_B)}{\mathbb{E}\left[R_A\right] \left(\sigma^2_B - \sigma_A\sigma_B \mathbb{C}orr(R_A, R_B)\right)+\mathbb{E}\left[R_B\right]\left(\sigma^2_A - \sigma_A\sigma_B \mathbb{C}orr(R_A, R_B)\right)} = \frac{\mathbb{E}\left[R_A\right] \sigma_B^2 - \mathbb{E}\left[R_B\right]\sigma_A\sigma_B \mathbb{C}orr(R_A, R_B)}{Z_A + Z_B}>0,\\ \end{align*} Now, adding the negative part of the fraction on both sides, we observe that:

\begin{align*} \frac{\mathbb{E}\left[R_A\right] \sigma_B^2}{Z_A + Z_B}&> \frac{\mathbb{E}\left[R_B\right]\sigma_A\sigma_B \mathbb{C}orr(R_A, R_B)}{Z_A + Z_B}\\ &\Updownarrow\\ \mathbb{E}\left[R_A\right] \sigma_B^2&> \mathbb{E}\left[R_B\right]\sigma_A\sigma_B \mathbb{C}orr(R_A, R_B)\\ &\Updownarrow\\ \frac{\mathbb{E}\left[R_A\right] \sigma_B^2}{\sigma_A}&> \mathbb{E}\left[R_B\right]\sigma_B \mathbb{C}orr(R_A, R_B)\\ &\Updownarrow\\ \frac{\mathbb{E}\left[R_A\right]}{\sigma_A}&> \frac{\mathbb{E}\left[R_B\right] \mathbb{C}orr(R_A, R_B)}{\sigma_B},\\ \end{align*} where we in the second inequality uses the fact that the denominator is positive, and in the third and fourth inequality have divided with $\sigma_A$ and $\sigma_B^2$ respecively.

## Answer by Kermittfrog (score 3)

https://quant.stackexchange.com/a/67916

Here, I am specifically answering your question in the comments:

No, there is no guarantee that the denominator will be positive. For simplicity, assume that $r_f$=0 and that $E(R_A)=E(R_B)\frac{\sigma_A}{\sigma_B}\rho+\frac{c}{\sigma_b^2}$ with $c>0$. This level of $E(R_A)$ satisfies the condition in your question. Then,

$$ W_A=\frac{c}{c\left(1-\frac{\sigma_A}{\sigma_B}\rho\right)+E(R_B)\sigma_A^2\left(1-\rho^2\right)} $$

and we can clearly have a situation where the optimal investment amount $W_A$ becomes negative, i.e. when

$$ c\left(1-\frac{\sigma_A}{\sigma_B}\rho\right)+E(R_B)\sigma_A^2\left(1-\rho^2\right)<0 $$

But if you start to play with the numbers, you will see that it may require some extreme combinations of $E(R_B), \sigma_A, \sigma_B, \rho, c$ in order for the quick-and-dirty rule not to work.