# Simple mean reversion strategy portfolio construction

I had a quick idea I wanted to test, but am not sure of the correct way to size bets. Basically, I think that for a given index (say S&P), I want to be long under performers and short over performers (hence the mean reversion). Here is my approach

1. To simplify things, lets take an approach where I simply sort from highest to lowest the past period return for all the tickers.

2. I then want to "pair off" each of the tickers in that index. So pairing most over-perform with most under-perform, second most, third most, etc.

3. For each pair, create a beta neutral position where I am long the under performer and short the over performer

4. Weight each pair in a linear ramp fashion. So what I mean is if I have 100 dollars and 50 pairs, I would have $\sum_{i=1}^{50}D_i = 100$, where $D_1$ would represent the dollar weight on the most under/over perform pair and the $D_i$ satisfy $D_i = (51-i) \times x$ so a linear ramp and if we solved for $50x + 49x + ... x = 100 \implies x = \frac{100}{51\times25}$ so $D_1 \approx \$3.92$or 3.92% in portfolio weight terms. 5. However, my question then lies on the right approach to breaking apart the long and short side for each pair so that combined exposure is 3.92% while ensuring the pairs beta is also zero. My initial thoughts on (5) is that for$s$short and$l$long, the positions need to satisfy$p_s \beta_s + p_l \beta_l = 0$and$p_s + p_l = D_i$, but in this case with two equations and two unknowns, there will only be one solution (assuming linear independence). However, that solution could mean that$p_l <0$which is taking a short position in the ticker I want to be long. How do I account for this? Is there anything else with my assumptions and thoughts for 1-4 that might be wrong? Any help would be much appreciated! Lets take a concrete example where I have 50 dollars to allocate to this pair$p_s\beta_s + p_l \beta_l = 0$and$p_s + p_l = 50$, where$p_l >0$and$p_s <0$. If the following happen: 1.$\beta_s = 2, \beta_l = 1$, then we have a clear solution where$p_s = -50$and$p_l = 100$2. If we flip betas instead to be$\beta_s = 1, \beta_l = 2$, then solving gives us that$p_l = -50 < 0$which is a short position. 3. Other signs combinations of$\beta_s,\beta_l$make it so that if$\beta_s$and$\beta_l\$ are opposite signs, then I cannot generate a beta neutral position (either it will be always beta positive or beta negative)