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The optimal entry/exit thresholds for mean reversion trading (assuming an underlying Ornstein-Uhlenbeck (OU) process) is derived in the paper "Optimal Mean Reversion Trading with Transaction Costs and Stop-Loss Exit" by Tim Leung and Xin Li, and these results are also included in the book "Optimal Mean Reversion Trading Mathematical Analysis and Practical Applications" by the same authors.

I would have expected that the values of these threshold levels ($b^*$ and $d^*$) relative to the OU mean $\theta$ would not change if you changed $\theta$. That is, for some arbitrary shift in the OU mean, $\triangle \theta$, it would always be true that $$b^*(\theta) - \theta = b^*(\theta + \triangle \theta) - (\theta + \triangle \theta)$$ and $$d^*(\theta) - \theta = d^*(\theta + \triangle \theta) - (\theta + \triangle \theta)$$

This must be true since buying the long-short pair is equal to selling the short-long pair, and the OU process parameters for the short-long pair are identical other than the sign of $\theta$.

However, this is clearly not the case when $\theta$ is around zero for some reason, as seen in Figures 2.3 and 2.5 in the book:

Figure 2.3 Figure 2.5

In all of the charts, there are curves for three different values of $\theta$ in increments of 0.3, and so I would expect that the three lines would be exactly the same, but shifted up by exactly 0.3 each time.

In Figure 2.3, this is more or less the case for the lines corresponding to $\theta=0.3$ and $\theta=0$, but the line corresponding to $\theta=-0.3$ seems to be trying to avoid the values around $b^*=0$? Meanwhile in Figure 2.5, the lines look like vertically shifted versions of each other, however the distance between them is not exactly 0.3.

The underlying OU process can go negative, as can the price of a long-short pair of assets being modelled, so it can't be anything to do with a positive price constraint. What am I missing here?

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  • $\begingroup$ To me it seems like the terminology "entry-" and "exit-levels" points to an asymmetry. When you decide to enter, you decide to pay both the entry and exit transaction costs in some future. When you decide to exit, you only decide the timing of the exit cost. I think you would retrieve symmetry if you set transaction costs at zero. $\endgroup$
    – Mats Lind
    Commented Oct 14, 2022 at 13:43
  • $\begingroup$ @MatsLind I generated the same charts as above for the case $c=0.000001$, and the results look similar to those above, with all the same features, so I don't think transaction costs can explain it. The results in the paper require $c>0$, which is also strange in itself, I tried ignoring this and set $c=0$ and found that the optimal entry levels were EQUAL to the optimal exit levels, which makes no sense either. $\endgroup$
    – mpeac
    Commented Oct 16, 2022 at 0:21
  • $\begingroup$ Setting $r=0$ and $c=0.000001$ also doesn't help. $\endgroup$
    – mpeac
    Commented Oct 16, 2022 at 0:27
  • $\begingroup$ That's interesting! What about the stop-loss? That could explain it but it doesn't seem like there is one in these cacluations? $\endgroup$
    – Mats Lind
    Commented Oct 17, 2022 at 8:39
  • $\begingroup$ @MatsLind There are two sets of results in the paper, with and without stop loss. I've only been considering the case without a stop loss. I'm starting to think that there is a mistake in the paper somewhere, but I can't spot it. $\endgroup$
    – mpeac
    Commented Oct 17, 2022 at 22:50

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