Math time! Let's generalize this with algebra rather than using specific numbers. Note that this may not be super accurate with the recent changes to the market rules regarding shorts prioritized by collateral.
I want to short x BitUSD at a price of p1 BTSX per BitUSD. I provide m BTSX margin as part of the collateral. This means if my short is fully matched, there will be c = (p1 * x + m) BTSX held in collateral in the short position which owes the network x BitUSD. Now let's say the price of BTSX changes and I want to cover. I purchase x BitUSD at a new price of p2 BTSX per BitUSD (meaning I pay p2 * x BTSX). I use the x BitUSD to cover my short position and that gives me the c BTSX held in collateral. I originally had (m + p2 * x) BTSX in my possession, now I have (p1 * x + m) BTSX in my possession. The net change in BTSX through this process is (p1 - p2) * x BTSX.
I had to keep some amount of BTSX sitting around in order to do this short. I had m BTSX locked up in collateral that I could not touch. But I also had to keep some amount of BTSX around to buy BitUSD to cover; I will define this amount to be s * p1 * x BTSX. Let us say I was fairly confident in the value of BTSX rising with respect to BitUSD. In that case I would be fine with s = 1 since it would be more than the p2 * x BTSX needed to buy the BitUSD to cover. If I want to be more conservative, I should choose s > 1, so that I can afford to cover if the value of BTSX with respect to BitUSD falls. The largest value of s should be (1 + m/(p1*x))/1.5, since if p2 rises high enough that I would need even more BTSX than that to buy BitUSD it would already be too late (the network would do a margin call since the collateral would become less than 150% of the value of the BitUSD debt owed). If BitShares X was modified to allow shorts to cover by bidding for BitUSD on the exchange, I wouldn't need to hold any extra BTSX (meaning s = 0). Whatever my choice, we can claim that I have (m + s * p1 * x) sitting around for the purposes of gaining from my short position. If the value of BTSX does rise with respect to BitUSD (meaning p2 < p1), then in terms of the fractional gain relative to the amount of BTSX I had to hold I gain f = [(p1-p2)*x]/[m + s*p1*x] = [1 - p2/p1]/[m/(p1*x) + s].
Let's define r to be the fractional increase over this period in the price of BTSX in terms of BitUSD. The price of BTSX was first 1/p1 BitUSD and then later became 1/p2 BitUSD. Therefore, r = (1/p2 - 1/p1)/(1/p1) = p1/p2 - 1. I can use this to simplify f above: f = r/[(1+r) * (s + m/(p1*x)]. Let's take the very conservative estimate for s, just to simplify the variables a little more: f = r/[(1+r) * ((2/3) + (5/3) * (m/(p1*x)))]. Finally, let us assume that you can get your short matched at the bare minimum collateral so that we can replace m/(p1*x) with 1. Then f = (3*r)/(7*(1+r)).
And there you have it. Under the above assumptions, if I short and the price of BTSX goes up by 10%, I can cover and increase my BTSX holdings by up to 3.9% (f = (3*0.1)/(7*(1+0.1)) = 0.03896). Notice that even if the value of the dollar drops to zero (meaning r approaches infinity), I can still only increase my BTSX holdings by up to 43% (again assuming I am using the more conservative assumption above).
I would really appreciate it if someone checks my math for errors.
Exercise for the reader
: Given an average steady rate of increase, a, in the price of BTSX with respect to BitUSD (or USD), what is my return on investment with compounding (meaning I short, cover, use gained BTSX to short again, repeat) as a function of a regular short-cover time period T (assuming the price at the moment of each short/cover instance falls on the average steady increase curve)? Solve for both linear and exponential growth curves.
Topic: Newbie Shorting Questions (Read 7792 times)
