Percent Calc Problem

+50%
You didn't label your answer... gott have them units on there.
if a short position is in the reds, say $-200, and then makes some gains so that now the position is worth $-100, then how much is this last PL change, expressed in percentage? Is it +100% or +50%?

PL change = fitty percent.

 
Last edited:
As others have said, you haven't provided enough information, even if you think you have.

Let me ask you this: suppose your PL at time 1 is $25 and your PL at time 2 is $100. What is your percent change in PL? Normally we would say it is 75 / X, where X is the total account value, the amount at risk, or the notional value of the contracts, etc.

Alternatively, you might tell me that it is 400%, representing the fact that your PL is now 4 times larger at time 2 than it was at time 1.
In such cases with both positive numbers I use this formula:
pct_chg = endval / startval * 100 - 100
ie. for the above case:
pct_chg = 100 / 25 * 100 - 100 = 300 (not 400%)

(ie: You want the percentage gain *in PL*, not the percentage gain on the account.)
I see. But I think in this simple case of just 1 position from t0 to t1 it's both the same:
"percentage gain in PL" and "percentage gain in the account".
I practically mean the account change in percent of such a single position,
but as said with negative numbers at both t0 and t1, where t1 > t0.

And: I wonder if there exist a math formula that covers all cases (incl. negative values) of such percent calculations.
The page(s) about "percentage" at Wikipedia don't even mention any negative numbers, so they were useless in this case. :-(
 
Last edited:
If you record your first measurment at -200 then take a measurment at-100, the negative value has decreased 50%, holmes.
That's the question, Watson! Are you sure? :)
My math says different: +100%, as said in prev posting.
 
Last edited:
Are you sure?
yep.
The water under which this trade exists (the negative value) is half as deep at the second measurement point as it was at the first measurement point.
fitty. lol
Secondly, of what value is knowing this redonkulous measure in the first place.
 
Back
Top