Determining the probability of your option position being profitable

oxusmorouz

Well-Known Member
#1
Probabilities are hard to define and quantify and hence subjective nature of the elements cannot be completely eliminated. I go with the assumption that the variables follow a normal distribution.
I shall take a simple linear payoff model to explain what I've understood :- Writing a naked call option.

1) Collect sufficient data of prices. [Eg - 500 bars of nifty data]
2) Determine the number of trading days remaining to maturity, and call it as "n".[If you're buying a call option today(22nd), oct expiry (25th), no of days remaining is 3]
3) Calculate "n" day rate of change, in % terms for the sample collected in step 1. [ Calculate 3 day ROC of the 500 bar data available]
4) Determine your break even point, and the % change in price to break even. [If you're selling a 5600 call, with CMP at 5184, % change to achieve break even is 8%]
5) Calculate average ROC for the data available [sum of "3" divided by count of 3, the value is .4971]
6) Calculate standard deviation for the data available in "5".[StDev = 2.71625]
7) z score is defined as (4 - 5)/6 [(8 - .4971)/2.71625] = 2.7622
8) Find the probability of z value, by looking up the normal distribution table http://www.isixsigma.com/library/content/zdistribution.asp.
9) For a z value of 2.7622, the probability is .4971 on the right side of the distribution. Since we are not affected by the fall in prices, probability that the price will be below value in step 5 is 0.5

We can, thus conclude, that if we write a 5600 nifty call, probability of our trade being successful would be .5 + .4971 = .9971 or 99.71%.

Although this example is for a linear payoff, probability of success for a non-linear payoff can be calculated using a two tailed test.
Suggested reading : http://en.wikipedia.org/wiki/Normal_distribution

Regards.

Metastock formula for automating the above example:
x:= Input("No of trading days to maturity",1,10000,25);
y:= Input("% to Breakeven",-1000,1000,0);
a:= ROC(C,x,%);
b:= ExtFml( "JRS_MSX.Stdev",a,Cum(1) - x,1);
d:= ExtFml( "JRS_MSX.MOV",a,Cum(1) - x,S);
f:= LastValue(d);
g:= LastValue(b);
i:= (y - f)/g;
i

Required download: JRS.MSX.dll (or forum20.dll) available in the metastock forum.
 

ghosh_ak34

Well-Known Member
#2
Too cumbersum method and still less chances for success.
Please provide some simple stuff. may be an excel withy macros will be of help.

Regards
 
#3
Too cumbersum method and still less chances for success.
Please provide some simple stuff. may be an excel withy macros will be of help.

Regards
Hello ghosh

How did you find out that the method has less chance of success??

Hello oxusmorouz

Not too mathemetically inclined, nither know much abt programming etc, so plz bear with me on that.

From what i could understand you are effectively calculating the probability of the index not being above a certain level, on the day of expiry.

In the above case the probability of index closing below 5600 is calculated to be 99.71%, similarly you could have also calculated the probability of lets say 5400/5500/5550 so on.

In effect can this method be used to forecast the index on expiry??

Could you calculate using the same data the probability for closing above/below 5500 and 5400 it would be interesting, now that we know where it closed.

Thanks
nb
 
Last edited:

chintan786

Well-Known Member
#5
excellent Oxy... dude u are too good ....... wht book u prefer..... is this Levin and robin...

those who really intersted can buy this book of QA ( qualitative Analysis)...


Oxy... great effort dear...... if possible plz through some light on ECONOMETRICS also...


chintan:)
 

oxusmorouz

Well-Known Member
#7
From what i could understand you are effectively calculating the probability of the index not being above a certain level, on the day of expiry.
Yes.

In effect can this method be used to forecast the index on expiry??
No, but it would give one an idea about to the chances of profitability, after the directional call is made.

Could you calculate using the same data the probability for closing above/below 5500 and 5400 it would be interesting, now that we know where it closed.
Sure. Taking the same data, the z score was 1.94, or a probability of 97.5% for a close above 5500 in 3 days time. For 5400, the probability of a close above it in 3 days time was approx 86%. I must point out that a single day gain of 5% made a big difference, and after the event, probability of a close below 5400 dwindled to 23.6% ;)
 

oxusmorouz

Well-Known Member
#8
excellent Oxy... dude u are too good ....... wht book u prefer..... is this Levin and robin...
those who really intersted can buy this book of QA ( qualitative Analysis)...
Oxy... great effort dear...... if possible plz through some light on ECONOMETRICS also...
chintan:)
Thanks Chintan. I'll post if I come across something good.
 
U

uasish

Guest
#10
Ajay,

How could i miss this Thread !! now-a-days i am getting sloopy.Thks for your effort.

Asish
 

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