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This question is for the math guru's

Discussion in 'Strategy Development' started by Sky123987, Nov 11, 2007.

1. Sky123987

Assumptions:
1)Each tick is .01 and it a 50/50 shot to do up vs down.

Problem 1.
1) If the average range is .25 for a 2 minute time span from 10:30 to 10:32, what is the average range from say 10:31 to 10:32, or ultimately I'd like to create an equation to return the average range from a time interval within a larger time interval that has a known average range.

New_AR = CalcTheAR(int SecondsRemaining, int Seconds, double AR){
}

So say I'd like to know the average range from 10:30:45 to 10:32.
My imputs would be...
New_ATR = CalcTheATR(75, 120, .25){
}

I'm really having a hard time coming up with an equation for this.

Problem 2.
If the average range from 10:30 to 10:32 is .25. Say at 10:30:00 the price is 60.00. I would like to know the percentage of the time price reaches 60.50 or greater from 10:30 to 10:32, or really an equation to find this out.

Percent_Reaching = CalcThePercent(double StartPrice, double AR, double Target){
}

or...
Percent_Reaching = CalcThePercent(60.00, .25, 60.50){
}

2. syswizard

To determine probability, You need the DISTRIBUTION of the range for the selected interval....not just the AVERAGE.
Best to get tons of historical data, and determine the distribution of the range. Once you do this, then FIT the distribution to a binomial, normal, guassian or other distribution curve.
Once you do this, then you can employe an equation that will provide the requisite probability.

3. gnome

Or, you could just look at a chart.

4. syswizard

True. But that would not be a "quant" approach.

5. Sky123987

thanks for the advice guys

6. neke

Since the moves at an infinitesimal time interval ( say 1sec) is random, the average over a time-frame is proportional to the square root of the interval (will require a lot of mathematics to prove, you could get something similar in a good options valuation paper). So for the first problem, this should solve your problem:

New_AR = sqrt(SecondsRemaining / Seconds) * AR

For the second problem ((60,0.25,60.50) , I probably need to spend more time on the distribution, but I believe a simple approximation would be to treat it like a normal distribution with a mean of 0.0 and standard deviation of 0.25, in which case you are looking at the probability of having a move of 2 standard deviations to the plus side of the mean. From normal distribution tables (which you can lookup from your program), that comes to about 2.5%. Like I side I think the normal distribution would be a good guess until one can look this more carefully.

7. neke

As an addition to what was earlier said, yes the second should be a normal distribution, but you need to calculate the standard deviation instead of the average you got. Your average is actually called the mean deviation: sigma(abs(d))/n
where d is difference between endpoint and startpoint for each sample data
n is the number of observations in your sample.

To get standard deviation do: sqrt(sigma(d*d)/n)

If it's 50/50 the average range is 0

9. lately

I must be confused.

Why wouldn't you just use the same equation you used to figure out the average range for 10:31 to 10:32? I don't understand why it matters if it's in a larger range if it's not going to affect anything.

10. MarkBrown

i have done this - you have to run the indicator on a tick based chart where the indicator is plotted on time basis.

#10     Nov 13, 2007
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