MONTE CARLO SIMULATION NEEDED
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- Last Post 06 April 2012
That is akin to saying a nail is needed, only worse.
There an infinite number monte carlo simulations that can be done and a near infinite number have been done.
To what end? I wrote my own spreadsheet monte carlo simulation that I call Random Cannon that simulates a rifle shooting with given degrees of variance and other factors.
But having a simulator and knowing how to use it, are two very different things. Using one properly requires an understanding of hypothesis testing and probability theory, along with statistical analyses.
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I'm not sure what “pairs of shots” are about.
Mean is average; if the average distance from center is zero, the mean is zero. Moreover, given variance = 1, the standard deviation is 1 -- because standard deviation is the square root of variance.
Ed Harris had something useful to say about, IIRC, dispersion -- which, i think, is what you're after. (Ed may have used a different term that's familiar to manufacturers; I remember dispersion because that's a machine gun term.)
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Ed, Looks like Joe is trying to reinvent statistics - again.
Joe, there are 100 shots - pair 'em up and go to it.
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We have X and Y distributed Normal with mean zero and variance =1.
I'm taking X and Y as horizontal and vertical dispersion.
Think of X. Shots average X = 0, some +1, some -1 some -.412, some + .006-the mean is zero. Same with Y. The 3 dimensional representation of probability is a mountain looking shape. Make sense?
"Frequency distribution” might be the term you're after --
like if you placed a penny on a target indicating where
each shot fell.
Simulate a pair of shots, 2 shots. Calculate the distance between them.
Shot 1 at (1,1). Shot 2 at (-3, 6). X1-X2 = 1-(-3) = -2.
Y1-Y2 = 1-6 = -5. Squares are -2^2 = 4, -5^2 = 25. Distance between = (4+25)^(1/2) = 29^(1/2).
I don't understand what's special about a pair of shots.
I hope you're not thinking “If the first shot is high and right, chances are the next shot will be low and left" or “Since the last two shots were far apart, the next two will be close together."
Do it again, and again. With a large number of results, calculate the mean and variance of the distribution of distances.
Make sense?
joe b.
If you take 100 shots as pairs, all you've got is 50
samples of 2 shots. You can calculate a measure of
central tendency for each sample -- but the average
of those 50 statistics is not the same as the statistic
for a sample of size 100.
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No, we discussed this before and mathematical statistics and the rules of independent events still applies now just as it did then.
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Joe, You have a habit of not listening. Sadly, you repeat that over and over. You can't reinvent a new math because math is what it is. It won't change the need for independence anymore than 2+2 will one day become 5. Why do you not understand that?
You NEED a class. Badly.
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Here are 50 pairs of values
(-0.64, -0.66), ( 1.02, -0.39) ( 2.12, 0.49), (-0.10, 0.43) (-0.40, 0.13), (-0.21, 0.72) (-0.37, 0.03), (-0.53, 0.07) ( 0.12, 1.10), ( 0.16, -0.10) ( 0.85, 0.25), ( 0.26, -0.10) (-0.53, 0.24), (-0.04, -0.62) ( 0.33, 1.27), (-0.87, -0.81) (-0.00, -0.01), (-0.36, 0.78) (-1.34, 0.36), ( 0.49, -0.40) ( 0.05, 0.72), ( 0.35, 0.69) ( 0.78, 0.20), (-0.61, -0.50) ( 0.34, -0.12), (-0.35, 0.44) ( 0.49, -0.04), (-1.27, 0.26) (-1.02, 0.38), (-0.59, 0.24) ( 0.96, -0.12), ( 0.45, 0.32) ( 0.83, 0.00), (-0.07, 0.18) (-0.31, -0.31), ( 0.02, -0.53) ( 0.22, -0.52), (-0.11, 0.25) ( 0.58, 0.52), ( 0.43, 1.13) ( 1.05, -0.13), ( 0.58, 0.31) (-0.11, 0.24), ( 0.34, 0.30) (-0.33, -0.68), (-0.17, 0.04) (-0.15, -1.03), ( 0.41, -0.48) ( 0.61, 1.16), (-0.04, -1.38) (-0.14, 0.17), (-0.15, -0.32) (-0.40, 0.90), ( 0.70, -0.54) ( 0.10, -0.88), (-0.60, 0.96) ( 0.20, 0.02), (-0.33, 0.56) ( 0.21, -0.97), ( 0.43, 0.19) ( 0.39, -0.28), (-0.48, -0.07) (-0.48, 0.31), (-0.47, -0.45) (-0.06, -0.84), (-0.68, 0.71) ( 0.63, 0.24), (-0.69, 0.35) (-0.57, -0.01), (-0.09, 0.53) ( 0.18, -0.37), (-0.50, -0.16) (-0.03, -0.45), (-0.95, 0.01) ( 0.29, 0.11), (-0.03, 0.57) ( 0.26, -0.73), ( 1.15, -0.51) ( 0.52, -0.21), (-0.63, 0.16) (-0.01, -0.12), (-0.39, -0.31) ( 0.27, -0.00), ( 0.07, -0.43) (-1.01, 0.25), ( 0.72, -0.24) (-0.29, -0.24), ( 0.19, 0.22) ( 0.55, 0.28), (-0.01, 0.93) (-0.42, -0.20), (-0.49, -0.73) ( 0.08, 0.76), ( 0.68, -0.53) ( 0.17, -0.14), ( 0.66, 0.72) (-0.32, -0.65), ( 0.43, -0.14) ( 0.89, 0.86), ( 0.58, -1.04)
Does that help?
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You have related the “expertise” of the dutchman before. That didn't work out well at all.
You new help won't either if you don't get around to understanding elementary concepts.
BTW, bullet holes are distributed in a Rayleigh distribution, not bivariate normal. Not that it matters much.
You are right that I do not know much about stats and probability theory. But it is still an infinite amount more than what you believe you know (just about all of which is completely wrong).
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I thought you said you wanted 100 shots with horizontal and vertical dispersion normally distributed about zero with a standard deviation of 1.
I gave you 50 pairs of those.
If you analyze these observations, you'll probably find that the means aren't exactly zero -- but they're close and will get closer as more observations are included in the sample. You will also probably find that the range of dispersion will increase.
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Rayleigh Distribution: good to know.
Not everything's normal, is it?
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Rayleigh Distribution: good to know.
Not everything's normal, is it?
If you ever find a way to normalize a Rayleigh variable, please pass it on. I searched around a bit at one time but never found anything. The bi-variate normal is probably way more than adequate however.
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No, this doesn't help. What I need are the answers to the questions.
Distribution-Normal?
Arithmetic mean
Standard deviation/variance
N = ?500,000? or some such number.
Thanks;
joe b.
I looked around a bit, trying to figure out why pairs of shots are significant,
and came across
Now I understand: measuring the distance between 10 shot pairs is much easier than measuring the size of a 20-shot group.
JoeB, are you trying to test BJD's approach by applying his technique to two samples having the same characteristics and testing whether a false positive results -- i.e., his technique incorrectly asserts significant difference between samples?
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