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Old 2010-03-22, 23:16   #1
CRGreathouse
 
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Default Interval calculations with a given alpha

A rectangular prism is measured to have side lengths 2 ± 1, 6 ± 1, and 8 ± 1. If the actual values are uniformly distributed in their respective ranges (independent of the other values), what is the 95% confidence interval for the volume of the prism?


I'd like to consider generalizations of the above problem -- though I must admit even the above is difficult for me. (I've largely avoided statistics and haven't taken a single [college] class in probability theory.)

Given measurements ([a_0, a_1], [b_0, b_1], ...) and tolerance α, find the interval [s_0, s_1] for the calculation f([a_0, a_1], [b_0, b_1], ...) where the probability that f is less than s_0 is α/2 and the probability that f is greater than s_1 is α/2.

I was thinking of
  • Simple calculations like f(x, y) = xy, f(x, y, z) = xyz, f(x, y) = x + y, f(x, y) = x^2 y, ...
  • Variables are nonnegative reals
  • Measurements uniformly distributed in their respective intervals
Other reasonable choices would be interesting as well.

Does anyone know how to work with a system like this? It seemed easy at first, but I'm quickly finding that it's more complicated than I expected. Is this well-known? Is there a name I can search for?

Last fiddled with by CRGreathouse on 2010-03-22 at 23:19
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Old 2010-03-23, 00:59   #2
davieddy
 
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I hope these help:

Mean of the sum is the sum of the means.

Variance of the sum is the sum of the variances.

Central Limit Theorem.

Small increments.

That is how physicists approach combining
random errors.

David
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Old 2010-03-23, 02:44   #3
CRGreathouse
 
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I was thinking more like integrals over the indifference curve between the variables. Something more refined than taking

[a0, a1] * [b0, b1] = [a0 * b0, a1 * b1]

which corresponds to α = 0 above.

Last fiddled with by CRGreathouse on 2010-03-23 at 02:46
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Old 2010-03-23, 11:31   #4
R.D. Silverman
 
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Quote:
Originally Posted by CRGreathouse View Post
A rectangular prism is measured to have side lengths 2 ± 1, 6 ± 1, and 8 ± 1. If the actual values are uniformly distributed in their respective ranges (independent of the other values), what is the 95% confidence interval for the volume of the prism?


I'd like to consider generalizations of the above problem -- though I must admit even the above is difficult for me. (I've largely avoided statistics and haven't taken a single [college] class in probability theory.)

Given measurements ([a_0, a_1], [b_0, b_1], ...) and tolerance α, find the interval [s_0, s_1] for the calculation f([a_0, a_1], [b_0, b_1], ...) where the probability that f is less than s_0 is α/2 and the probability that f is greater than s_1 is α/2.

I was thinking of
  • Simple calculations like f(x, y) = xy, f(x, y, z) = xyz, f(x, y) = x + y, f(x, y) = x^2 y, ...
  • Variables are nonnegative reals
  • Measurements uniformly distributed in their respective intervals
Other reasonable choices would be interesting as well.

Does anyone know how to work with a system like this? It seemed easy at first, but I'm quickly finding that it's more complicated than I expected. Is this well-known? Is there a name I can search for?

Let x be an r.v. uniformly distributed on [x1, x2], y be a u.r.v. on [y1, y2],
and z ba a u.r.v. on [z1,z2].

Start by determining the following: if u and v are uniform r.v.'s what is the
density function for u*v? Use this twice to determine the density function for
x*y*z. Now a simple integration will yield a confidence interval. I don't
recall ever seeing a pdf for the product of two uniform rv's. I would look
in Feller, Hogg & Craig, or Raiffa, Schleiffer. You might find it in there.
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Old 2010-03-23, 13:49   #5
CRGreathouse
 
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Thanks. That's the path I was following. I may look up those references; those might be helpful.
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Old 2010-03-24, 20:56   #6
davieddy
 
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I think the simplest example is also the most Eddyfying.

Let u,v,w be uniformly distributed between -1 and +1.
(variance 1/3).

The pdf for x=u+v is
1/2 - x/4 for 0<x<2 and zero for x>2 and obviously symmetrical
about x = 0.

The pdf for x=u+v+w is
(3 - x2)/8 for 0<x<1
(x-3)2/16 for 1<x<3

This bears a remarkable resemblance to e-x[sup]2/2[/sup]/sqrt(2pi)
(both have variance 1 and points of inflection at x=+/- 1).

probability of -1<x<1 = 2/3
probability of -2<x<2 = 23/24

Is it any wonder that we "believe" in the Normal Distribution?

David
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Old 2010-03-28, 14:09   #7
R.D. Silverman
 
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Quote:
Originally Posted by CRGreathouse View Post
A rectangular prism is measured to have side lengths 2 ± 1, 6 ± 1, and 8 ± 1. If the actual values are uniformly distributed in their respective ranges (independent of the other values), what is the 95% confidence interval for the volume of the prism?


I'd like to consider generalizations of the above problem -- though I must admit even the above is difficult for me. (I've largely avoided statistics and haven't taken a single [college] class in probability theory.)

Given measurements ([a_0, a_1], [b_0, b_1], ...) and tolerance α, find the interval [s_0, s_1] for the calculation f([a_0, a_1], [b_0, b_1], ...) where the probability that f is less than s_0 is α/2 and the probability that f is greater than s_1 is α/2.

I was thinking of
  • Simple calculations like f(x, y) = xy, f(x, y, z) = xyz, f(x, y) = x + y, f(x, y) = x^2 y, ...
  • Variables are nonnegative reals
  • Measurements uniformly distributed in their respective intervals
Other reasonable choices would be interesting as well.

Does anyone know how to work with a system like this? It seemed easy at first, but I'm quickly finding that it's more complicated than I expected. Is this well-known? Is there a name I can search for?
I found this:

> A web search turns up that there's no neat way of obtaining the
> probability density function of a the product of several uniformly-
> distributed random variables. Basically you have to go to log space
> and treat is as a sum (which approximates the normal distribution as N
> increases).
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Old 2010-03-28, 15:05   #8
davieddy
 
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Quote:
Originally Posted by R.D. Silverman View Post
I found this:

> A web search turns up that there's no neat way of obtaining the
> probability density function of a the product of several uniformly-
> distributed random variables. Basically you have to go to log space
> and treat is as a sum (which approximates the normal distribution as N
> increases).
Hmm..
Terrible echo around here.

David
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Old 2010-03-28, 23:07   #9
CRGreathouse
 
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Quote:
Originally Posted by R.D. Silverman View Post
> A web search turns up that there's no neat way of obtaining the
> probability density function of a the product of several uniformly-
> distributed random variables. Basically you have to go to log space
> and treat is as a sum (which approximates the normal distribution as N
> increases).
Yeah, I was afraid of that.
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Old 2010-03-29, 14:43   #10
R.D. Silverman
 
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Quote:
Originally Posted by CRGreathouse View Post
Yeah, I was afraid of that.
More info (from Prof. Ray Vickson)

"Interestingly, getting the
distribution of the square is easy, but not of the product; the latter
can be done but is not easy and is the subject of research papers. "


"square" refers to x^2 where x is uniform.
"product" refers to "xy", where x and y are independent uniforms.
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Old 2010-04-09, 06:23   #11
davieddy
 
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Quote:
Originally Posted by R.D. Silverman View Post
More info (from Prof. Ray Vickson)

"Interestingly, getting the
distribution of the square is easy, but not of the product; the latter
can be done but is not easy and is the subject of research papers. "


"square" refers to x^2 where x is uniform.
"product" refers to "xy", where x and y are independent uniforms.
Interesting?
The "problems" are distinct and both trivial:

x uniform between a1 and a2
y uniform between b1 and b2

1) s=x^2
pdf(s) = (dx/ds)*pdf(x) = 1/(2*sqrt(s)*(a2-a1)) for a1^2 < s < a2^2

2) s'=xy
Probability(s'<s) is the fraction of the rectangle lying
under the curve y = s/x.

David
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