20100322, 23:16  #1 
Aug 2006
5987_{10} Posts 
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
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 wellknown? Is there a name I can search for? Last fiddled with by CRGreathouse on 20100322 at 23:19 
20100323, 00:59  #2 
"Lucan"
Dec 2006
England
194A_{16} Posts 
Hints
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 
20100323, 02:44  #3 
Aug 2006
5,987 Posts 
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 20100323 at 02:46 
20100323, 11:31  #4  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}·1,877 Posts 
Quote:
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. 

20100323, 13:49  #5 
Aug 2006
1763_{16} Posts 
Thanks. That's the path I was following. I may look up those references; those might be helpful.

20100324, 20:56  #6 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
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  x^{2})/8 for 0<x<1 (x3)^{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 
20100328, 14:09  #7  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}×1,877 Posts 
Quote:
> 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). 

20100328, 15:05  #8  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
Terrible echo around here. David 

20100328, 23:07  #9  
Aug 2006
5,987 Posts 
Quote:


20100329, 14:43  #10 
"Bob Silverman"
Nov 2003
North of Boston
2^{2}×1,877 Posts 
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. 
20100409, 06:23  #11  
"Lucan"
Dec 2006
England
1100101001010_{2} Posts 
Quote:
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)*(a2a1)) 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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