Math question, crosspost from Probability

[Magic_Man]

I posted this in probability, but no one answered yet, so I thought I'd try here.

Given three normally distributed random variables, A, B, and C, with means uA, uB, uC, and standard deviations sA, sB, sC:

What is the probability that A=a is the smallest? I tried:
p(A is smallest) = p(A<B) AND p(A<C),
where p(A<B) is cdf((uB-uA)/sqrt(sA^2 + sB^2)),
cdf being the normal cumulative distribution function. That is, p(A<B) is the probability that (A-B) is less than 0.

However, I don't think that p(A<B) and p(A<C) are independent, so the AND above would be:
p(A is smallest) = p( (A<B)|(A<C)) * p(A<C)

and I don't know how to calculate the first term above. Am I going about this all wrong? Thanks for your help.

~MagicMan

[arahant]

[ QUOTE ]
I posted this in probability, but no one answered yet, so I thought I'd try here.

Given three normally distributed random variables, A, B, and C, with means uA, uB, uC, and standard deviations sA, sB, sC:

What is the probability that A=a is the smallest? I tried:
p(A is smallest) = p(A<B) AND p(A<C),
where p(A<B) is cdf((uB-uA)/sqrt(sA^2 + sB^2)),
cdf being the normal cumulative distribution function. That is, p(A<B) is the probability that (A-B) is less than 0.

However, I don't think that p(A<B) and p(A<C) are independent, so the AND above would be:
p(A is smallest) = p( (A<B)|(A<C)) * p(A<C)

and I don't know how to calculate the first term above. Am I going about this all wrong? Thanks for your help.

~MagicMan

[/ QUOTE ]

Just to be clear on the question...a is an instance of a variable with distribution A? And are you looking for P(a<b and a<c), where b and c are instances of variables with the other distributions?

You're confusing me by putting in a single lowercase...

[Magic_Man]

[ QUOTE ]
[ QUOTE ]
I posted this in probability, but no one answered yet, so I thought I'd try here.

Given three normally distributed random variables, A, B, and C, with means uA, uB, uC, and standard deviations sA, sB, sC:

What is the probability that A=a is the smallest? I tried:
p(A is smallest) = p(A<B) AND p(A<C),
where p(A<B) is cdf((uB-uA)/sqrt(sA^2 + sB^2)),
cdf being the normal cumulative distribution function. That is, p(A<B) is the probability that (A-B) is less than 0.

However, I don't think that p(A<B) and p(A<C) are independent, so the AND above would be:
p(A is smallest) = p( (A<B)|(A<C)) * p(A<C)

and I don't know how to calculate the first term above. Am I going about this all wrong? Thanks for your help.

~MagicMan

[/ QUOTE ]

Just to be clear on the question...a is an instance of a variable with distribution A? And are you looking for P(a<b and a<c), where b and c are instances of variables with the other distributions?

You're confusing me by putting in a single lowercase...

[/ QUOTE ]

Yeah, I knew that would be confusing. I'm not really very experienced at writing cogent probability questions. What I mean is, if I pick one instance of each of the variables, what is the probability that the instance from A is the smallest? I think this translates to:

p(A=a is smallest) = p ( (A=a) < (B=b) ) AND p( (A=a) <(C=c) )

Is that right?

~MagicMan

[jason1990]

http://forumserver.twoplustwo.com/showflat.php?Cat=0&Number=8398520&Main=8392180#Pos t8398520