So it sounds like we could rephrase this as: For A and B as defined previously what is the distribution of (A | A > B)
Is this correct?
Suppose I have two independent random variables A, and B, which are distributed Normally as follows:
A: N~(MA,VA)
B: N~(MB,VB)
Now consider that I draw many A's and B's and arrange these randomly and pair-wise (A's next to B's).
Having done that, I then identify all those values of A drawn, which are greater than their partnering value of B. Let us call these identified values C.
My question is now this: How is C distributed?
So it sounds like we could rephrase this as: For A and B as defined previously what is the distribution of (A | A > B)
Is this correct?
I don't have emotions and sometimes that makes me very sad.
Dason, that is correct.
Lets hope that someone else chimes in, because I am slightly out of my element here, but my inclination would be:
Bin(0.5,n)
I guess it maybe called a "random-truncated normal distribution".
The cumulative distribution function can be calculated by definition as usual:
Since and they are independent,
we have
and thus the denominator
For the numerator,
which at least can be computed numerically.
If you seek the probability density function, by differentiation,
BigBugBuzzz (09-18-2012), Dason (09-18-2012), hlsmith (09-18-2012)
Thank you BGM for taking the time. I have a few questions about your solution:
1) What is a? Dason noted that my problem was to find the distribution of (A | A > B) which doesn't include a.
2) What is phi?
3) What is capital phi?
(Is this a new distribution?!)
Cheers
a is the value of A under consideration. Since he was specifying a density "a" is the input value for the pdf.
phi is the density of the standard normal distribution. Capital phi is the CDF of the standard normal distribution. Is the result a new distribution? It doesn't have a name (that I know of) but I'm sure somebody else has thought of it before.
I don't have emotions and sometimes that makes me very sad.
BigBugBuzzz (09-18-2012)
Okay, that makes more sense. However, then my question is, do phi and PHI have particular means and variances?
(Oh, perhaps I already know the answer... Mean = muA - muB, and VAR = sig^2A + sig^2B ?)
Looking at the first part of fc(a), if muA and muB are equal, then we get division by zero, no? Can that be right?
I am not so sure about BGM's solution. I have tried to examine the result using Mathematica, but I am probably doing something wrong. My Mathematica code is shown below:
\[Mu]1 = 11
\[Mu]2 = 10
\[Mu] = \[Mu]1 - \[Mu]2
\[Sigma]1 = 1
\[Sigma]2 = 5
\[Sigma] = \[Sigma]1 + \[Sigma]2
\[CapitalEta] = (\[Mu] /(Sqrt[\[Sigma]1^2 + \[Sigma]2^2]))^-1
\[CapitalRho] = ((x - \[Mu]2)/\[Sigma]2)*(1/\[Sigma]1)
\[CapitalNu] = ((x - \[Mu]1)/\[Sigma]1 )
Plot[CDF[NormalDistribution[\[Mu], \[Sigma]], x]*\[CapitalEta]*
CDF[NormalDistribution[\[Mu], \[Sigma]], x]*\[CapitalRho]*
PDF[NormalDistribution[\[Mu], \[Sigma]], x]*\[CapitalNu], {x, -100,
100}]
Any ideas where I am making a mistake?
if muA and muB are equal, then we get [0/Sqrt(sig^2A+sig^2B)]^-1
that is, 0^-1 ...
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