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Re: MultinormalDistribution Question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg120177] Re: MultinormalDistribution Question
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Mon, 11 Jul 2011 06:58:36 -0400 (EDT)
  • Reply-to: hanlonr at cox.net

Clear[x, y];

mean1 = 58/10;
sigma1 = 2/10;

mean2 = 53/10;
sigma2 = 2/10;

rho = 6/10;

dist = MultinormalDistribution[{mean1, mean2},
   {{sigma1^2, rho*sigma1*sigma2},
    {rho*sigma1*sigma2, sigma2^2}}];

plot1 = Plot3D[PDF[dist, {x, y}],
  {x, mean1 - 3 sigma1, mean1 + 3 sigma1},
  {y, mean2 - 3 sigma2, mean2 + 3 sigma2},
  PlotRange -> All]

m = Integrate[y*PDF[dist, {63/10, y}],
  {y, -Infinity, Infinity}]

(14*Sqrt[2/Pi])/E^(25/8)

m // N

0.490792

Alternatively,

m == Expectation[y*DiracDelta[x - 63/10], 
  Distributed[{x, y}, dist]]

True

s = Sqrt[Integrate[(y - m)^2*PDF[dist, {63/10, y}],
    {y, -Infinity, Infinity}]];

s // N

1.51329

Alternatively,

s == N[Sqrt[Expectation[(y - m)^2*DiracDelta[x - 63/10], 
    Distributed[{x, y}, dist]]]]

True


Bob Hanlon

---- Steve <s123 at epix.net> wrote: 

=============
Hello,

Can someone help me with this ?

I have 2 normal distributions; dist1 describes x and dist2 describes
y. Each are fully defined and are correlated to one another by the
correlation coefficient. How can I detemine the mean and standard
deviation of the expected normal distribution that is associated with
a given x value from dist1 ?

An example:
mean1 = 5.8
sigma1 =0 .2

mean2 = 5.3
sigma2 = 0.2

Correlation Coefficient, rho = 0.6

Given an x value of 6.3 (from dist1) what is the corresponding mean
and standard deviation of y ?

I can view the combined density function from the following:

Mu = {mean1, mean2}
CapSigma = {{sigma1^2, rho*sigma1*sigma2} , {rho,  rho*sigma1*sigma2}
dist = MultinormalDistribution[Mu,CapSigma]
pdf = PDF[dist,{x,y}]
plot1 = Plot3D[pdf, {x,4,7},{y,4,7}, PlotRange->All]

but can't see how to determine the mean and the standard deviation of
y for a given value of x, like 6.3

Any help would be appreciated.

Thanks,

--Steve




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