       Re: MultinormalDistribution Question

• To: mathgroup at smc.vnet.net
• Subject: [mg120178] Re: MultinormalDistribution Question
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Mon, 11 Jul 2011 06:58:46 -0400 (EDT)
• References: <201107100901.FAA24634@smc.vnet.net>

```First, your covariance matrix is not symmetric definite. I think you probably meant:

CapSigma = {{sigma1^2, rho*sigma1*sigma2}, {rho*sigma1*sigma2,
sigma2^2}}

Assuming that, you can get the conditional expectation from the definition:

Integrate[y PDF[dist, {6.3, y}], {y, -Infinity, Infinity}]/
PDF[MarginalDistribution[dist, 1], 6.3]

5.6

Alternatively, you can use Mathematica 8 built in NExpectation function:

NExpectation[y \[Conditioned] 6.299 <= x <= 6.301, {x, y} \[Distributed] dist]
5.6

Once you have the conditional expectation, you can compute the conditional variance, e.g.

Chop[Integrate[(y - 5.6)^2*PDF[dist, {6.3, y}], {y, -Infinity, Infinity}]/PDF[MarginalDistribution[dist, 1], 6.3]]
0.0256

Andrzej Kozlowski

On 10 Jul 2011, at 11:01, Steve 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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