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Science Forum Index » Statistics - Math Forum » A multivariate normal distribution question
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Message |
| smartnose |
 Posted: Sun Apr 27, 2008 5:25 am |
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Guest
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Question: x1, x2, x3 are I.I.D. samples from normal distribution
N(0,cosi^2).
y1=x1+x2, y2=x2+x3, then what is the conditional distribution of y1
given y2, i.e. p(y1|y2)?
I work out two solutions, but they contradicts. Should you have time
and
interest, please help me with this.
Solution1:
Given y2=a, then x2=a-x3, so y1=x1+a-x3. It's easy to
verify that y1 is still a normal distribution
N(a,2*cosi^2).
Solution2:
vector [x1 x2 x3]' is a multivariate normal distribution and its mean
is [0, 0, 0]' and covariance:
| cosi^2 0 0 |
S= | 0 cosi^2 0 |
| 0 0 cosi^2 |
vector [y1 y2]'= A * [x1 x2 x3]'
A is a matrix
| 1 1 0 |
| 0 1 1 |
So, according to the theorem (Mardia, Multivariate Analysis, 3.2.1,
3.2.4) , a linear transformation of multivariate normal distribution
is still a normal distribution and it's mean and variance matrix can
be obtained by:
mean=A* [0,0,0]'=[0,0,0]';
variance=A*S*A'
| 2* cosi^2 cosi^2 |
= | cosi^2 2*cosi^2|
And given y2=a, y1 is still a normal distribution, and the mean and
covariance is
mean = u1 + Sigma21 * invert(Sigma 22 ) * (a -u 2) = 0 + 1/2 * (a
-0)=1/2 a?????? which is not equal to solution 1???
And the same problem happened with the variance matrix.
How could this happen? And which solution is right? |
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| Paul Rubin |
| Posted: Sun Apr 27, 2008 12:57 pm |
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smartnose wrote:
Quote: Question: x1, x2, x3 are I.I.D. samples from normal distribution
N(0,cosi^2).
y1=x1+x2, y2=x2+x3, then what is the conditional distribution of y1
given y2, i.e. p(y1|y2)?
I work out two solutions, but they contradicts. Should you have time
and
interest, please help me with this.
Solution1:
Given y2=a, then x2=a-x3, so y1=x1+a-x3. It's easy to
verify that y1 is still a normal distribution
N(a,2*cosi^2).
Solution2:
vector [x1 x2 x3]' is a multivariate normal distribution and its mean
is [0, 0, 0]' and covariance:
| cosi^2 0 0 |
S= | 0 cosi^2 0 |
| 0 0 cosi^2 |
vector [y1 y2]'= A * [x1 x2 x3]'
A is a matrix
| 1 1 0 |
| 0 1 1 |
So, according to the theorem (Mardia, Multivariate Analysis, 3.2.1,
3.2.4) , a linear transformation of multivariate normal distribution
is still a normal distribution and it's mean and variance matrix can
be obtained by:
mean=A* [0,0,0]'=[0,0,0]';
variance=A*S*A'
| 2* cosi^2 cosi^2 |
= | cosi^2 2*cosi^2|
And given y2=a, y1 is still a normal distribution, and the mean and
covariance is
mean = u1 + Sigma21 * invert(Sigma 22 ) * (a -u 2) = 0 + 1/2 * (a
-0)=1/2 a?????? which is not equal to solution 1???
And the same problem happened with the variance matrix.
How could this happen? And which solution is right?
You might want to think about the following question: What is the
conditional distribution of x3 given that x2 + x3 = a?
/Paul |
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| Jack Tomsky |
| Posted: Mon Apr 28, 2008 7:21 pm |
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Guest
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Quote: Question: x1, x2, x3 are I.I.D. samples from normal
distribution
N(0,cosi^2).
y1=x1+x2, y2=x2+x3, then what is the conditional
distribution of y1
given y2, i.e. p(y1|y2)?
I work out two solutions, but they contradicts.
Should you have time
and
interest, please help me with this.
Solution1:
Given y2=a, then x2=a-x3, so y1=x1+a-x3. It's easy to
verify that y1 is still a normal distribution
N(a,2*cosi^2).
Solution2:
vector [x1 x2 x3]' is a multivariate normal
distribution and its mean
is [0, 0, 0]' and covariance:
| cosi^2 0 0 |
S= | 0 cosi^2 0 |
| 0 0 cosi^2 |
vector [y1 y2]'= A * [x1 x2 x3]'
A is a matrix
| 1 1 0 |
| 0 1 1 |
So, according to the theorem (Mardia, Multivariate
Analysis, 3.2.1,
3.2.4) , a linear transformation of multivariate
normal distribution
is still a normal distribution and it's mean and
variance matrix can
be obtained by:
mean=A* [0,0,0]'=[0,0,0]';
variance=A*S*A'
| 2* cosi^2 cosi^2 |
= | cosi^2 2*cosi^2|
And given y2=a, y1 is still a normal distribution,
and the mean and
covariance is
mean = u1 + Sigma21 * invert(Sigma 22 ) * (a -u 2) =
0 + 1/2 * (a
-0)=1/2 a?????? which is not equal to solution 1???
And the same problem happened with the variance
matrix.
How could this happen? And which solution is right?
We start off with x' = (x1, x2, x3) ~ N[O, S].
S = (cosi^2)*I, where I is the 3 by 3 identity matrix.
Let y' = (y1, y2).
Then y = Ax, where A is 2 by 3 and the first row is (1, 1, 0) and the second row is (0, 1, 1).
y ~ N[0, (cosi^2)*AA'].
AA' is 2 by 2, whose first row is (2, 1) and whose second row is (1, 2). Call AA' = B.
So far, this matches up with your second approach. Note that b12 and b22 are each scalars.
y1|y2 is univariate normal with mean and variance,
E(y1|y2) = mu1 + (b12/b22)*(y2-mu2) = y2/2
Var(y1|y2) = (cosi^2)*[b11 - b12*b12/b22] = (3/2)*(cosi^2).
Jack |
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