Estimate overall joint probabilitiy given pairwise joint probabilities
$begingroup$
The true random variables are continuous. But I only have discrete probability distributions as estimation of the continuous random variables.
The following are known.
- $P(X_{i}=x_{i})$
- $P(X_{i}=x_{i}, X_{j}=x_{j})$
Suppose the overall joint probability $P(vec{X} = vec{x})$ is a multivariate normal distribution.
How to estimate the overall joint probability as a discrete probability distribution?
probability probability-theory probability-distributions
asked Dec 29 '18 at 16:35
R zuR zu
33910
$endgroup$
add a comment |
$begingroup$
The true random variables are continuous. But I only have discrete probability distributions as estimation of the continuous random variables.
The following are known.
- $P(X_{i}=x_{i})$
- $P(X_{i}=x_{i}, X_{j}=x_{j})$
Suppose the overall joint probability $P(vec{X} = vec{x})$ is a multivariate normal distribution.
How to estimate the overall joint probability as a discrete probability distribution?
probability probability-theory probability-distributions
asked Dec 29 '18 at 16:35
R zuR zu
33910
$endgroup$
$begingroup$
I can get the estimation by expectation maximization. I want to know a more direct way though. Can moment-generating function do this?
$endgroup$
– R zu
Dec 29 '18 at 16:39
$begingroup$
Only the 1st and 2nd moments of single variable normal distribution are nonzero. I guess multivariate normal is the same.
$endgroup$
– R zu
Dec 29 '18 at 16:44
add a comment |
$begingroup$
The true random variables are continuous. But I only have discrete probability distributions as estimation of the continuous random variables.
The following are known.
- $P(X_{i}=x_{i})$
- $P(X_{i}=x_{i}, X_{j}=x_{j})$
Suppose the overall joint probability $P(vec{X} = vec{x})$ is a multivariate normal distribution.
How to estimate the overall joint probability as a discrete probability distribution?
probability probability-theory probability-distributions
asked Dec 29 '18 at 16:35
R zuR zu
33910
$endgroup$
The true random variables are continuous. But I only have discrete probability distributions as estimation of the continuous random variables.
The following are known.
- $P(X_{i}=x_{i})$
- $P(X_{i}=x_{i}, X_{j}=x_{j})$
Suppose the overall joint probability $P(vec{X} = vec{x})$ is a multivariate normal distribution.
How to estimate the overall joint probability as a discrete probability distribution?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Dec 29 '18 at 16:35
R zuR zu
33910
asked Dec 29 '18 at 16:35
R zuR zu
33910
edited Dec 29 '18 at 19:25
R zu
asked Dec 29 '18 at 16:35
R zuR zu
33910
asked Dec 29 '18 at 16:35
R zuR zu
33910
asked Dec 29 '18 at 16:35
R zuR zu
33910
33910
$begingroup$
I can get the estimation by expectation maximization. I want to know a more direct way though. Can moment-generating function do this?
$endgroup$
– R zu
Dec 29 '18 at 16:39
$begingroup$
Only the 1st and 2nd moments of single variable normal distribution are nonzero. I guess multivariate normal is the same.
$endgroup$
– R zu
Dec 29 '18 at 16:44
add a comment |
$begingroup$
I can get the estimation by expectation maximization. I want to know a more direct way though. Can moment-generating function do this?
$endgroup$
– R zu
Dec 29 '18 at 16:39
$begingroup$
Only the 1st and 2nd moments of single variable normal distribution are nonzero. I guess multivariate normal is the same.
$endgroup$
– R zu
Dec 29 '18 at 16:44
$begingroup$
I can get the estimation by expectation maximization. I want to know a more direct way though. Can moment-generating function do this?
$endgroup$
– R zu
Dec 29 '18 at 16:39
$begingroup$
I can get the estimation by expectation maximization. I want to know a more direct way though. Can moment-generating function do this?
$endgroup$
– R zu
Dec 29 '18 at 16:39
$begingroup$
Only the 1st and 2nd moments of single variable normal distribution are nonzero. I guess multivariate normal is the same.
$endgroup$
– R zu
Dec 29 '18 at 16:44
$begingroup$
Only the 1st and 2nd moments of single variable normal distribution are nonzero. I guess multivariate normal is the same.
$endgroup$
– R zu
Dec 29 '18 at 16:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
From wikipedia:
$boldsymbolmu$ is the vector of means.
$boldsymbolSigma$ is the covariance matrix.
$$begin{align}
f_{mathbf X}(x_1,ldots,x_k)
& = frac{expleft(-frac 1 2 ({mathbf x}-{boldsymbolmu})^mathrm{T}{boldsymbolSigma}^{-1}({mathbf x}-{boldsymbolmu})right)}{sqrt{(2pi)^k|boldsymbolSigma|}}
end{align}$$
answered Dec 29 '18 at 21:40
R zuR zu
33910
$endgroup$
$begingroup$
derivation: faculty.math.illinois.edu/~r-ash/Stat/StatLec21-25.pdf
$endgroup$
– R zu
Dec 29 '18 at 21:41
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From wikipedia:
$boldsymbolmu$ is the vector of means.
$boldsymbolSigma$ is the covariance matrix.
$$begin{align}
f_{mathbf X}(x_1,ldots,x_k)
& = frac{expleft(-frac 1 2 ({mathbf x}-{boldsymbolmu})^mathrm{T}{boldsymbolSigma}^{-1}({mathbf x}-{boldsymbolmu})right)}{sqrt{(2pi)^k|boldsymbolSigma|}}
end{align}$$
answered Dec 29 '18 at 21:40
R zuR zu
33910
$endgroup$
$begingroup$
derivation: faculty.math.illinois.edu/~r-ash/Stat/StatLec21-25.pdf
$endgroup$
– R zu
Dec 29 '18 at 21:41
add a comment |
$begingroup$
From wikipedia:
$boldsymbolmu$ is the vector of means.
$boldsymbolSigma$ is the covariance matrix.
$$begin{align}
f_{mathbf X}(x_1,ldots,x_k)
& = frac{expleft(-frac 1 2 ({mathbf x}-{boldsymbolmu})^mathrm{T}{boldsymbolSigma}^{-1}({mathbf x}-{boldsymbolmu})right)}{sqrt{(2pi)^k|boldsymbolSigma|}}
end{align}$$
answered Dec 29 '18 at 21:40
R zuR zu
33910
$endgroup$
$begingroup$
derivation: faculty.math.illinois.edu/~r-ash/Stat/StatLec21-25.pdf
$endgroup$
– R zu
Dec 29 '18 at 21:41
add a comment |
$begingroup$
From wikipedia:
$boldsymbolmu$ is the vector of means.
$boldsymbolSigma$ is the covariance matrix.
$$begin{align}
f_{mathbf X}(x_1,ldots,x_k)
& = frac{expleft(-frac 1 2 ({mathbf x}-{boldsymbolmu})^mathrm{T}{boldsymbolSigma}^{-1}({mathbf x}-{boldsymbolmu})right)}{sqrt{(2pi)^k|boldsymbolSigma|}}
end{align}$$
answered Dec 29 '18 at 21:40
R zuR zu
33910
$endgroup$
From wikipedia:
$boldsymbolmu$ is the vector of means.
$boldsymbolSigma$ is the covariance matrix.
$$begin{align}
f_{mathbf X}(x_1,ldots,x_k)
& = frac{expleft(-frac 1 2 ({mathbf x}-{boldsymbolmu})^mathrm{T}{boldsymbolSigma}^{-1}({mathbf x}-{boldsymbolmu})right)}{sqrt{(2pi)^k|boldsymbolSigma|}}
end{align}$$
answered Dec 29 '18 at 21:40
R zuR zu
33910
answered Dec 29 '18 at 21:40
R zuR zu
33910
answered Dec 29 '18 at 21:40
R zuR zu
33910
answered Dec 29 '18 at 21:40
R zuR zu
33910
33910
$begingroup$
derivation: faculty.math.illinois.edu/~r-ash/Stat/StatLec21-25.pdf
$endgroup$
– R zu
Dec 29 '18 at 21:41
add a comment |
$begingroup$
derivation: faculty.math.illinois.edu/~r-ash/Stat/StatLec21-25.pdf
$endgroup$
– R zu
Dec 29 '18 at 21:41
$begingroup$
derivation: faculty.math.illinois.edu/~r-ash/Stat/StatLec21-25.pdf
$endgroup$
– R zu
Dec 29 '18 at 21:41
$begingroup$
derivation: faculty.math.illinois.edu/~r-ash/Stat/StatLec21-25.pdf
$endgroup$
– R zu
Dec 29 '18 at 21:41
add a comment |
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$begingroup$
I can get the estimation by expectation maximization. I want to know a more direct way though. Can moment-generating function do this?
$endgroup$
– R zu
Dec 29 '18 at 16:39
$begingroup$
Only the 1st and 2nd moments of single variable normal distribution are nonzero. I guess multivariate normal is the same.
$endgroup$
– R zu
Dec 29 '18 at 16:44