Compute probability of a particular ordering of Multivariate normal distribution
Multi tool use
$begingroup$
There are $m$ normally distributed, independent random variables $N_0,...,N_m$ with distinct means $mu_1, ldots mu_m$ and standard deviations $sigma_1, ldots, sigma_m$. Then, we get a permutation of the numbers ${1, ldots, m}$. How can we efficiently compute, numerically, the (log) probability of observing the random variables in same ordering as this permutation?
An example:
- we have four independent random variables $N1,N2,N3,N4$, all with different means and variances.
- We are given the permutation $(3, 1, 2, 4)$.
- What's $Pr(N3>N1>N2>N4)$?
I can get the answer by calculating the following formula.
$$int_{-infty}^infty int_{n_4}^infty int_{n_2}^infty int_{n_1}^infty p(n_1)p(n_2)p(n_3)p(n_4) dn_3 dn_1 dn_2 dn_4$$
However, it is very difficult to analytically calculate this formula.
How do you think that it is better to calculate such multivariate normal CDF with higher accuracy?
I feel that having variables in this integration range is very troublesome.
I welcome any hints on calculation methods and means.
I do not know the details of the quasi-Monte Carlo method,,, Is it possible to apply quasi-Monte Carlo integration even when including variables in the integration interval like this?
probability integration combinatorics probability-distributions normal-distribution
asked Jan 10 at 7:38
dotcomuserdotcomuser
1
$endgroup$
add a comment |
$begingroup$
There are $m$ normally distributed, independent random variables $N_0,...,N_m$ with distinct means $mu_1, ldots mu_m$ and standard deviations $sigma_1, ldots, sigma_m$. Then, we get a permutation of the numbers ${1, ldots, m}$. How can we efficiently compute, numerically, the (log) probability of observing the random variables in same ordering as this permutation?
An example:
- we have four independent random variables $N1,N2,N3,N4$, all with different means and variances.
- We are given the permutation $(3, 1, 2, 4)$.
- What's $Pr(N3>N1>N2>N4)$?
I can get the answer by calculating the following formula.
$$int_{-infty}^infty int_{n_4}^infty int_{n_2}^infty int_{n_1}^infty p(n_1)p(n_2)p(n_3)p(n_4) dn_3 dn_1 dn_2 dn_4$$
However, it is very difficult to analytically calculate this formula.
How do you think that it is better to calculate such multivariate normal CDF with higher accuracy?
I feel that having variables in this integration range is very troublesome.
I welcome any hints on calculation methods and means.
I do not know the details of the quasi-Monte Carlo method,,, Is it possible to apply quasi-Monte Carlo integration even when including variables in the integration interval like this?
probability integration combinatorics probability-distributions normal-distribution
asked Jan 10 at 7:38
dotcomuserdotcomuser
1
$endgroup$
$begingroup$
One way would be to use a Monte Carlo method (en.wikipedia.org/wiki/Monte_Carlo_method): draw a large number of samples of $(N_0, dots, N_m)$ and then calculate the proportion of these draws that are in the order you're interested in.
$endgroup$
– Alex
Jan 12 at 22:01
add a comment |
$begingroup$
There are $m$ normally distributed, independent random variables $N_0,...,N_m$ with distinct means $mu_1, ldots mu_m$ and standard deviations $sigma_1, ldots, sigma_m$. Then, we get a permutation of the numbers ${1, ldots, m}$. How can we efficiently compute, numerically, the (log) probability of observing the random variables in same ordering as this permutation?
An example:
- we have four independent random variables $N1,N2,N3,N4$, all with different means and variances.
- We are given the permutation $(3, 1, 2, 4)$.
- What's $Pr(N3>N1>N2>N4)$?
I can get the answer by calculating the following formula.
$$int_{-infty}^infty int_{n_4}^infty int_{n_2}^infty int_{n_1}^infty p(n_1)p(n_2)p(n_3)p(n_4) dn_3 dn_1 dn_2 dn_4$$
However, it is very difficult to analytically calculate this formula.
How do you think that it is better to calculate such multivariate normal CDF with higher accuracy?
I feel that having variables in this integration range is very troublesome.
I welcome any hints on calculation methods and means.
I do not know the details of the quasi-Monte Carlo method,,, Is it possible to apply quasi-Monte Carlo integration even when including variables in the integration interval like this?
probability integration combinatorics probability-distributions normal-distribution
asked Jan 10 at 7:38
dotcomuserdotcomuser
1
$endgroup$
There are $m$ normally distributed, independent random variables $N_0,...,N_m$ with distinct means $mu_1, ldots mu_m$ and standard deviations $sigma_1, ldots, sigma_m$. Then, we get a permutation of the numbers ${1, ldots, m}$. How can we efficiently compute, numerically, the (log) probability of observing the random variables in same ordering as this permutation?
An example:
- we have four independent random variables $N1,N2,N3,N4$, all with different means and variances.
- We are given the permutation $(3, 1, 2, 4)$.
- What's $Pr(N3>N1>N2>N4)$?
I can get the answer by calculating the following formula.
$$int_{-infty}^infty int_{n_4}^infty int_{n_2}^infty int_{n_1}^infty p(n_1)p(n_2)p(n_3)p(n_4) dn_3 dn_1 dn_2 dn_4$$
However, it is very difficult to analytically calculate this formula.
How do you think that it is better to calculate such multivariate normal CDF with higher accuracy?
I feel that having variables in this integration range is very troublesome.
I welcome any hints on calculation methods and means.
I do not know the details of the quasi-Monte Carlo method,,, Is it possible to apply quasi-Monte Carlo integration even when including variables in the integration interval like this?
probability integration combinatorics probability-distributions normal-distribution
probability integration combinatorics probability-distributions normal-distribution
asked Jan 10 at 7:38
dotcomuserdotcomuser
1
asked Jan 10 at 7:38
dotcomuserdotcomuser
1
asked Jan 10 at 7:38
dotcomuserdotcomuser
1
asked Jan 10 at 7:38
dotcomuserdotcomuser
1
asked Jan 10 at 7:38
dotcomuserdotcomuser
1
1
$begingroup$
One way would be to use a Monte Carlo method (en.wikipedia.org/wiki/Monte_Carlo_method): draw a large number of samples of $(N_0, dots, N_m)$ and then calculate the proportion of these draws that are in the order you're interested in.
$endgroup$
– Alex
Jan 12 at 22:01
add a comment |
$begingroup$
One way would be to use a Monte Carlo method (en.wikipedia.org/wiki/Monte_Carlo_method): draw a large number of samples of $(N_0, dots, N_m)$ and then calculate the proportion of these draws that are in the order you're interested in.
$endgroup$
– Alex
Jan 12 at 22:01
$begingroup$
One way would be to use a Monte Carlo method (en.wikipedia.org/wiki/Monte_Carlo_method): draw a large number of samples of $(N_0, dots, N_m)$ and then calculate the proportion of these draws that are in the order you're interested in.
$endgroup$
– Alex
Jan 12 at 22:01
$begingroup$
One way would be to use a Monte Carlo method (en.wikipedia.org/wiki/Monte_Carlo_method): draw a large number of samples of $(N_0, dots, N_m)$ and then calculate the proportion of these draws that are in the order you're interested in.
$endgroup$
– Alex
Jan 12 at 22:01
add a comment |
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One way would be to use a Monte Carlo method (en.wikipedia.org/wiki/Monte_Carlo_method): draw a large number of samples of $(N_0, dots, N_m)$ and then calculate the proportion of these draws that are in the order you're interested in.
$endgroup$
– Alex
Jan 12 at 22:01