Mixing
Correlated Chance Constraint
$begingroup$
If we want to satisfy the following constraint with the probability of 1-$epsilon$:
$$
mathcal{P}(a leq x) ge 1 - epsilon
$$
where $x$ is the decision variable, $a$ is a random variable with a normal distribution $ N(mu,sigma^2)$, it is enough to have
$$
x geq mu + sigma Phi^{-1}(1-epsilon)
$$
My question is what we can say about the same problem in a higher dimension, for example 2, where $a_1$ and $a_2$ are correlated.
$$
mathcal{P}(a_1 leq x_1 , a_2 leq x_2 ) ge 1 - epsilon
$$
optimization correlation
asked Jan 29 at 21:16
MahrazMahraz
103
$endgroup$
add a comment |
$begingroup$
If we want to satisfy the following constraint with the probability of 1-$epsilon$:
$$
mathcal{P}(a leq x) ge 1 - epsilon
$$
where $x$ is the decision variable, $a$ is a random variable with a normal distribution $ N(mu,sigma^2)$, it is enough to have
$$
x geq mu + sigma Phi^{-1}(1-epsilon)
$$
My question is what we can say about the same problem in a higher dimension, for example 2, where $a_1$ and $a_2$ are correlated.
$$
mathcal{P}(a_1 leq x_1 , a_2 leq x_2 ) ge 1 - epsilon
$$
optimization correlation
asked Jan 29 at 21:16
MahrazMahraz
103
$endgroup$
$begingroup$
1. where did $mu$ go? 2. can you apply a linear transformation to eliminte the correlation?
$endgroup$
– LinAlg
Jan 29 at 21:29
$begingroup$
@LinAlg $mu$ was missed, it is fixed now.
$endgroup$
– Mahraz
Jan 29 at 22:06
$begingroup$
This problem won't have a closed-form solution in dimension $dgeq2$, even in the uncorrelated case, I don't think. It can be evaluated numerically.
$endgroup$
– David M.
Feb 2 at 14:52
$begingroup$
@DavidM. in uncorrelated case, I think we can simply decouple the problems into several one dimension problem and solve them respectively.
$endgroup$
– Mahraz
Feb 4 at 20:37
$begingroup$
I think you would still end up with a product of CDFs, which I don't think you could invert analytically.
$endgroup$
– David M.
Feb 4 at 20:45
add a comment |
$begingroup$
If we want to satisfy the following constraint with the probability of 1-$epsilon$:
$$
mathcal{P}(a leq x) ge 1 - epsilon
$$
where $x$ is the decision variable, $a$ is a random variable with a normal distribution $ N(mu,sigma^2)$, it is enough to have
$$
x geq mu + sigma Phi^{-1}(1-epsilon)
$$
My question is what we can say about the same problem in a higher dimension, for example 2, where $a_1$ and $a_2$ are correlated.
$$
mathcal{P}(a_1 leq x_1 , a_2 leq x_2 ) ge 1 - epsilon
$$
optimization correlation
asked Jan 29 at 21:16
MahrazMahraz
103
$endgroup$
If we want to satisfy the following constraint with the probability of 1-$epsilon$:
$$
mathcal{P}(a leq x) ge 1 - epsilon
$$
where $x$ is the decision variable, $a$ is a random variable with a normal distribution $ N(mu,sigma^2)$, it is enough to have
$$
x geq mu + sigma Phi^{-1}(1-epsilon)
$$
My question is what we can say about the same problem in a higher dimension, for example 2, where $a_1$ and $a_2$ are correlated.
$$
mathcal{P}(a_1 leq x_1 , a_2 leq x_2 ) ge 1 - epsilon
$$
optimization correlation
optimization correlation
asked Jan 29 at 21:16
MahrazMahraz
103
asked Jan 29 at 21:16
MahrazMahraz
103
edited Jan 29 at 22:05
Mahraz
asked Jan 29 at 21:16
MahrazMahraz
103
asked Jan 29 at 21:16
MahrazMahraz
103
asked Jan 29 at 21:16
MahrazMahraz
103
103
$begingroup$
1. where did $mu$ go? 2. can you apply a linear transformation to eliminte the correlation?
$endgroup$
– LinAlg
Jan 29 at 21:29
$begingroup$
@LinAlg $mu$ was missed, it is fixed now.
$endgroup$
– Mahraz
Jan 29 at 22:06
$begingroup$
This problem won't have a closed-form solution in dimension $dgeq2$, even in the uncorrelated case, I don't think. It can be evaluated numerically.
$endgroup$
– David M.
Feb 2 at 14:52
$begingroup$
@DavidM. in uncorrelated case, I think we can simply decouple the problems into several one dimension problem and solve them respectively.
$endgroup$
– Mahraz
Feb 4 at 20:37
$begingroup$
I think you would still end up with a product of CDFs, which I don't think you could invert analytically.
$endgroup$
– David M.
Feb 4 at 20:45
add a comment |
$begingroup$
1. where did $mu$ go? 2. can you apply a linear transformation to eliminte the correlation?
$endgroup$
– LinAlg
Jan 29 at 21:29
$begingroup$
@LinAlg $mu$ was missed, it is fixed now.
$endgroup$
– Mahraz
Jan 29 at 22:06
$begingroup$
This problem won't have a closed-form solution in dimension $dgeq2$, even in the uncorrelated case, I don't think. It can be evaluated numerically.
$endgroup$
– David M.
Feb 2 at 14:52
$begingroup$
@DavidM. in uncorrelated case, I think we can simply decouple the problems into several one dimension problem and solve them respectively.
$endgroup$
– Mahraz
Feb 4 at 20:37
$begingroup$
I think you would still end up with a product of CDFs, which I don't think you could invert analytically.
$endgroup$
– David M.
Feb 4 at 20:45
$begingroup$
1. where did $mu$ go? 2. can you apply a linear transformation to eliminte the correlation?
$endgroup$
– LinAlg
Jan 29 at 21:29
$begingroup$
1. where did $mu$ go? 2. can you apply a linear transformation to eliminte the correlation?
$endgroup$
– LinAlg
Jan 29 at 21:29
$begingroup$
@LinAlg $mu$ was missed, it is fixed now.
$endgroup$
– Mahraz
Jan 29 at 22:06
$begingroup$
@LinAlg $mu$ was missed, it is fixed now.
$endgroup$
– Mahraz
Jan 29 at 22:06
$begingroup$
This problem won't have a closed-form solution in dimension $dgeq2$, even in the uncorrelated case, I don't think. It can be evaluated numerically.
$endgroup$
– David M.
Feb 2 at 14:52
$begingroup$
This problem won't have a closed-form solution in dimension $dgeq2$, even in the uncorrelated case, I don't think. It can be evaluated numerically.
$endgroup$
– David M.
Feb 2 at 14:52
$begingroup$
@DavidM. in uncorrelated case, I think we can simply decouple the problems into several one dimension problem and solve them respectively.
$endgroup$
– Mahraz
Feb 4 at 20:37
$begingroup$
@DavidM. in uncorrelated case, I think we can simply decouple the problems into several one dimension problem and solve them respectively.
$endgroup$
– Mahraz
Feb 4 at 20:37
$begingroup$
I think you would still end up with a product of CDFs, which I don't think you could invert analytically.
$endgroup$
– David M.
Feb 4 at 20:45
$begingroup$
I think you would still end up with a product of CDFs, which I don't think you could invert analytically.
$endgroup$
– David M.
Feb 4 at 20:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Partial answer (too long for comment). You can evaluate the bivariate normal CDF numerically (using e.g. Python and Scipy, which is what I did). We see that, even in the correlated case, the feasible region is convex, which is a hopeful sign, computationally (basically the CDF of the bivariate normal distribution is quasi-convex).
For these numerical examples, I took $mu_1=mu_2=0$, $sigma_1=0.2$, $sigma_2=0.4$ and covariance matrix
$$
Sigma=begin{bmatrix}
sigma_1^2 & rhosigma_1sigma_2\rhosigma_1sigma_2 & sigma_2^2
end{bmatrix}
$$
where $rhoin[0,1]$ is the correlation parameter. Here's $rho=0.5$, with $varepsilon=0.4$.
The simplest naive approximation to this region would be
$$
x_igeqmu_i+Phi^{-1}(1-varepsilon)sigma_i.
$$
In general, this approximation isn't great:
But, counter-intuitively (to me), this approximation seems to be exact in the limit $rhoto1$. Here's $rho=0.99$:
answered Feb 2 at 15:19
David M.David M.
2,188421
$endgroup$
$begingroup$
Thanks for your explanation.
$endgroup$
– Mahraz
Feb 4 at 20:42
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
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oldest
votes
$begingroup$
Partial answer (too long for comment). You can evaluate the bivariate normal CDF numerically (using e.g. Python and Scipy, which is what I did). We see that, even in the correlated case, the feasible region is convex, which is a hopeful sign, computationally (basically the CDF of the bivariate normal distribution is quasi-convex).
For these numerical examples, I took $mu_1=mu_2=0$, $sigma_1=0.2$, $sigma_2=0.4$ and covariance matrix
$$
Sigma=begin{bmatrix}
sigma_1^2 & rhosigma_1sigma_2\rhosigma_1sigma_2 & sigma_2^2
end{bmatrix}
$$
where $rhoin[0,1]$ is the correlation parameter. Here's $rho=0.5$, with $varepsilon=0.4$.
The simplest naive approximation to this region would be
$$
x_igeqmu_i+Phi^{-1}(1-varepsilon)sigma_i.
$$
In general, this approximation isn't great:
But, counter-intuitively (to me), this approximation seems to be exact in the limit $rhoto1$. Here's $rho=0.99$:
answered Feb 2 at 15:19
David M.David M.
2,188421
$endgroup$
$begingroup$
Thanks for your explanation.
$endgroup$
– Mahraz
Feb 4 at 20:42
add a comment |
$begingroup$
Partial answer (too long for comment). You can evaluate the bivariate normal CDF numerically (using e.g. Python and Scipy, which is what I did). We see that, even in the correlated case, the feasible region is convex, which is a hopeful sign, computationally (basically the CDF of the bivariate normal distribution is quasi-convex).
For these numerical examples, I took $mu_1=mu_2=0$, $sigma_1=0.2$, $sigma_2=0.4$ and covariance matrix
$$
Sigma=begin{bmatrix}
sigma_1^2 & rhosigma_1sigma_2\rhosigma_1sigma_2 & sigma_2^2
end{bmatrix}
$$
where $rhoin[0,1]$ is the correlation parameter. Here's $rho=0.5$, with $varepsilon=0.4$.
The simplest naive approximation to this region would be
$$
x_igeqmu_i+Phi^{-1}(1-varepsilon)sigma_i.
$$
In general, this approximation isn't great:
But, counter-intuitively (to me), this approximation seems to be exact in the limit $rhoto1$. Here's $rho=0.99$:
answered Feb 2 at 15:19
David M.David M.
2,188421
$endgroup$
$begingroup$
Thanks for your explanation.
$endgroup$
– Mahraz
Feb 4 at 20:42
add a comment |
$begingroup$
Partial answer (too long for comment). You can evaluate the bivariate normal CDF numerically (using e.g. Python and Scipy, which is what I did). We see that, even in the correlated case, the feasible region is convex, which is a hopeful sign, computationally (basically the CDF of the bivariate normal distribution is quasi-convex).
For these numerical examples, I took $mu_1=mu_2=0$, $sigma_1=0.2$, $sigma_2=0.4$ and covariance matrix
$$
Sigma=begin{bmatrix}
sigma_1^2 & rhosigma_1sigma_2\rhosigma_1sigma_2 & sigma_2^2
end{bmatrix}
$$
where $rhoin[0,1]$ is the correlation parameter. Here's $rho=0.5$, with $varepsilon=0.4$.
The simplest naive approximation to this region would be
$$
x_igeqmu_i+Phi^{-1}(1-varepsilon)sigma_i.
$$
In general, this approximation isn't great:
But, counter-intuitively (to me), this approximation seems to be exact in the limit $rhoto1$. Here's $rho=0.99$:
answered Feb 2 at 15:19
David M.David M.
2,188421
$endgroup$
Partial answer (too long for comment). You can evaluate the bivariate normal CDF numerically (using e.g. Python and Scipy, which is what I did). We see that, even in the correlated case, the feasible region is convex, which is a hopeful sign, computationally (basically the CDF of the bivariate normal distribution is quasi-convex).
For these numerical examples, I took $mu_1=mu_2=0$, $sigma_1=0.2$, $sigma_2=0.4$ and covariance matrix
$$
Sigma=begin{bmatrix}
sigma_1^2 & rhosigma_1sigma_2\rhosigma_1sigma_2 & sigma_2^2
end{bmatrix}
$$
where $rhoin[0,1]$ is the correlation parameter. Here's $rho=0.5$, with $varepsilon=0.4$.
The simplest naive approximation to this region would be
$$
x_igeqmu_i+Phi^{-1}(1-varepsilon)sigma_i.
$$
In general, this approximation isn't great:
But, counter-intuitively (to me), this approximation seems to be exact in the limit $rhoto1$. Here's $rho=0.99$:
answered Feb 2 at 15:19
David M.David M.
2,188421
answered Feb 2 at 15:19
David M.David M.
2,188421
answered Feb 2 at 15:19
David M.David M.
2,188421
answered Feb 2 at 15:19
David M.David M.
2,188421
2,188421
$begingroup$
Thanks for your explanation.
$endgroup$
– Mahraz
Feb 4 at 20:42
add a comment |
$begingroup$
Thanks for your explanation.
$endgroup$
– Mahraz
Feb 4 at 20:42
$begingroup$
Thanks for your explanation.
$endgroup$
– Mahraz
Feb 4 at 20:42
$begingroup$
Thanks for your explanation.
$endgroup$
– Mahraz
Feb 4 at 20:42
add a comment |
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$begingroup$
1. where did $mu$ go? 2. can you apply a linear transformation to eliminte the correlation?
$endgroup$
– LinAlg
Jan 29 at 21:29
$begingroup$
@LinAlg $mu$ was missed, it is fixed now.
$endgroup$
– Mahraz
Jan 29 at 22:06
$begingroup$
This problem won't have a closed-form solution in dimension $dgeq2$, even in the uncorrelated case, I don't think. It can be evaluated numerically.
$endgroup$
– David M.
Feb 2 at 14:52
$begingroup$
@DavidM. in uncorrelated case, I think we can simply decouple the problems into several one dimension problem and solve them respectively.
$endgroup$
– Mahraz
Feb 4 at 20:37
$begingroup$
I think you would still end up with a product of CDFs, which I don't think you could invert analytically.
$endgroup$
– David M.
Feb 4 at 20:45