What is the covariance matrix?
2
$begingroup$
I need to draw samples from a bivariate normal distribution. I was told that the means are some $(mu_1, mu_2)$ and the std is $sigma$.
I wasn't given the covariance matrix.
My question is, was I suppose to infer the covariance matrix from this info? How?
Thanks
EDIT:
If we assume the marginal distributions are $(mu_i, sigma)$, then would it be correct to say that the covariance matrix of bivariate distribution is $[[sigma, 0], [[0, sigma]]$? (+ assuming independence)
probability probability-distributions normal-distribution
asked Dec 26 '18 at 21:10
yasecoyaseco
1113
$endgroup$
|
show 3 more comments
2
$begingroup$
I need to draw samples from a bivariate normal distribution. I was told that the means are some $(mu_1, mu_2)$ and the std is $sigma$.
I wasn't given the covariance matrix.
My question is, was I suppose to infer the covariance matrix from this info? How?
Thanks
EDIT:
If we assume the marginal distributions are $(mu_i, sigma)$, then would it be correct to say that the covariance matrix of bivariate distribution is $[[sigma, 0], [[0, sigma]]$? (+ assuming independence)
probability probability-distributions normal-distribution
asked Dec 26 '18 at 21:10
yasecoyaseco
1113
$endgroup$
$begingroup$
If $X_1,X_2$ are two independent Gaussian with mean $0,0$ and variance $v_1,v_2$ (and covariance $0$) then for any $a,b,c,d$, $Y_1 = aX_1+bX_2,Y_2 = cX_1+dX_2$ are non independent Gaussian with mean $0$, variance $a^2v_1+b^2v_2, c^2v_1+d^2v_2$ and covariance $acv_1+bdv_2$. The converse is that you can always find a linear transformatiom making the pair of Gaussian decorrelated.
$endgroup$
– reuns
Dec 26 '18 at 21:26
$begingroup$
@reuns, can you refer to my edit please?
$endgroup$
– yaseco
Dec 26 '18 at 21:38
$begingroup$
Yes, if you assume independence, then it would be correct. Actually you can assume "only" that the variables are uncorrelated for the covariance matrix to look that way. I put "only" in quotes because for jointly normal random variables, being uncorrelated is the same as being independent, so that is not really a weaker assumption
$endgroup$
– Ant
Dec 26 '18 at 21:39
$begingroup$
Thank you Ant! but when I think about it, shouldn't it be $sigma^2$ on the diagonal? (instead of $sigma$)
$endgroup$
– yaseco
Dec 26 '18 at 21:45
$begingroup$
Right, Ant gives an important point : the assumption that the pdf of $(X_1,X_2)$ is a $2$-dimensional Gaussian (much stronger than $X_1,X_2$ are separately Gaussian) implies that uncorrelated is equivalent to independent
$endgroup$
– reuns
Dec 26 '18 at 21:48
|
show 3 more comments
2
2
2
$begingroup$
I need to draw samples from a bivariate normal distribution. I was told that the means are some $(mu_1, mu_2)$ and the std is $sigma$.
I wasn't given the covariance matrix.
My question is, was I suppose to infer the covariance matrix from this info? How?
Thanks
EDIT:
If we assume the marginal distributions are $(mu_i, sigma)$, then would it be correct to say that the covariance matrix of bivariate distribution is $[[sigma, 0], [[0, sigma]]$? (+ assuming independence)
probability probability-distributions normal-distribution
asked Dec 26 '18 at 21:10
yasecoyaseco
1113
$endgroup$
I need to draw samples from a bivariate normal distribution. I was told that the means are some $(mu_1, mu_2)$ and the std is $sigma$.
I wasn't given the covariance matrix.
My question is, was I suppose to infer the covariance matrix from this info? How?
Thanks
EDIT:
If we assume the marginal distributions are $(mu_i, sigma)$, then would it be correct to say that the covariance matrix of bivariate distribution is $[[sigma, 0], [[0, sigma]]$? (+ assuming independence)
probability probability-distributions normal-distribution
probability probability-distributions normal-distribution
asked Dec 26 '18 at 21:10
yasecoyaseco
1113
asked Dec 26 '18 at 21:10
yasecoyaseco
1113
edited Dec 26 '18 at 21:37
yaseco
asked Dec 26 '18 at 21:10
yasecoyaseco
1113
asked Dec 26 '18 at 21:10
yasecoyaseco
1113
asked Dec 26 '18 at 21:10
yasecoyaseco
1113
1113
$begingroup$
If $X_1,X_2$ are two independent Gaussian with mean $0,0$ and variance $v_1,v_2$ (and covariance $0$) then for any $a,b,c,d$, $Y_1 = aX_1+bX_2,Y_2 = cX_1+dX_2$ are non independent Gaussian with mean $0$, variance $a^2v_1+b^2v_2, c^2v_1+d^2v_2$ and covariance $acv_1+bdv_2$. The converse is that you can always find a linear transformatiom making the pair of Gaussian decorrelated.
$endgroup$
– reuns
Dec 26 '18 at 21:26
$begingroup$
@reuns, can you refer to my edit please?
$endgroup$
– yaseco
Dec 26 '18 at 21:38
$begingroup$
Yes, if you assume independence, then it would be correct. Actually you can assume "only" that the variables are uncorrelated for the covariance matrix to look that way. I put "only" in quotes because for jointly normal random variables, being uncorrelated is the same as being independent, so that is not really a weaker assumption
$endgroup$
– Ant
Dec 26 '18 at 21:39
$begingroup$
Thank you Ant! but when I think about it, shouldn't it be $sigma^2$ on the diagonal? (instead of $sigma$)
$endgroup$
– yaseco
Dec 26 '18 at 21:45
$begingroup$
Right, Ant gives an important point : the assumption that the pdf of $(X_1,X_2)$ is a $2$-dimensional Gaussian (much stronger than $X_1,X_2$ are separately Gaussian) implies that uncorrelated is equivalent to independent
$endgroup$
– reuns
Dec 26 '18 at 21:48
|
show 3 more comments
$begingroup$
If $X_1,X_2$ are two independent Gaussian with mean $0,0$ and variance $v_1,v_2$ (and covariance $0$) then for any $a,b,c,d$, $Y_1 = aX_1+bX_2,Y_2 = cX_1+dX_2$ are non independent Gaussian with mean $0$, variance $a^2v_1+b^2v_2, c^2v_1+d^2v_2$ and covariance $acv_1+bdv_2$. The converse is that you can always find a linear transformatiom making the pair of Gaussian decorrelated.
$endgroup$
– reuns
Dec 26 '18 at 21:26
$begingroup$
@reuns, can you refer to my edit please?
$endgroup$
– yaseco
Dec 26 '18 at 21:38
$begingroup$
Yes, if you assume independence, then it would be correct. Actually you can assume "only" that the variables are uncorrelated for the covariance matrix to look that way. I put "only" in quotes because for jointly normal random variables, being uncorrelated is the same as being independent, so that is not really a weaker assumption
$endgroup$
– Ant
Dec 26 '18 at 21:39
$begingroup$
Thank you Ant! but when I think about it, shouldn't it be $sigma^2$ on the diagonal? (instead of $sigma$)
$endgroup$
– yaseco
Dec 26 '18 at 21:45
$begingroup$
Right, Ant gives an important point : the assumption that the pdf of $(X_1,X_2)$ is a $2$-dimensional Gaussian (much stronger than $X_1,X_2$ are separately Gaussian) implies that uncorrelated is equivalent to independent
$endgroup$
– reuns
Dec 26 '18 at 21:48
$begingroup$
If $X_1,X_2$ are two independent Gaussian with mean $0,0$ and variance $v_1,v_2$ (and covariance $0$) then for any $a,b,c,d$, $Y_1 = aX_1+bX_2,Y_2 = cX_1+dX_2$ are non independent Gaussian with mean $0$, variance $a^2v_1+b^2v_2, c^2v_1+d^2v_2$ and covariance $acv_1+bdv_2$. The converse is that you can always find a linear transformatiom making the pair of Gaussian decorrelated.
$endgroup$
– reuns
Dec 26 '18 at 21:26
$begingroup$
If $X_1,X_2$ are two independent Gaussian with mean $0,0$ and variance $v_1,v_2$ (and covariance $0$) then for any $a,b,c,d$, $Y_1 = aX_1+bX_2,Y_2 = cX_1+dX_2$ are non independent Gaussian with mean $0$, variance $a^2v_1+b^2v_2, c^2v_1+d^2v_2$ and covariance $acv_1+bdv_2$. The converse is that you can always find a linear transformatiom making the pair of Gaussian decorrelated.
$endgroup$
– reuns
Dec 26 '18 at 21:26
$begingroup$
@reuns, can you refer to my edit please?
$endgroup$
– yaseco
Dec 26 '18 at 21:38
$begingroup$
@reuns, can you refer to my edit please?
$endgroup$
– yaseco
Dec 26 '18 at 21:38
$begingroup$
Yes, if you assume independence, then it would be correct. Actually you can assume "only" that the variables are uncorrelated for the covariance matrix to look that way. I put "only" in quotes because for jointly normal random variables, being uncorrelated is the same as being independent, so that is not really a weaker assumption
$endgroup$
– Ant
Dec 26 '18 at 21:39
$begingroup$
Yes, if you assume independence, then it would be correct. Actually you can assume "only" that the variables are uncorrelated for the covariance matrix to look that way. I put "only" in quotes because for jointly normal random variables, being uncorrelated is the same as being independent, so that is not really a weaker assumption
$endgroup$
– Ant
Dec 26 '18 at 21:39
$begingroup$
Thank you Ant! but when I think about it, shouldn't it be $sigma^2$ on the diagonal? (instead of $sigma$)
$endgroup$
– yaseco
Dec 26 '18 at 21:45
$begingroup$
Thank you Ant! but when I think about it, shouldn't it be $sigma^2$ on the diagonal? (instead of $sigma$)
$endgroup$
– yaseco
Dec 26 '18 at 21:45
$begingroup$
Right, Ant gives an important point : the assumption that the pdf of $(X_1,X_2)$ is a $2$-dimensional Gaussian (much stronger than $X_1,X_2$ are separately Gaussian) implies that uncorrelated is equivalent to independent
$endgroup$
– reuns
Dec 26 '18 at 21:48
$begingroup$
Right, Ant gives an important point : the assumption that the pdf of $(X_1,X_2)$ is a $2$-dimensional Gaussian (much stronger than $X_1,X_2$ are separately Gaussian) implies that uncorrelated is equivalent to independent
$endgroup$
– reuns
Dec 26 '18 at 21:48
|
show 3 more comments
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$begingroup$
If $X_1,X_2$ are two independent Gaussian with mean $0,0$ and variance $v_1,v_2$ (and covariance $0$) then for any $a,b,c,d$, $Y_1 = aX_1+bX_2,Y_2 = cX_1+dX_2$ are non independent Gaussian with mean $0$, variance $a^2v_1+b^2v_2, c^2v_1+d^2v_2$ and covariance $acv_1+bdv_2$. The converse is that you can always find a linear transformatiom making the pair of Gaussian decorrelated.
$endgroup$
– reuns
Dec 26 '18 at 21:26
$begingroup$
@reuns, can you refer to my edit please?
$endgroup$
– yaseco
Dec 26 '18 at 21:38
$begingroup$
Yes, if you assume independence, then it would be correct. Actually you can assume "only" that the variables are uncorrelated for the covariance matrix to look that way. I put "only" in quotes because for jointly normal random variables, being uncorrelated is the same as being independent, so that is not really a weaker assumption
$endgroup$
– Ant
Dec 26 '18 at 21:39
$begingroup$
Thank you Ant! but when I think about it, shouldn't it be $sigma^2$ on the diagonal? (instead of $sigma$)
$endgroup$
– yaseco
Dec 26 '18 at 21:45
$begingroup$
Right, Ant gives an important point : the assumption that the pdf of $(X_1,X_2)$ is a $2$-dimensional Gaussian (much stronger than $X_1,X_2$ are separately Gaussian) implies that uncorrelated is equivalent to independent
$endgroup$
– reuns
Dec 26 '18 at 21:48