Is drawing a random vector from normal distribution is the same as sampling every coordinate from normal...
$begingroup$
So say I have a matrix where each row is sampled from the normal distribution with mean 0 and identity covariance matrix. Is it equivalent to simply sampling every matrix entry from standard normal distribution with mean 0 and std 1?
E.g is sampling vectors from multi dimensional normal distribution is equivalent to sampling every vector coordinate according to standard normal.
Why or why not?
Thanks!
linear-algebra
asked Jan 22 at 23:26
YohanRothYohanRoth
6381715
$endgroup$
add a comment |
$begingroup$
So say I have a matrix where each row is sampled from the normal distribution with mean 0 and identity covariance matrix. Is it equivalent to simply sampling every matrix entry from standard normal distribution with mean 0 and std 1?
E.g is sampling vectors from multi dimensional normal distribution is equivalent to sampling every vector coordinate according to standard normal.
Why or why not?
Thanks!
linear-algebra
asked Jan 22 at 23:26
YohanRothYohanRoth
6381715
$endgroup$
add a comment |
$begingroup$
So say I have a matrix where each row is sampled from the normal distribution with mean 0 and identity covariance matrix. Is it equivalent to simply sampling every matrix entry from standard normal distribution with mean 0 and std 1?
E.g is sampling vectors from multi dimensional normal distribution is equivalent to sampling every vector coordinate according to standard normal.
Why or why not?
Thanks!
linear-algebra
asked Jan 22 at 23:26
YohanRothYohanRoth
6381715
$endgroup$
So say I have a matrix where each row is sampled from the normal distribution with mean 0 and identity covariance matrix. Is it equivalent to simply sampling every matrix entry from standard normal distribution with mean 0 and std 1?
E.g is sampling vectors from multi dimensional normal distribution is equivalent to sampling every vector coordinate according to standard normal.
Why or why not?
Thanks!
linear-algebra
linear-algebra
asked Jan 22 at 23:26
YohanRothYohanRoth
6381715
asked Jan 22 at 23:26
YohanRothYohanRoth
6381715
asked Jan 22 at 23:26
YohanRothYohanRoth
6381715
asked Jan 22 at 23:26
YohanRothYohanRoth
6381715
asked Jan 22 at 23:26
YohanRothYohanRoth
6381715
6381715
add a comment |
add a comment |
1 Answer
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Yes, if you sample the coordinates independently. You can see this by writing out the joint pdf for the vector coordinates, and noting that it is the product of the individual pdfs for the coordinates. Basically because $$exp(-sum_i x_i^2/2) = prod_i exp(-x_i^2/2).$$
answered Jan 22 at 23:31
kimchi loverkimchi lover
10.9k31128
$endgroup$
$begingroup$
"pdf" means probability distribution function. Does this answer your question? If not, what does "normal distribution" mean to you, in your own terms. (It's sometimes hard to understand how much background knowledge a questioner here has: assume to much and an answer might be meaningless, assume too little, and an answer might be belittling.)
$endgroup$
– kimchi lover
Jan 22 at 23:52
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
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$begingroup$
Yes, if you sample the coordinates independently. You can see this by writing out the joint pdf for the vector coordinates, and noting that it is the product of the individual pdfs for the coordinates. Basically because $$exp(-sum_i x_i^2/2) = prod_i exp(-x_i^2/2).$$
answered Jan 22 at 23:31
kimchi loverkimchi lover
10.9k31128
$endgroup$
$begingroup$
"pdf" means probability distribution function. Does this answer your question? If not, what does "normal distribution" mean to you, in your own terms. (It's sometimes hard to understand how much background knowledge a questioner here has: assume to much and an answer might be meaningless, assume too little, and an answer might be belittling.)
$endgroup$
– kimchi lover
Jan 22 at 23:52
add a comment |
$begingroup$
Yes, if you sample the coordinates independently. You can see this by writing out the joint pdf for the vector coordinates, and noting that it is the product of the individual pdfs for the coordinates. Basically because $$exp(-sum_i x_i^2/2) = prod_i exp(-x_i^2/2).$$
answered Jan 22 at 23:31
kimchi loverkimchi lover
10.9k31128
$endgroup$
$begingroup$
"pdf" means probability distribution function. Does this answer your question? If not, what does "normal distribution" mean to you, in your own terms. (It's sometimes hard to understand how much background knowledge a questioner here has: assume to much and an answer might be meaningless, assume too little, and an answer might be belittling.)
$endgroup$
– kimchi lover
Jan 22 at 23:52
add a comment |
$begingroup$
Yes, if you sample the coordinates independently. You can see this by writing out the joint pdf for the vector coordinates, and noting that it is the product of the individual pdfs for the coordinates. Basically because $$exp(-sum_i x_i^2/2) = prod_i exp(-x_i^2/2).$$
answered Jan 22 at 23:31
kimchi loverkimchi lover
10.9k31128
$endgroup$
Yes, if you sample the coordinates independently. You can see this by writing out the joint pdf for the vector coordinates, and noting that it is the product of the individual pdfs for the coordinates. Basically because $$exp(-sum_i x_i^2/2) = prod_i exp(-x_i^2/2).$$
answered Jan 22 at 23:31
kimchi loverkimchi lover
10.9k31128
answered Jan 22 at 23:31
kimchi loverkimchi lover
10.9k31128
answered Jan 22 at 23:31
kimchi loverkimchi lover
10.9k31128
answered Jan 22 at 23:31
kimchi loverkimchi lover
10.9k31128
10.9k31128
$begingroup$
"pdf" means probability distribution function. Does this answer your question? If not, what does "normal distribution" mean to you, in your own terms. (It's sometimes hard to understand how much background knowledge a questioner here has: assume to much and an answer might be meaningless, assume too little, and an answer might be belittling.)
$endgroup$
– kimchi lover
Jan 22 at 23:52
add a comment |
$begingroup$
"pdf" means probability distribution function. Does this answer your question? If not, what does "normal distribution" mean to you, in your own terms. (It's sometimes hard to understand how much background knowledge a questioner here has: assume to much and an answer might be meaningless, assume too little, and an answer might be belittling.)
$endgroup$
– kimchi lover
Jan 22 at 23:52
$begingroup$
"pdf" means probability distribution function. Does this answer your question? If not, what does "normal distribution" mean to you, in your own terms. (It's sometimes hard to understand how much background knowledge a questioner here has: assume to much and an answer might be meaningless, assume too little, and an answer might be belittling.)
$endgroup$
– kimchi lover
Jan 22 at 23:52
$begingroup$
"pdf" means probability distribution function. Does this answer your question? If not, what does "normal distribution" mean to you, in your own terms. (It's sometimes hard to understand how much background knowledge a questioner here has: assume to much and an answer might be meaningless, assume too little, and an answer might be belittling.)
$endgroup$
– kimchi lover
Jan 22 at 23:52
add a comment |
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