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Generating a number belonging to N(0,1) using *m* numbers from U(0,1) using central limit theorem
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I was going through a blog which details how to generate a multivariate Gaussian vector, given a mean vector μ and co-variance matrix σ.
As a starting point, author uses generated uniform random numbers to simulate Gaussian random numbers using central limit theorem. The relevant extract from the blog is below:
Let's say you generate m uniform random numbers (each between 0 and 1) and you use the variable xi to denote each of these. The Central Limit Theorem allows us to convert these m numbers belonging to U(0,1) into a single number that belongs to the Guassian distribution N(0,1).
$$x = frac{sum_ix_i-frac{m}{2}}{sqrt{frac{m}{12}}}$$
Here, x is a one dimensional Gaussian random number - produced using the help of m uniform random variables. The $frac{m}{2}$ is derived from the term mμu (where μu is the mean of the uniform distribution - $frac{1-0}{2} = 0.5$). The denominator is derived from the term σ$sqrt{m}$ where $sigma^2$ is the variance of the uniform distribution between 0 and 1 (comes to exactly $frac{1}{12}$).
Link to the article : http://www.aishack.in/tutorials/generating-multivariate-gaussian-random/
I wanted to understand the details of the underlying math as I am not sure how central limit theorem is used to generate N(0,1) number from m U(0,1) numbers. I googled the same but was not able to find an explanation for this.
normal-distribution uniform-distribution central-limit-theorem
asked Jan 30 at 10:04
confused_certaintiesconfused_certainties
82
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add a comment |
$begingroup$
I was going through a blog which details how to generate a multivariate Gaussian vector, given a mean vector μ and co-variance matrix σ.
As a starting point, author uses generated uniform random numbers to simulate Gaussian random numbers using central limit theorem. The relevant extract from the blog is below:
Let's say you generate m uniform random numbers (each between 0 and 1) and you use the variable xi to denote each of these. The Central Limit Theorem allows us to convert these m numbers belonging to U(0,1) into a single number that belongs to the Guassian distribution N(0,1).
$$x = frac{sum_ix_i-frac{m}{2}}{sqrt{frac{m}{12}}}$$
Here, x is a one dimensional Gaussian random number - produced using the help of m uniform random variables. The $frac{m}{2}$ is derived from the term mμu (where μu is the mean of the uniform distribution - $frac{1-0}{2} = 0.5$). The denominator is derived from the term σ$sqrt{m}$ where $sigma^2$ is the variance of the uniform distribution between 0 and 1 (comes to exactly $frac{1}{12}$).
Link to the article : http://www.aishack.in/tutorials/generating-multivariate-gaussian-random/
I wanted to understand the details of the underlying math as I am not sure how central limit theorem is used to generate N(0,1) number from m U(0,1) numbers. I googled the same but was not able to find an explanation for this.
normal-distribution uniform-distribution central-limit-theorem
asked Jan 30 at 10:04
confused_certaintiesconfused_certainties
82
$endgroup$
add a comment |
$begingroup$
I was going through a blog which details how to generate a multivariate Gaussian vector, given a mean vector μ and co-variance matrix σ.
As a starting point, author uses generated uniform random numbers to simulate Gaussian random numbers using central limit theorem. The relevant extract from the blog is below:
Let's say you generate m uniform random numbers (each between 0 and 1) and you use the variable xi to denote each of these. The Central Limit Theorem allows us to convert these m numbers belonging to U(0,1) into a single number that belongs to the Guassian distribution N(0,1).
$$x = frac{sum_ix_i-frac{m}{2}}{sqrt{frac{m}{12}}}$$
Here, x is a one dimensional Gaussian random number - produced using the help of m uniform random variables. The $frac{m}{2}$ is derived from the term mμu (where μu is the mean of the uniform distribution - $frac{1-0}{2} = 0.5$). The denominator is derived from the term σ$sqrt{m}$ where $sigma^2$ is the variance of the uniform distribution between 0 and 1 (comes to exactly $frac{1}{12}$).
Link to the article : http://www.aishack.in/tutorials/generating-multivariate-gaussian-random/
I wanted to understand the details of the underlying math as I am not sure how central limit theorem is used to generate N(0,1) number from m U(0,1) numbers. I googled the same but was not able to find an explanation for this.
normal-distribution uniform-distribution central-limit-theorem
asked Jan 30 at 10:04
confused_certaintiesconfused_certainties
82
$endgroup$
I was going through a blog which details how to generate a multivariate Gaussian vector, given a mean vector μ and co-variance matrix σ.
As a starting point, author uses generated uniform random numbers to simulate Gaussian random numbers using central limit theorem. The relevant extract from the blog is below:
Let's say you generate m uniform random numbers (each between 0 and 1) and you use the variable xi to denote each of these. The Central Limit Theorem allows us to convert these m numbers belonging to U(0,1) into a single number that belongs to the Guassian distribution N(0,1).
$$x = frac{sum_ix_i-frac{m}{2}}{sqrt{frac{m}{12}}}$$
Here, x is a one dimensional Gaussian random number - produced using the help of m uniform random variables. The $frac{m}{2}$ is derived from the term mμu (where μu is the mean of the uniform distribution - $frac{1-0}{2} = 0.5$). The denominator is derived from the term σ$sqrt{m}$ where $sigma^2$ is the variance of the uniform distribution between 0 and 1 (comes to exactly $frac{1}{12}$).
Link to the article : http://www.aishack.in/tutorials/generating-multivariate-gaussian-random/
I wanted to understand the details of the underlying math as I am not sure how central limit theorem is used to generate N(0,1) number from m U(0,1) numbers. I googled the same but was not able to find an explanation for this.
normal-distribution uniform-distribution central-limit-theorem
normal-distribution uniform-distribution central-limit-theorem
asked Jan 30 at 10:04
confused_certaintiesconfused_certainties
82
asked Jan 30 at 10:04
confused_certaintiesconfused_certainties
82
asked Jan 30 at 10:04
confused_certaintiesconfused_certainties
82
asked Jan 30 at 10:04
confused_certaintiesconfused_certainties
82
asked Jan 30 at 10:04
confused_certaintiesconfused_certainties
82
82
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1 Answer
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The idea is that the if there are $m$ independent random variables $X_i$ each uniformly distributed on $[0,1]$, so with mean $frac12$ and variance $frac1{12}$, then their sum $S_m$ has mean $frac{m}2$ and variance $frac{m}{12}$ while their average $frac1m S_m$ has mean $frac{1}2$ and variance $frac{1}{12m}$
The Central Limit theorem says that $sqrt{m}left(frac1m S_m - frac{1}2right)$ converges in distribution to $mathcal Nleft(0,frac1{12}right)$, i.e. $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ converges in distribution to $mathcal N(0,1)$ as $m$ increases
So for sufficiently large $m$, you might use $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ to generate a random variable which has a distribution close to that of a standard normal random variable
answered Jan 30 at 10:19
HenryHenry
101k482170
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1 Answer
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$begingroup$
The idea is that the if there are $m$ independent random variables $X_i$ each uniformly distributed on $[0,1]$, so with mean $frac12$ and variance $frac1{12}$, then their sum $S_m$ has mean $frac{m}2$ and variance $frac{m}{12}$ while their average $frac1m S_m$ has mean $frac{1}2$ and variance $frac{1}{12m}$
The Central Limit theorem says that $sqrt{m}left(frac1m S_m - frac{1}2right)$ converges in distribution to $mathcal Nleft(0,frac1{12}right)$, i.e. $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ converges in distribution to $mathcal N(0,1)$ as $m$ increases
So for sufficiently large $m$, you might use $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ to generate a random variable which has a distribution close to that of a standard normal random variable
answered Jan 30 at 10:19
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
The idea is that the if there are $m$ independent random variables $X_i$ each uniformly distributed on $[0,1]$, so with mean $frac12$ and variance $frac1{12}$, then their sum $S_m$ has mean $frac{m}2$ and variance $frac{m}{12}$ while their average $frac1m S_m$ has mean $frac{1}2$ and variance $frac{1}{12m}$
The Central Limit theorem says that $sqrt{m}left(frac1m S_m - frac{1}2right)$ converges in distribution to $mathcal Nleft(0,frac1{12}right)$, i.e. $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ converges in distribution to $mathcal N(0,1)$ as $m$ increases
So for sufficiently large $m$, you might use $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ to generate a random variable which has a distribution close to that of a standard normal random variable
answered Jan 30 at 10:19
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
The idea is that the if there are $m$ independent random variables $X_i$ each uniformly distributed on $[0,1]$, so with mean $frac12$ and variance $frac1{12}$, then their sum $S_m$ has mean $frac{m}2$ and variance $frac{m}{12}$ while their average $frac1m S_m$ has mean $frac{1}2$ and variance $frac{1}{12m}$
The Central Limit theorem says that $sqrt{m}left(frac1m S_m - frac{1}2right)$ converges in distribution to $mathcal Nleft(0,frac1{12}right)$, i.e. $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ converges in distribution to $mathcal N(0,1)$ as $m$ increases
So for sufficiently large $m$, you might use $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ to generate a random variable which has a distribution close to that of a standard normal random variable
answered Jan 30 at 10:19
HenryHenry
101k482170
$endgroup$
The idea is that the if there are $m$ independent random variables $X_i$ each uniformly distributed on $[0,1]$, so with mean $frac12$ and variance $frac1{12}$, then their sum $S_m$ has mean $frac{m}2$ and variance $frac{m}{12}$ while their average $frac1m S_m$ has mean $frac{1}2$ and variance $frac{1}{12m}$
The Central Limit theorem says that $sqrt{m}left(frac1m S_m - frac{1}2right)$ converges in distribution to $mathcal Nleft(0,frac1{12}right)$, i.e. $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ converges in distribution to $mathcal N(0,1)$ as $m$ increases
So for sufficiently large $m$, you might use $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ to generate a random variable which has a distribution close to that of a standard normal random variable
answered Jan 30 at 10:19
HenryHenry
101k482170
answered Jan 30 at 10:19
HenryHenry
101k482170
answered Jan 30 at 10:19
HenryHenry
101k482170
answered Jan 30 at 10:19
HenryHenry
101k482170
101k482170
add a comment |
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