Simulating rnorm() using runif()
$begingroup$
I am trying to 'simulate' rnorm() using only runif().
I don't know if I should do:
sqrt(-2*log(U1))*cos(U2)
or
sqrt(-2*log(U1))*sin(U2)
Where U1 is a runif(0,1) and U2 runif(0,6.28)
I do not know if I should do it using cos or sin, or is it that I need to sample from one and the other consecutively? What is the mathematical logic behind it?
r normalization
edited Mar 14 at 16:51
masoud
675
asked Mar 14 at 14:13
Chicago1988Chicago1988
162
$endgroup$
add a comment |
$begingroup$
I am trying to 'simulate' rnorm() using only runif().
I don't know if I should do:
sqrt(-2*log(U1))*cos(U2)
or
sqrt(-2*log(U1))*sin(U2)
Where U1 is a runif(0,1) and U2 runif(0,6.28)
I do not know if I should do it using cos or sin, or is it that I need to sample from one and the other consecutively? What is the mathematical logic behind it?
r normalization
edited Mar 14 at 16:51
masoud
675
asked Mar 14 at 14:13
Chicago1988Chicago1988
162
$endgroup$
$begingroup$
You will want to investigate the Probability Integral Transform.
$endgroup$
– StatsStudent
Mar 14 at 15:28
1
$begingroup$
Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
$endgroup$
– whuber♦
Mar 14 at 18:44
add a comment |
$begingroup$
I am trying to 'simulate' rnorm() using only runif().
I don't know if I should do:
sqrt(-2*log(U1))*cos(U2)
or
sqrt(-2*log(U1))*sin(U2)
Where U1 is a runif(0,1) and U2 runif(0,6.28)
I do not know if I should do it using cos or sin, or is it that I need to sample from one and the other consecutively? What is the mathematical logic behind it?
r normalization
edited Mar 14 at 16:51
masoud
675
asked Mar 14 at 14:13
Chicago1988Chicago1988
162
$endgroup$
I am trying to 'simulate' rnorm() using only runif().
I don't know if I should do:
sqrt(-2*log(U1))*cos(U2)
or
sqrt(-2*log(U1))*sin(U2)
Where U1 is a runif(0,1) and U2 runif(0,6.28)
I do not know if I should do it using cos or sin, or is it that I need to sample from one and the other consecutively? What is the mathematical logic behind it?
r normalization
r normalization
edited Mar 14 at 16:51
masoud
675
asked Mar 14 at 14:13
Chicago1988Chicago1988
162
edited Mar 14 at 16:51
masoud
675
asked Mar 14 at 14:13
Chicago1988Chicago1988
162
edited Mar 14 at 16:51
masoud
675
edited Mar 14 at 16:51
masoud
675
edited Mar 14 at 16:51
masoud
675
675
asked Mar 14 at 14:13
Chicago1988Chicago1988
162
asked Mar 14 at 14:13
Chicago1988Chicago1988
162
asked Mar 14 at 14:13
Chicago1988Chicago1988
162
162
$begingroup$
You will want to investigate the Probability Integral Transform.
$endgroup$
– StatsStudent
Mar 14 at 15:28
1
$begingroup$
Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
$endgroup$
– whuber♦
Mar 14 at 18:44
add a comment |
$begingroup$
You will want to investigate the Probability Integral Transform.
$endgroup$
– StatsStudent
Mar 14 at 15:28
1
$begingroup$
Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
$endgroup$
– whuber♦
Mar 14 at 18:44
$begingroup$
You will want to investigate the Probability Integral Transform.
$endgroup$
– StatsStudent
Mar 14 at 15:28
$begingroup$
You will want to investigate the Probability Integral Transform.
$endgroup$
– StatsStudent
Mar 14 at 15:28
1
1
$begingroup$
Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
$endgroup$
– whuber♦
Mar 14 at 18:44
$begingroup$
Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
$endgroup$
– whuber♦
Mar 14 at 18:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It looks like you are trying to use the Box-Muller transform.
The method is to use $U_1, U_2$ which are independently distribution Uniform(0,1). Then
$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2) $
and
$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2) $
are a pair of independent $N(0,1)$.
The derivation comes from using polar coordinates, but to be honest the exact steps escape me at the moment (see link if you are interested).
answered Mar 14 at 14:58
Cliff ABCliff AB
13.6k12567
$endgroup$
add a comment |
$begingroup$
Let $N_1,N_2$ be independent $N(0,1)$ random variates. $(N_1,N_2)$ defines a point in Cartesian coordinates. We transform to polar by
$$ N_1 = R cos theta$$
$$ N_2 = R sin theta$$
This transformation is one to one and has continuous derivatives. So we can derive the joint distribution $f_{R,theta}(r,theta)$ by using
$$f_{R,theta}(r,theta) = f_{N_1,N_2}(n_1,n_2)|J^{-1}|$$
where $J$ is the Jacobian of the transformation and
$$ J^{-1}(r,theta )=
{begin{bmatrix}{dfrac {partial n_1}{partial r}}{dfrac {partial n_1}{partial theta }}\{dfrac {partial n_2}{partial r}}{dfrac {partial n_2}{partial theta }}end{bmatrix}}
= begin{bmatrix}cos theta & -rsin theta \sin theta & rcos theta end{bmatrix}$$
Hence $$ |J^{-1}| = r cos^2 theta + r sin^2 theta = r $$
and
$$ f_{R,theta}(r,theta) = r frac{1}{2pi} e^{-frac{1}{2}(n_1^2+n_2^2)} = boxed{frac{r}{2 pi}e^{-frac{1}{2}(r^2)}} qquad 0 leq theta leq 2 pi, 0 leq r leq infty $$
And we see $theta sim unif[0,2pi]$ (since $f_{Theta}(theta) = frac{1}{2pi}$ ) and $R^2 sim exp(1/2)$
Hence, to simulate $Theta$ we simply take $2 pi U_2$ where $$U_2 sim unif[0,1]$$ and to simulate $R$ we can take $$-2ln U_1$$ where $U_1 sim unif[0,1]$. ( to see why find the density of $X=-ln U, U sim unif[0,1]$).
So, Box and Muller simply inverted $N_1= R cos theta$, $N_2= R sin theta$ and moved from $(R,Theta)$ to $(N_1,N_2)$ by simulating $Theta$ from $2 pi U_2$, and an independent $R$ from $ sqrt{- 2 ln U_1}$
explicitly
$$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2)$$
$$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2)$$
And that is that is the mathematical logic behind it.
p.s. As I don't know whether it is clear through the mathematical justification (I believe it is, but you may have been lost in details), note you need a pair of drawings from the uniform to get a pair of normals
answered Mar 14 at 16:14
StatsStats
668210
$endgroup$
add a comment |
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$begingroup$
It looks like you are trying to use the Box-Muller transform.
The method is to use $U_1, U_2$ which are independently distribution Uniform(0,1). Then
$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2) $
and
$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2) $
are a pair of independent $N(0,1)$.
The derivation comes from using polar coordinates, but to be honest the exact steps escape me at the moment (see link if you are interested).
answered Mar 14 at 14:58
Cliff ABCliff AB
13.6k12567
$endgroup$
add a comment |
$begingroup$
It looks like you are trying to use the Box-Muller transform.
The method is to use $U_1, U_2$ which are independently distribution Uniform(0,1). Then
$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2) $
and
$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2) $
are a pair of independent $N(0,1)$.
The derivation comes from using polar coordinates, but to be honest the exact steps escape me at the moment (see link if you are interested).
answered Mar 14 at 14:58
Cliff ABCliff AB
13.6k12567
$endgroup$
add a comment |
$begingroup$
It looks like you are trying to use the Box-Muller transform.
The method is to use $U_1, U_2$ which are independently distribution Uniform(0,1). Then
$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2) $
and
$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2) $
are a pair of independent $N(0,1)$.
The derivation comes from using polar coordinates, but to be honest the exact steps escape me at the moment (see link if you are interested).
answered Mar 14 at 14:58
Cliff ABCliff AB
13.6k12567
$endgroup$
It looks like you are trying to use the Box-Muller transform.
The method is to use $U_1, U_2$ which are independently distribution Uniform(0,1). Then
$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2) $
and
$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2) $
are a pair of independent $N(0,1)$.
The derivation comes from using polar coordinates, but to be honest the exact steps escape me at the moment (see link if you are interested).
answered Mar 14 at 14:58
Cliff ABCliff AB
13.6k12567
edited Mar 14 at 18:15
answered Mar 14 at 14:58
Cliff ABCliff AB
13.6k12567
answered Mar 14 at 14:58
Cliff ABCliff AB
13.6k12567
answered Mar 14 at 14:58
Cliff ABCliff AB
13.6k12567
13.6k12567
add a comment |
add a comment |
$begingroup$
Let $N_1,N_2$ be independent $N(0,1)$ random variates. $(N_1,N_2)$ defines a point in Cartesian coordinates. We transform to polar by
$$ N_1 = R cos theta$$
$$ N_2 = R sin theta$$
This transformation is one to one and has continuous derivatives. So we can derive the joint distribution $f_{R,theta}(r,theta)$ by using
$$f_{R,theta}(r,theta) = f_{N_1,N_2}(n_1,n_2)|J^{-1}|$$
where $J$ is the Jacobian of the transformation and
$$ J^{-1}(r,theta )=
{begin{bmatrix}{dfrac {partial n_1}{partial r}}{dfrac {partial n_1}{partial theta }}\{dfrac {partial n_2}{partial r}}{dfrac {partial n_2}{partial theta }}end{bmatrix}}
= begin{bmatrix}cos theta & -rsin theta \sin theta & rcos theta end{bmatrix}$$
Hence $$ |J^{-1}| = r cos^2 theta + r sin^2 theta = r $$
and
$$ f_{R,theta}(r,theta) = r frac{1}{2pi} e^{-frac{1}{2}(n_1^2+n_2^2)} = boxed{frac{r}{2 pi}e^{-frac{1}{2}(r^2)}} qquad 0 leq theta leq 2 pi, 0 leq r leq infty $$
And we see $theta sim unif[0,2pi]$ (since $f_{Theta}(theta) = frac{1}{2pi}$ ) and $R^2 sim exp(1/2)$
Hence, to simulate $Theta$ we simply take $2 pi U_2$ where $$U_2 sim unif[0,1]$$ and to simulate $R$ we can take $$-2ln U_1$$ where $U_1 sim unif[0,1]$. ( to see why find the density of $X=-ln U, U sim unif[0,1]$).
So, Box and Muller simply inverted $N_1= R cos theta$, $N_2= R sin theta$ and moved from $(R,Theta)$ to $(N_1,N_2)$ by simulating $Theta$ from $2 pi U_2$, and an independent $R$ from $ sqrt{- 2 ln U_1}$
explicitly
$$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2)$$
$$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2)$$
And that is that is the mathematical logic behind it.
p.s. As I don't know whether it is clear through the mathematical justification (I believe it is, but you may have been lost in details), note you need a pair of drawings from the uniform to get a pair of normals
answered Mar 14 at 16:14
StatsStats
668210
$endgroup$
add a comment |
$begingroup$
Let $N_1,N_2$ be independent $N(0,1)$ random variates. $(N_1,N_2)$ defines a point in Cartesian coordinates. We transform to polar by
$$ N_1 = R cos theta$$
$$ N_2 = R sin theta$$
This transformation is one to one and has continuous derivatives. So we can derive the joint distribution $f_{R,theta}(r,theta)$ by using
$$f_{R,theta}(r,theta) = f_{N_1,N_2}(n_1,n_2)|J^{-1}|$$
where $J$ is the Jacobian of the transformation and
$$ J^{-1}(r,theta )=
{begin{bmatrix}{dfrac {partial n_1}{partial r}}{dfrac {partial n_1}{partial theta }}\{dfrac {partial n_2}{partial r}}{dfrac {partial n_2}{partial theta }}end{bmatrix}}
= begin{bmatrix}cos theta & -rsin theta \sin theta & rcos theta end{bmatrix}$$
Hence $$ |J^{-1}| = r cos^2 theta + r sin^2 theta = r $$
and
$$ f_{R,theta}(r,theta) = r frac{1}{2pi} e^{-frac{1}{2}(n_1^2+n_2^2)} = boxed{frac{r}{2 pi}e^{-frac{1}{2}(r^2)}} qquad 0 leq theta leq 2 pi, 0 leq r leq infty $$
And we see $theta sim unif[0,2pi]$ (since $f_{Theta}(theta) = frac{1}{2pi}$ ) and $R^2 sim exp(1/2)$
Hence, to simulate $Theta$ we simply take $2 pi U_2$ where $$U_2 sim unif[0,1]$$ and to simulate $R$ we can take $$-2ln U_1$$ where $U_1 sim unif[0,1]$. ( to see why find the density of $X=-ln U, U sim unif[0,1]$).
So, Box and Muller simply inverted $N_1= R cos theta$, $N_2= R sin theta$ and moved from $(R,Theta)$ to $(N_1,N_2)$ by simulating $Theta$ from $2 pi U_2$, and an independent $R$ from $ sqrt{- 2 ln U_1}$
explicitly
$$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2)$$
$$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2)$$
And that is that is the mathematical logic behind it.
p.s. As I don't know whether it is clear through the mathematical justification (I believe it is, but you may have been lost in details), note you need a pair of drawings from the uniform to get a pair of normals
answered Mar 14 at 16:14
StatsStats
668210
$endgroup$
add a comment |
$begingroup$
Let $N_1,N_2$ be independent $N(0,1)$ random variates. $(N_1,N_2)$ defines a point in Cartesian coordinates. We transform to polar by
$$ N_1 = R cos theta$$
$$ N_2 = R sin theta$$
This transformation is one to one and has continuous derivatives. So we can derive the joint distribution $f_{R,theta}(r,theta)$ by using
$$f_{R,theta}(r,theta) = f_{N_1,N_2}(n_1,n_2)|J^{-1}|$$
where $J$ is the Jacobian of the transformation and
$$ J^{-1}(r,theta )=
{begin{bmatrix}{dfrac {partial n_1}{partial r}}{dfrac {partial n_1}{partial theta }}\{dfrac {partial n_2}{partial r}}{dfrac {partial n_2}{partial theta }}end{bmatrix}}
= begin{bmatrix}cos theta & -rsin theta \sin theta & rcos theta end{bmatrix}$$
Hence $$ |J^{-1}| = r cos^2 theta + r sin^2 theta = r $$
and
$$ f_{R,theta}(r,theta) = r frac{1}{2pi} e^{-frac{1}{2}(n_1^2+n_2^2)} = boxed{frac{r}{2 pi}e^{-frac{1}{2}(r^2)}} qquad 0 leq theta leq 2 pi, 0 leq r leq infty $$
And we see $theta sim unif[0,2pi]$ (since $f_{Theta}(theta) = frac{1}{2pi}$ ) and $R^2 sim exp(1/2)$
Hence, to simulate $Theta$ we simply take $2 pi U_2$ where $$U_2 sim unif[0,1]$$ and to simulate $R$ we can take $$-2ln U_1$$ where $U_1 sim unif[0,1]$. ( to see why find the density of $X=-ln U, U sim unif[0,1]$).
So, Box and Muller simply inverted $N_1= R cos theta$, $N_2= R sin theta$ and moved from $(R,Theta)$ to $(N_1,N_2)$ by simulating $Theta$ from $2 pi U_2$, and an independent $R$ from $ sqrt{- 2 ln U_1}$
explicitly
$$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2)$$
$$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2)$$
And that is that is the mathematical logic behind it.
p.s. As I don't know whether it is clear through the mathematical justification (I believe it is, but you may have been lost in details), note you need a pair of drawings from the uniform to get a pair of normals
answered Mar 14 at 16:14
StatsStats
668210
$endgroup$
Let $N_1,N_2$ be independent $N(0,1)$ random variates. $(N_1,N_2)$ defines a point in Cartesian coordinates. We transform to polar by
$$ N_1 = R cos theta$$
$$ N_2 = R sin theta$$
This transformation is one to one and has continuous derivatives. So we can derive the joint distribution $f_{R,theta}(r,theta)$ by using
$$f_{R,theta}(r,theta) = f_{N_1,N_2}(n_1,n_2)|J^{-1}|$$
where $J$ is the Jacobian of the transformation and
$$ J^{-1}(r,theta )=
{begin{bmatrix}{dfrac {partial n_1}{partial r}}{dfrac {partial n_1}{partial theta }}\{dfrac {partial n_2}{partial r}}{dfrac {partial n_2}{partial theta }}end{bmatrix}}
= begin{bmatrix}cos theta & -rsin theta \sin theta & rcos theta end{bmatrix}$$
Hence $$ |J^{-1}| = r cos^2 theta + r sin^2 theta = r $$
and
$$ f_{R,theta}(r,theta) = r frac{1}{2pi} e^{-frac{1}{2}(n_1^2+n_2^2)} = boxed{frac{r}{2 pi}e^{-frac{1}{2}(r^2)}} qquad 0 leq theta leq 2 pi, 0 leq r leq infty $$
And we see $theta sim unif[0,2pi]$ (since $f_{Theta}(theta) = frac{1}{2pi}$ ) and $R^2 sim exp(1/2)$
Hence, to simulate $Theta$ we simply take $2 pi U_2$ where $$U_2 sim unif[0,1]$$ and to simulate $R$ we can take $$-2ln U_1$$ where $U_1 sim unif[0,1]$. ( to see why find the density of $X=-ln U, U sim unif[0,1]$).
So, Box and Muller simply inverted $N_1= R cos theta$, $N_2= R sin theta$ and moved from $(R,Theta)$ to $(N_1,N_2)$ by simulating $Theta$ from $2 pi U_2$, and an independent $R$ from $ sqrt{- 2 ln U_1}$
explicitly
$$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2)$$
$$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2)$$
And that is that is the mathematical logic behind it.
p.s. As I don't know whether it is clear through the mathematical justification (I believe it is, but you may have been lost in details), note you need a pair of drawings from the uniform to get a pair of normals
answered Mar 14 at 16:14
StatsStats
668210
edited Mar 14 at 21:11
answered Mar 14 at 16:14
StatsStats
668210
answered Mar 14 at 16:14
StatsStats
668210
answered Mar 14 at 16:14
StatsStats
668210
668210
add a comment |
add a comment |
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$begingroup$
You will want to investigate the Probability Integral Transform.
$endgroup$
– StatsStudent
Mar 14 at 15:28
1
$begingroup$
Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
$endgroup$
– whuber♦
Mar 14 at 18:44