Find Random Number Generator following the density $f (x) = frac{1 + alpha x}{2}$, $ −1 ≤ x ≤ 1$, $−1...
$begingroup$
How could random variables with the following density function be generated
from a uniform random number generator? $f (x) = frac{1 + alpha x}{2}$, $ −1 ≤ x ≤ 1$, $−1 ≤alpha ≤ 1$.
To find the Random number generator one first needs to find the distribution function and then the quantile function. However, I have troubles solving this on my own:
begin{align}
F(x) & = int_{-1}^x frac{1+alpha u}{2} du\
& = frac 12 Big( (x+frac{alpha x^2}{2})-(-1+frac{alpha}{2})Big)\
& = alpha x^2 + 2x + 2 - alpha
end{align}
I am not sure how to handle the $alpha$. For getting the quantile(=inverse) function i need to solve this last expression for x.
probability probability-theory statistics probability-distributions inverse-function
edited Dec 11 '18 at 6:04
user1101010
9011830
asked Dec 10 '18 at 10:57
user1607user1607
1718
$endgroup$
add a comment |
$begingroup$
How could random variables with the following density function be generated
from a uniform random number generator? $f (x) = frac{1 + alpha x}{2}$, $ −1 ≤ x ≤ 1$, $−1 ≤alpha ≤ 1$.
To find the Random number generator one first needs to find the distribution function and then the quantile function. However, I have troubles solving this on my own:
begin{align}
F(x) & = int_{-1}^x frac{1+alpha u}{2} du\
& = frac 12 Big( (x+frac{alpha x^2}{2})-(-1+frac{alpha}{2})Big)\
& = alpha x^2 + 2x + 2 - alpha
end{align}
I am not sure how to handle the $alpha$. For getting the quantile(=inverse) function i need to solve this last expression for x.
probability probability-theory statistics probability-distributions inverse-function
edited Dec 11 '18 at 6:04
user1101010
9011830
asked Dec 10 '18 at 10:57
user1607user1607
1718
$endgroup$
$begingroup$
I think you could use the quadratic formula to solve for x, treating a as a constant!
$endgroup$
– Zach
Dec 10 '18 at 12:08
add a comment |
$begingroup$
How could random variables with the following density function be generated
from a uniform random number generator? $f (x) = frac{1 + alpha x}{2}$, $ −1 ≤ x ≤ 1$, $−1 ≤alpha ≤ 1$.
To find the Random number generator one first needs to find the distribution function and then the quantile function. However, I have troubles solving this on my own:
begin{align}
F(x) & = int_{-1}^x frac{1+alpha u}{2} du\
& = frac 12 Big( (x+frac{alpha x^2}{2})-(-1+frac{alpha}{2})Big)\
& = alpha x^2 + 2x + 2 - alpha
end{align}
I am not sure how to handle the $alpha$. For getting the quantile(=inverse) function i need to solve this last expression for x.
probability probability-theory statistics probability-distributions inverse-function
edited Dec 11 '18 at 6:04
user1101010
9011830
asked Dec 10 '18 at 10:57
user1607user1607
1718
$endgroup$
How could random variables with the following density function be generated
from a uniform random number generator? $f (x) = frac{1 + alpha x}{2}$, $ −1 ≤ x ≤ 1$, $−1 ≤alpha ≤ 1$.
To find the Random number generator one first needs to find the distribution function and then the quantile function. However, I have troubles solving this on my own:
begin{align}
F(x) & = int_{-1}^x frac{1+alpha u}{2} du\
& = frac 12 Big( (x+frac{alpha x^2}{2})-(-1+frac{alpha}{2})Big)\
& = alpha x^2 + 2x + 2 - alpha
end{align}
I am not sure how to handle the $alpha$. For getting the quantile(=inverse) function i need to solve this last expression for x.
probability probability-theory statistics probability-distributions inverse-function
probability probability-theory statistics probability-distributions inverse-function
edited Dec 11 '18 at 6:04
user1101010
9011830
asked Dec 10 '18 at 10:57
user1607user1607
1718
edited Dec 11 '18 at 6:04
user1101010
9011830
asked Dec 10 '18 at 10:57
user1607user1607
1718
edited Dec 11 '18 at 6:04
user1101010
9011830
edited Dec 11 '18 at 6:04
user1101010
9011830
edited Dec 11 '18 at 6:04
user1101010
9011830
9011830
asked Dec 10 '18 at 10:57
user1607user1607
1718
asked Dec 10 '18 at 10:57
user1607user1607
1718
asked Dec 10 '18 at 10:57
user1607user1607
1718
1718
$begingroup$
I think you could use the quadratic formula to solve for x, treating a as a constant!
$endgroup$
– Zach
Dec 10 '18 at 12:08
add a comment |
$begingroup$
I think you could use the quadratic formula to solve for x, treating a as a constant!
$endgroup$
– Zach
Dec 10 '18 at 12:08
$begingroup$
I think you could use the quadratic formula to solve for x, treating a as a constant!
$endgroup$
– Zach
Dec 10 '18 at 12:08
$begingroup$
I think you could use the quadratic formula to solve for x, treating a as a constant!
$endgroup$
– Zach
Dec 10 '18 at 12:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Not sure how you get from your second line to the third, there is clearly error somewhere
begin{equation}
F(x) = frac{alpha}{4}(x^2 - 1) + frac{1}{2}(x+1)
end{equation}
with obvious properties $F(x=-1)=0$ and $F(x=1)=1$. Now assign $F(x)$ to RV $xi$ in the [0...1] range and find inverse:
begin{equation}
x = F^{-1}(xi)
end{equation}
Simple quadratic equation with solution:
begin{equation}
x = frac{2 ; sqrt{frac{1}{4} + alpha(frac{alpha}{4} + xi - frac{1}{2})} - 1}{alpha}
end{equation}
For the case when $alpha=0$, $x=2xi-1$
Easy to check:
begin{equation}
xi=0, ;then; x=-1
end{equation}
begin{equation}
xi=1, ;then; x=+1
end{equation}
Python code
import numpy as np
import matplotlib.pyplot as plt
def PDF(x, alpha):
return (1.0 + alpha*x) / 2
def sampleAlpha(n, alpha):
U01 = np.random.uniform(size=n)
if alpha == 0.0:
return 2.0*U01 - 1.0
D = 0.25 + alpha*(U01 + 0.25*alpha - 0.5)
return (2.0*np.sqrt(D) - 1.0) / alpha
nbins = 100
xmin = -1.0
xmax = 1.0
alpha = 0.3
t = np.linspace(xmin, xmax, nbins + 1)
q = sampleAlpha(100000, alpha)
h, bedges = np.histogram(q, bins = t, density = True)
p = PDF(t, alpha)
plt.bar(bedges[:-1], h, width = (xmax - xmin)/float(nbins), align='edge')
plt.plot(t, p, 'r') # plotting PDF
plt.xlim(xmin, xmax)
plt.show()
generates graphs like one below for sampled vs PDF
answered Dec 11 '18 at 5:11
Severin PappadeuxSeverin Pappadeux
15318
$endgroup$
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Not sure how you get from your second line to the third, there is clearly error somewhere
begin{equation}
F(x) = frac{alpha}{4}(x^2 - 1) + frac{1}{2}(x+1)
end{equation}
with obvious properties $F(x=-1)=0$ and $F(x=1)=1$. Now assign $F(x)$ to RV $xi$ in the [0...1] range and find inverse:
begin{equation}
x = F^{-1}(xi)
end{equation}
Simple quadratic equation with solution:
begin{equation}
x = frac{2 ; sqrt{frac{1}{4} + alpha(frac{alpha}{4} + xi - frac{1}{2})} - 1}{alpha}
end{equation}
For the case when $alpha=0$, $x=2xi-1$
Easy to check:
begin{equation}
xi=0, ;then; x=-1
end{equation}
begin{equation}
xi=1, ;then; x=+1
end{equation}
Python code
import numpy as np
import matplotlib.pyplot as plt
def PDF(x, alpha):
return (1.0 + alpha*x) / 2
def sampleAlpha(n, alpha):
U01 = np.random.uniform(size=n)
if alpha == 0.0:
return 2.0*U01 - 1.0
D = 0.25 + alpha*(U01 + 0.25*alpha - 0.5)
return (2.0*np.sqrt(D) - 1.0) / alpha
nbins = 100
xmin = -1.0
xmax = 1.0
alpha = 0.3
t = np.linspace(xmin, xmax, nbins + 1)
q = sampleAlpha(100000, alpha)
h, bedges = np.histogram(q, bins = t, density = True)
p = PDF(t, alpha)
plt.bar(bedges[:-1], h, width = (xmax - xmin)/float(nbins), align='edge')
plt.plot(t, p, 'r') # plotting PDF
plt.xlim(xmin, xmax)
plt.show()
generates graphs like one below for sampled vs PDF
answered Dec 11 '18 at 5:11
Severin PappadeuxSeverin Pappadeux
15318
$endgroup$
add a comment |
$begingroup$
Not sure how you get from your second line to the third, there is clearly error somewhere
begin{equation}
F(x) = frac{alpha}{4}(x^2 - 1) + frac{1}{2}(x+1)
end{equation}
with obvious properties $F(x=-1)=0$ and $F(x=1)=1$. Now assign $F(x)$ to RV $xi$ in the [0...1] range and find inverse:
begin{equation}
x = F^{-1}(xi)
end{equation}
Simple quadratic equation with solution:
begin{equation}
x = frac{2 ; sqrt{frac{1}{4} + alpha(frac{alpha}{4} + xi - frac{1}{2})} - 1}{alpha}
end{equation}
For the case when $alpha=0$, $x=2xi-1$
Easy to check:
begin{equation}
xi=0, ;then; x=-1
end{equation}
begin{equation}
xi=1, ;then; x=+1
end{equation}
Python code
import numpy as np
import matplotlib.pyplot as plt
def PDF(x, alpha):
return (1.0 + alpha*x) / 2
def sampleAlpha(n, alpha):
U01 = np.random.uniform(size=n)
if alpha == 0.0:
return 2.0*U01 - 1.0
D = 0.25 + alpha*(U01 + 0.25*alpha - 0.5)
return (2.0*np.sqrt(D) - 1.0) / alpha
nbins = 100
xmin = -1.0
xmax = 1.0
alpha = 0.3
t = np.linspace(xmin, xmax, nbins + 1)
q = sampleAlpha(100000, alpha)
h, bedges = np.histogram(q, bins = t, density = True)
p = PDF(t, alpha)
plt.bar(bedges[:-1], h, width = (xmax - xmin)/float(nbins), align='edge')
plt.plot(t, p, 'r') # plotting PDF
plt.xlim(xmin, xmax)
plt.show()
generates graphs like one below for sampled vs PDF
answered Dec 11 '18 at 5:11
Severin PappadeuxSeverin Pappadeux
15318
$endgroup$
add a comment |
$begingroup$
Not sure how you get from your second line to the third, there is clearly error somewhere
begin{equation}
F(x) = frac{alpha}{4}(x^2 - 1) + frac{1}{2}(x+1)
end{equation}
with obvious properties $F(x=-1)=0$ and $F(x=1)=1$. Now assign $F(x)$ to RV $xi$ in the [0...1] range and find inverse:
begin{equation}
x = F^{-1}(xi)
end{equation}
Simple quadratic equation with solution:
begin{equation}
x = frac{2 ; sqrt{frac{1}{4} + alpha(frac{alpha}{4} + xi - frac{1}{2})} - 1}{alpha}
end{equation}
For the case when $alpha=0$, $x=2xi-1$
Easy to check:
begin{equation}
xi=0, ;then; x=-1
end{equation}
begin{equation}
xi=1, ;then; x=+1
end{equation}
Python code
import numpy as np
import matplotlib.pyplot as plt
def PDF(x, alpha):
return (1.0 + alpha*x) / 2
def sampleAlpha(n, alpha):
U01 = np.random.uniform(size=n)
if alpha == 0.0:
return 2.0*U01 - 1.0
D = 0.25 + alpha*(U01 + 0.25*alpha - 0.5)
return (2.0*np.sqrt(D) - 1.0) / alpha
nbins = 100
xmin = -1.0
xmax = 1.0
alpha = 0.3
t = np.linspace(xmin, xmax, nbins + 1)
q = sampleAlpha(100000, alpha)
h, bedges = np.histogram(q, bins = t, density = True)
p = PDF(t, alpha)
plt.bar(bedges[:-1], h, width = (xmax - xmin)/float(nbins), align='edge')
plt.plot(t, p, 'r') # plotting PDF
plt.xlim(xmin, xmax)
plt.show()
generates graphs like one below for sampled vs PDF
answered Dec 11 '18 at 5:11
Severin PappadeuxSeverin Pappadeux
15318
$endgroup$
Not sure how you get from your second line to the third, there is clearly error somewhere
begin{equation}
F(x) = frac{alpha}{4}(x^2 - 1) + frac{1}{2}(x+1)
end{equation}
with obvious properties $F(x=-1)=0$ and $F(x=1)=1$. Now assign $F(x)$ to RV $xi$ in the [0...1] range and find inverse:
begin{equation}
x = F^{-1}(xi)
end{equation}
Simple quadratic equation with solution:
begin{equation}
x = frac{2 ; sqrt{frac{1}{4} + alpha(frac{alpha}{4} + xi - frac{1}{2})} - 1}{alpha}
end{equation}
For the case when $alpha=0$, $x=2xi-1$
Easy to check:
begin{equation}
xi=0, ;then; x=-1
end{equation}
begin{equation}
xi=1, ;then; x=+1
end{equation}
Python code
import numpy as np
import matplotlib.pyplot as plt
def PDF(x, alpha):
return (1.0 + alpha*x) / 2
def sampleAlpha(n, alpha):
U01 = np.random.uniform(size=n)
if alpha == 0.0:
return 2.0*U01 - 1.0
D = 0.25 + alpha*(U01 + 0.25*alpha - 0.5)
return (2.0*np.sqrt(D) - 1.0) / alpha
nbins = 100
xmin = -1.0
xmax = 1.0
alpha = 0.3
t = np.linspace(xmin, xmax, nbins + 1)
q = sampleAlpha(100000, alpha)
h, bedges = np.histogram(q, bins = t, density = True)
p = PDF(t, alpha)
plt.bar(bedges[:-1], h, width = (xmax - xmin)/float(nbins), align='edge')
plt.plot(t, p, 'r') # plotting PDF
plt.xlim(xmin, xmax)
plt.show()
generates graphs like one below for sampled vs PDF
answered Dec 11 '18 at 5:11
Severin PappadeuxSeverin Pappadeux
15318
answered Dec 11 '18 at 5:11
Severin PappadeuxSeverin Pappadeux
15318
answered Dec 11 '18 at 5:11
Severin PappadeuxSeverin Pappadeux
15318
answered Dec 11 '18 at 5:11
Severin PappadeuxSeverin Pappadeux
15318
15318
add a comment |
add a comment |
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$begingroup$
I think you could use the quadratic formula to solve for x, treating a as a constant!
$endgroup$
– Zach
Dec 10 '18 at 12:08