Generate random variable with known PDF expression in MATLAB
I know the probability density function (PDF) expression of random variable r as 2r/R^2 where 0<=r<=R. Then, its CDF is r^2/R^2.
Can someone help me to generate random variable r in MATLAB following above distribution?
matlab random probability-density inverse-transform
edited Nov 10 '18 at 22:56
SecretAgentMan
591313
asked Jun 4 '18 at 6:23
FreyFrey
203311
add a comment |
I know the probability density function (PDF) expression of random variable r as 2r/R^2 where 0<=r<=R. Then, its CDF is r^2/R^2.
Can someone help me to generate random variable r in MATLAB following above distribution?
matlab random probability-density inverse-transform
edited Nov 10 '18 at 22:56
SecretAgentMan
591313
asked Jun 4 '18 at 6:23
FreyFrey
203311
Possible duplicate of Generate random numbers according to distributions
– Gelliant
Jun 4 '18 at 8:30
2
I thinksz = [1 1e6]; R = 5; x = sqrt(rand(sz))*R; histogram(x)does what you want
– Luis Mendo
Jun 4 '18 at 8:42
add a comment |
I know the probability density function (PDF) expression of random variable r as 2r/R^2 where 0<=r<=R. Then, its CDF is r^2/R^2.
Can someone help me to generate random variable r in MATLAB following above distribution?
matlab random probability-density inverse-transform
edited Nov 10 '18 at 22:56
SecretAgentMan
591313
asked Jun 4 '18 at 6:23
FreyFrey
203311
I know the probability density function (PDF) expression of random variable r as 2r/R^2 where 0<=r<=R. Then, its CDF is r^2/R^2.
Can someone help me to generate random variable r in MATLAB following above distribution?
matlab random probability-density inverse-transform
matlab random probability-density inverse-transform
edited Nov 10 '18 at 22:56
SecretAgentMan
591313
asked Jun 4 '18 at 6:23
FreyFrey
203311
edited Nov 10 '18 at 22:56
SecretAgentMan
591313
asked Jun 4 '18 at 6:23
FreyFrey
203311
edited Nov 10 '18 at 22:56
SecretAgentMan
591313
edited Nov 10 '18 at 22:56
SecretAgentMan
591313
edited Nov 10 '18 at 22:56
SecretAgentMan
591313
591313
asked Jun 4 '18 at 6:23
FreyFrey
203311
asked Jun 4 '18 at 6:23
FreyFrey
203311
asked Jun 4 '18 at 6:23
FreyFrey
203311
203311
Possible duplicate of Generate random numbers according to distributions
– Gelliant
Jun 4 '18 at 8:30
2
I thinksz = [1 1e6]; R = 5; x = sqrt(rand(sz))*R; histogram(x)does what you want
– Luis Mendo
Jun 4 '18 at 8:42
add a comment |
Possible duplicate of Generate random numbers according to distributions
– Gelliant
Jun 4 '18 at 8:30
2
I thinksz = [1 1e6]; R = 5; x = sqrt(rand(sz))*R; histogram(x)does what you want
– Luis Mendo
Jun 4 '18 at 8:42
Possible duplicate of Generate random numbers according to distributions
– Gelliant
Jun 4 '18 at 8:30
Possible duplicate of Generate random numbers according to distributions
– Gelliant
Jun 4 '18 at 8:30
2
2
I think
sz = [1 1e6]; R = 5; x = sqrt(rand(sz))*R; histogram(x) does what you want– Luis Mendo
Jun 4 '18 at 8:42
I think
sz = [1 1e6]; R = 5; x = sqrt(rand(sz))*R; histogram(x) does what you want– Luis Mendo
Jun 4 '18 at 8:42
add a comment |
1 Answer
1
active
oldest
votes
https://blogs.sas.com/content/iml/2013/07/22/the-inverse-cdf-method.html
I use the same variables as they use
f(x) = 2x/R^2
F(x) = x^2/R^2
solving for x in the equation F(x) = u
u*R^2 = x^2
x = sqrt(u * R^2) v -sqrt(u * R^2)
in Matlab:
N=1E5;
R=1;
u = rand(1,N);
x = sqrt(u*R^2);
histogram(x)
answered Jun 4 '18 at 8:40
GelliantGelliant
1,5731419
Your distribution is not entirely correct as he has defined it from 0 to R, (thus the 2 in the PDF). So the negative part of the squareroot should be "ignored".
– Nicky Mattsson
Jun 4 '18 at 9:15
Thanks, I will correct it.
– Gelliant
Jun 4 '18 at 11:56
You are welcome - but please remove the incorrect code rather than commenting it out.
– Nicky Mattsson
Jun 5 '18 at 13:57
More on the Inverse Transform at stats.stackexchange.com/questions/184325/… and here.
– SecretAgentMan
Nov 10 '18 at 4:07
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
https://blogs.sas.com/content/iml/2013/07/22/the-inverse-cdf-method.html
I use the same variables as they use
f(x) = 2x/R^2
F(x) = x^2/R^2
solving for x in the equation F(x) = u
u*R^2 = x^2
x = sqrt(u * R^2) v -sqrt(u * R^2)
in Matlab:
N=1E5;
R=1;
u = rand(1,N);
x = sqrt(u*R^2);
histogram(x)
answered Jun 4 '18 at 8:40
GelliantGelliant
1,5731419
Your distribution is not entirely correct as he has defined it from 0 to R, (thus the 2 in the PDF). So the negative part of the squareroot should be "ignored".
– Nicky Mattsson
Jun 4 '18 at 9:15
Thanks, I will correct it.
– Gelliant
Jun 4 '18 at 11:56
You are welcome - but please remove the incorrect code rather than commenting it out.
– Nicky Mattsson
Jun 5 '18 at 13:57
More on the Inverse Transform at stats.stackexchange.com/questions/184325/… and here.
– SecretAgentMan
Nov 10 '18 at 4:07
add a comment |
https://blogs.sas.com/content/iml/2013/07/22/the-inverse-cdf-method.html
I use the same variables as they use
f(x) = 2x/R^2
F(x) = x^2/R^2
solving for x in the equation F(x) = u
u*R^2 = x^2
x = sqrt(u * R^2) v -sqrt(u * R^2)
in Matlab:
N=1E5;
R=1;
u = rand(1,N);
x = sqrt(u*R^2);
histogram(x)
answered Jun 4 '18 at 8:40
GelliantGelliant
1,5731419
Your distribution is not entirely correct as he has defined it from 0 to R, (thus the 2 in the PDF). So the negative part of the squareroot should be "ignored".
– Nicky Mattsson
Jun 4 '18 at 9:15
Thanks, I will correct it.
– Gelliant
Jun 4 '18 at 11:56
You are welcome - but please remove the incorrect code rather than commenting it out.
– Nicky Mattsson
Jun 5 '18 at 13:57
More on the Inverse Transform at stats.stackexchange.com/questions/184325/… and here.
– SecretAgentMan
Nov 10 '18 at 4:07
add a comment |
https://blogs.sas.com/content/iml/2013/07/22/the-inverse-cdf-method.html
I use the same variables as they use
f(x) = 2x/R^2
F(x) = x^2/R^2
solving for x in the equation F(x) = u
u*R^2 = x^2
x = sqrt(u * R^2) v -sqrt(u * R^2)
in Matlab:
N=1E5;
R=1;
u = rand(1,N);
x = sqrt(u*R^2);
histogram(x)
answered Jun 4 '18 at 8:40
GelliantGelliant
1,5731419
https://blogs.sas.com/content/iml/2013/07/22/the-inverse-cdf-method.html
I use the same variables as they use
f(x) = 2x/R^2
F(x) = x^2/R^2
solving for x in the equation F(x) = u
u*R^2 = x^2
x = sqrt(u * R^2) v -sqrt(u * R^2)
in Matlab:
N=1E5;
R=1;
u = rand(1,N);
x = sqrt(u*R^2);
histogram(x)
answered Jun 4 '18 at 8:40
GelliantGelliant
1,5731419
edited Jun 5 '18 at 14:47
answered Jun 4 '18 at 8:40
GelliantGelliant
1,5731419
answered Jun 4 '18 at 8:40
GelliantGelliant
1,5731419
answered Jun 4 '18 at 8:40
GelliantGelliant
1,5731419
1,5731419
Your distribution is not entirely correct as he has defined it from 0 to R, (thus the 2 in the PDF). So the negative part of the squareroot should be "ignored".
– Nicky Mattsson
Jun 4 '18 at 9:15
Thanks, I will correct it.
– Gelliant
Jun 4 '18 at 11:56
You are welcome - but please remove the incorrect code rather than commenting it out.
– Nicky Mattsson
Jun 5 '18 at 13:57
More on the Inverse Transform at stats.stackexchange.com/questions/184325/… and here.
– SecretAgentMan
Nov 10 '18 at 4:07
add a comment |
Your distribution is not entirely correct as he has defined it from 0 to R, (thus the 2 in the PDF). So the negative part of the squareroot should be "ignored".
– Nicky Mattsson
Jun 4 '18 at 9:15
Thanks, I will correct it.
– Gelliant
Jun 4 '18 at 11:56
You are welcome - but please remove the incorrect code rather than commenting it out.
– Nicky Mattsson
Jun 5 '18 at 13:57
More on the Inverse Transform at stats.stackexchange.com/questions/184325/… and here.
– SecretAgentMan
Nov 10 '18 at 4:07
Your distribution is not entirely correct as he has defined it from 0 to R, (thus the 2 in the PDF). So the negative part of the squareroot should be "ignored".
– Nicky Mattsson
Jun 4 '18 at 9:15
Your distribution is not entirely correct as he has defined it from 0 to R, (thus the 2 in the PDF). So the negative part of the squareroot should be "ignored".
– Nicky Mattsson
Jun 4 '18 at 9:15
Thanks, I will correct it.
– Gelliant
Jun 4 '18 at 11:56
Thanks, I will correct it.
– Gelliant
Jun 4 '18 at 11:56
You are welcome - but please remove the incorrect code rather than commenting it out.
– Nicky Mattsson
Jun 5 '18 at 13:57
You are welcome - but please remove the incorrect code rather than commenting it out.
– Nicky Mattsson
Jun 5 '18 at 13:57
More on the Inverse Transform at stats.stackexchange.com/questions/184325/… and here.
– SecretAgentMan
Nov 10 '18 at 4:07
More on the Inverse Transform at stats.stackexchange.com/questions/184325/… and here.
– SecretAgentMan
Nov 10 '18 at 4:07
add a comment |
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Possible duplicate of Generate random numbers according to distributions
– Gelliant
Jun 4 '18 at 8:30
2
I think
sz = [1 1e6]; R = 5; x = sqrt(rand(sz))*R; histogram(x)does what you want– Luis Mendo
Jun 4 '18 at 8:42