Why is Monte Carlo integration randomly sampled?
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As I understand, Monte Carlo integration uses stochastic sampling to sample points.
Obviously, you would need many samples to reach an accurate result, but why does this process have to be random?
Would using a symmetrical grid of very dense samples (e.g. a 1 million by 1 million grid) achieve the same goal?
Are there any benefits to random sampling?
calculus integration numerical-methods stochastic-integrals monte-carlo
asked Oct 19 '18 at 9:39
Prithvi BoinpallyPrithvi Boinpally
234
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|
show 1 more comment
$begingroup$
As I understand, Monte Carlo integration uses stochastic sampling to sample points.
Obviously, you would need many samples to reach an accurate result, but why does this process have to be random?
Would using a symmetrical grid of very dense samples (e.g. a 1 million by 1 million grid) achieve the same goal?
Are there any benefits to random sampling?
calculus integration numerical-methods stochastic-integrals monte-carlo
asked Oct 19 '18 at 9:39
Prithvi BoinpallyPrithvi Boinpally
234
$endgroup$
$begingroup$
Yes, using a symmetrical grid would be a solution also. But it's very very very time consuming to operate on a million times a million points, even on a powerful computer.
$endgroup$
– Matti P.
Oct 19 '18 at 9:42
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So how is random sampling faster?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:45
1
$begingroup$
Because you need much fewer points. And due to the random nature of the process, the answer will still be relatively reliable.
$endgroup$
– Matti P.
Oct 19 '18 at 9:54
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Could you elaborate on the math behind why there are fewer samples needed? Does it have anything to do with convergence?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:55
1
$begingroup$
Monte-Carlo with $N$ points has an error of $O(1/sqrt{N})$. Sampling over a grid in $d$ dimensions of $n$ sample coordinates per dimension gives a total of $N=n^d$ points. Thus you would need an integration method with error order $d/2$ to beat Monte-Carlo. Low-discrepancy sequences, also called quasi-random or sub-random, provide a Monte-Carlo error of even $O(1/N)$ which makes beating it with grid-sampling even harder.
$endgroup$
– LutzL
Oct 19 '18 at 10:06
|
show 1 more comment
$begingroup$
As I understand, Monte Carlo integration uses stochastic sampling to sample points.
Obviously, you would need many samples to reach an accurate result, but why does this process have to be random?
Would using a symmetrical grid of very dense samples (e.g. a 1 million by 1 million grid) achieve the same goal?
Are there any benefits to random sampling?
calculus integration numerical-methods stochastic-integrals monte-carlo
asked Oct 19 '18 at 9:39
Prithvi BoinpallyPrithvi Boinpally
234
$endgroup$
As I understand, Monte Carlo integration uses stochastic sampling to sample points.
Obviously, you would need many samples to reach an accurate result, but why does this process have to be random?
Would using a symmetrical grid of very dense samples (e.g. a 1 million by 1 million grid) achieve the same goal?
Are there any benefits to random sampling?
calculus integration numerical-methods stochastic-integrals monte-carlo
calculus integration numerical-methods stochastic-integrals monte-carlo
asked Oct 19 '18 at 9:39
Prithvi BoinpallyPrithvi Boinpally
234
asked Oct 19 '18 at 9:39
Prithvi BoinpallyPrithvi Boinpally
234
edited Dec 8 '18 at 22:53
Prithvi Boinpally
asked Oct 19 '18 at 9:39
Prithvi BoinpallyPrithvi Boinpally
234
asked Oct 19 '18 at 9:39
Prithvi BoinpallyPrithvi Boinpally
234
asked Oct 19 '18 at 9:39
Prithvi BoinpallyPrithvi Boinpally
234
234
$begingroup$
Yes, using a symmetrical grid would be a solution also. But it's very very very time consuming to operate on a million times a million points, even on a powerful computer.
$endgroup$
– Matti P.
Oct 19 '18 at 9:42
$begingroup$
So how is random sampling faster?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:45
1
$begingroup$
Because you need much fewer points. And due to the random nature of the process, the answer will still be relatively reliable.
$endgroup$
– Matti P.
Oct 19 '18 at 9:54
$begingroup$
Could you elaborate on the math behind why there are fewer samples needed? Does it have anything to do with convergence?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:55
1
$begingroup$
Monte-Carlo with $N$ points has an error of $O(1/sqrt{N})$. Sampling over a grid in $d$ dimensions of $n$ sample coordinates per dimension gives a total of $N=n^d$ points. Thus you would need an integration method with error order $d/2$ to beat Monte-Carlo. Low-discrepancy sequences, also called quasi-random or sub-random, provide a Monte-Carlo error of even $O(1/N)$ which makes beating it with grid-sampling even harder.
$endgroup$
– LutzL
Oct 19 '18 at 10:06
|
show 1 more comment
$begingroup$
Yes, using a symmetrical grid would be a solution also. But it's very very very time consuming to operate on a million times a million points, even on a powerful computer.
$endgroup$
– Matti P.
Oct 19 '18 at 9:42
$begingroup$
So how is random sampling faster?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:45
1
$begingroup$
Because you need much fewer points. And due to the random nature of the process, the answer will still be relatively reliable.
$endgroup$
– Matti P.
Oct 19 '18 at 9:54
$begingroup$
Could you elaborate on the math behind why there are fewer samples needed? Does it have anything to do with convergence?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:55
1
$begingroup$
Monte-Carlo with $N$ points has an error of $O(1/sqrt{N})$. Sampling over a grid in $d$ dimensions of $n$ sample coordinates per dimension gives a total of $N=n^d$ points. Thus you would need an integration method with error order $d/2$ to beat Monte-Carlo. Low-discrepancy sequences, also called quasi-random or sub-random, provide a Monte-Carlo error of even $O(1/N)$ which makes beating it with grid-sampling even harder.
$endgroup$
– LutzL
Oct 19 '18 at 10:06
$begingroup$
Yes, using a symmetrical grid would be a solution also. But it's very very very time consuming to operate on a million times a million points, even on a powerful computer.
$endgroup$
– Matti P.
Oct 19 '18 at 9:42
$begingroup$
Yes, using a symmetrical grid would be a solution also. But it's very very very time consuming to operate on a million times a million points, even on a powerful computer.
$endgroup$
– Matti P.
Oct 19 '18 at 9:42
$begingroup$
So how is random sampling faster?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:45
$begingroup$
So how is random sampling faster?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:45
1
1
$begingroup$
Because you need much fewer points. And due to the random nature of the process, the answer will still be relatively reliable.
$endgroup$
– Matti P.
Oct 19 '18 at 9:54
$begingroup$
Because you need much fewer points. And due to the random nature of the process, the answer will still be relatively reliable.
$endgroup$
– Matti P.
Oct 19 '18 at 9:54
$begingroup$
Could you elaborate on the math behind why there are fewer samples needed? Does it have anything to do with convergence?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:55
$begingroup$
Could you elaborate on the math behind why there are fewer samples needed? Does it have anything to do with convergence?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:55
1
1
$begingroup$
Monte-Carlo with $N$ points has an error of $O(1/sqrt{N})$. Sampling over a grid in $d$ dimensions of $n$ sample coordinates per dimension gives a total of $N=n^d$ points. Thus you would need an integration method with error order $d/2$ to beat Monte-Carlo. Low-discrepancy sequences, also called quasi-random or sub-random, provide a Monte-Carlo error of even $O(1/N)$ which makes beating it with grid-sampling even harder.
$endgroup$
– LutzL
Oct 19 '18 at 10:06
$begingroup$
Monte-Carlo with $N$ points has an error of $O(1/sqrt{N})$. Sampling over a grid in $d$ dimensions of $n$ sample coordinates per dimension gives a total of $N=n^d$ points. Thus you would need an integration method with error order $d/2$ to beat Monte-Carlo. Low-discrepancy sequences, also called quasi-random or sub-random, provide a Monte-Carlo error of even $O(1/N)$ which makes beating it with grid-sampling even harder.
$endgroup$
– LutzL
Oct 19 '18 at 10:06
|
show 1 more comment
1 Answer
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It doesn't have to be completely random. In fact, using semirandom sample points is an active area of research. For instance the Latin Hypercube or Sobol' sequence.
The reason it often is random is that nonrandom sample points can magnify certain biases present in the function to be estimated. And also, of course, because it's easier.
answered Oct 19 '18 at 13:10
dbxdbx
1,622312
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$begingroup$
It doesn't have to be completely random. In fact, using semirandom sample points is an active area of research. For instance the Latin Hypercube or Sobol' sequence.
The reason it often is random is that nonrandom sample points can magnify certain biases present in the function to be estimated. And also, of course, because it's easier.
answered Oct 19 '18 at 13:10
dbxdbx
1,622312
$endgroup$
add a comment |
$begingroup$
It doesn't have to be completely random. In fact, using semirandom sample points is an active area of research. For instance the Latin Hypercube or Sobol' sequence.
The reason it often is random is that nonrandom sample points can magnify certain biases present in the function to be estimated. And also, of course, because it's easier.
answered Oct 19 '18 at 13:10
dbxdbx
1,622312
$endgroup$
add a comment |
$begingroup$
It doesn't have to be completely random. In fact, using semirandom sample points is an active area of research. For instance the Latin Hypercube or Sobol' sequence.
The reason it often is random is that nonrandom sample points can magnify certain biases present in the function to be estimated. And also, of course, because it's easier.
answered Oct 19 '18 at 13:10
dbxdbx
1,622312
$endgroup$
It doesn't have to be completely random. In fact, using semirandom sample points is an active area of research. For instance the Latin Hypercube or Sobol' sequence.
The reason it often is random is that nonrandom sample points can magnify certain biases present in the function to be estimated. And also, of course, because it's easier.
answered Oct 19 '18 at 13:10
dbxdbx
1,622312
answered Oct 19 '18 at 13:10
dbxdbx
1,622312
answered Oct 19 '18 at 13:10
dbxdbx
1,622312
answered Oct 19 '18 at 13:10
dbxdbx
1,622312
1,622312
add a comment |
add a comment |
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$begingroup$
Yes, using a symmetrical grid would be a solution also. But it's very very very time consuming to operate on a million times a million points, even on a powerful computer.
$endgroup$
– Matti P.
Oct 19 '18 at 9:42
$begingroup$
So how is random sampling faster?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:45
1
$begingroup$
Because you need much fewer points. And due to the random nature of the process, the answer will still be relatively reliable.
$endgroup$
– Matti P.
Oct 19 '18 at 9:54
$begingroup$
Could you elaborate on the math behind why there are fewer samples needed? Does it have anything to do with convergence?
$endgroup$
– Prithvi Boinpally
Oct 19 '18 at 9:55
1
$begingroup$
Monte-Carlo with $N$ points has an error of $O(1/sqrt{N})$. Sampling over a grid in $d$ dimensions of $n$ sample coordinates per dimension gives a total of $N=n^d$ points. Thus you would need an integration method with error order $d/2$ to beat Monte-Carlo. Low-discrepancy sequences, also called quasi-random or sub-random, provide a Monte-Carlo error of even $O(1/N)$ which makes beating it with grid-sampling even harder.
$endgroup$
– LutzL
Oct 19 '18 at 10:06