Why is the Monte Carlo integration dimensionally independent?
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Suppose that the random variables $x_1, x_2, ... x_N$ are drawn independently from the probability density function $p(x)$.
Now the convergence rate of a deterministic numerical integration method is $O(N^{-2/d})$.
Why does the Monte Carlo integration method yield a convergence rate of $O(N^{-1/2})$ ?
integration convergence monte-carlo
asked Nov 18 at 21:33
user1511417
451414
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Suppose that the random variables $x_1, x_2, ... x_N$ are drawn independently from the probability density function $p(x)$.
Now the convergence rate of a deterministic numerical integration method is $O(N^{-2/d})$.
Why does the Monte Carlo integration method yield a convergence rate of $O(N^{-1/2})$ ?
integration convergence monte-carlo
asked Nov 18 at 21:33
user1511417
451414
The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 at 22:35
This statement does not help me.
– user1511417
Nov 19 at 10:31
There is a basic idea in sampling that the standard deviation of an average is $O(frac{1}{sqrt{N}})$. en.wikipedia.org/wiki/Standard_deviation may help.
– herb steinberg
Nov 19 at 22:59
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 at 23:33
1
For Monte Carlo the sample size is the key factor. The dimension will effect the standard deviation of of one sample, but it does not effect the deviation of the average ,except as a constant multiplier. When doing a deterministic integration, the grid refinement depends on the dimension.
– herb steinberg
Nov 22 at 2:40
add a comment |
up vote
0
down vote
favorite
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up vote
0
down vote
favorite
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Suppose that the random variables $x_1, x_2, ... x_N$ are drawn independently from the probability density function $p(x)$.
Now the convergence rate of a deterministic numerical integration method is $O(N^{-2/d})$.
Why does the Monte Carlo integration method yield a convergence rate of $O(N^{-1/2})$ ?
integration convergence monte-carlo
asked Nov 18 at 21:33
user1511417
451414
Suppose that the random variables $x_1, x_2, ... x_N$ are drawn independently from the probability density function $p(x)$.
Now the convergence rate of a deterministic numerical integration method is $O(N^{-2/d})$.
Why does the Monte Carlo integration method yield a convergence rate of $O(N^{-1/2})$ ?
integration convergence monte-carlo
integration convergence monte-carlo
asked Nov 18 at 21:33
user1511417
451414
asked Nov 18 at 21:33
user1511417
451414
edited Nov 22 at 11:20
asked Nov 18 at 21:33
user1511417
451414
asked Nov 18 at 21:33
user1511417
451414
asked Nov 18 at 21:33
user1511417
451414
451414
The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 at 22:35
This statement does not help me.
– user1511417
Nov 19 at 10:31
There is a basic idea in sampling that the standard deviation of an average is $O(frac{1}{sqrt{N}})$. en.wikipedia.org/wiki/Standard_deviation may help.
– herb steinberg
Nov 19 at 22:59
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 at 23:33
1
For Monte Carlo the sample size is the key factor. The dimension will effect the standard deviation of of one sample, but it does not effect the deviation of the average ,except as a constant multiplier. When doing a deterministic integration, the grid refinement depends on the dimension.
– herb steinberg
Nov 22 at 2:40
add a comment |
The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 at 22:35
This statement does not help me.
– user1511417
Nov 19 at 10:31
There is a basic idea in sampling that the standard deviation of an average is $O(frac{1}{sqrt{N}})$. en.wikipedia.org/wiki/Standard_deviation may help.
– herb steinberg
Nov 19 at 22:59
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 at 23:33
1
For Monte Carlo the sample size is the key factor. The dimension will effect the standard deviation of of one sample, but it does not effect the deviation of the average ,except as a constant multiplier. When doing a deterministic integration, the grid refinement depends on the dimension.
– herb steinberg
Nov 22 at 2:40
The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 at 22:35
The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 at 22:35
This statement does not help me.
– user1511417
Nov 19 at 10:31
This statement does not help me.
– user1511417
Nov 19 at 10:31
There is a basic idea in sampling that the standard deviation of an average is $O(frac{1}{sqrt{N}})$. en.wikipedia.org/wiki/Standard_deviation may help.
– herb steinberg
Nov 19 at 22:59
There is a basic idea in sampling that the standard deviation of an average is $O(frac{1}{sqrt{N}})$. en.wikipedia.org/wiki/Standard_deviation may help.
– herb steinberg
Nov 19 at 22:59
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 at 23:33
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 at 23:33
1
1
For Monte Carlo the sample size is the key factor. The dimension will effect the standard deviation of of one sample, but it does not effect the deviation of the average ,except as a constant multiplier. When doing a deterministic integration, the grid refinement depends on the dimension.
– herb steinberg
Nov 22 at 2:40
For Monte Carlo the sample size is the key factor. The dimension will effect the standard deviation of of one sample, but it does not effect the deviation of the average ,except as a constant multiplier. When doing a deterministic integration, the grid refinement depends on the dimension.
– herb steinberg
Nov 22 at 2:40
add a comment |
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The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 at 22:35
This statement does not help me.
– user1511417
Nov 19 at 10:31
There is a basic idea in sampling that the standard deviation of an average is $O(frac{1}{sqrt{N}})$. en.wikipedia.org/wiki/Standard_deviation may help.
– herb steinberg
Nov 19 at 22:59
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 at 23:33
1
For Monte Carlo the sample size is the key factor. The dimension will effect the standard deviation of of one sample, but it does not effect the deviation of the average ,except as a constant multiplier. When doing a deterministic integration, the grid refinement depends on the dimension.
– herb steinberg
Nov 22 at 2:40