Mixing
Why is the Monte Carlo integration dimensionally independent?
0
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 '18 at 21:33
user1511417user1511417
451414
add a comment |
0
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 '18 at 21:33
user1511417user1511417
451414
The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 '18 at 22:35
This statement does not help me.
– user1511417
Nov 19 '18 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 '18 at 22:59
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 '18 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 '18 at 2:40
add a comment |
0
0
0
2
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 '18 at 21:33
user1511417user1511417
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 '18 at 21:33
user1511417user1511417
451414
asked Nov 18 '18 at 21:33
user1511417user1511417
451414
edited Nov 22 '18 at 11:20
user1511417
asked Nov 18 '18 at 21:33
user1511417user1511417
451414
asked Nov 18 '18 at 21:33
user1511417user1511417
451414
asked Nov 18 '18 at 21:33
user1511417user1511417
451414
451414
The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 '18 at 22:35
This statement does not help me.
– user1511417
Nov 19 '18 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 '18 at 22:59
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 '18 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 '18 at 2:40
add a comment |
The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 '18 at 22:35
This statement does not help me.
– user1511417
Nov 19 '18 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 '18 at 22:59
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 '18 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 '18 at 2:40
The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 '18 at 22:35
The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 '18 at 22:35
This statement does not help me.
– user1511417
Nov 19 '18 at 10:31
This statement does not help me.
– user1511417
Nov 19 '18 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 '18 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 '18 at 22:59
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 '18 at 23:33
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 '18 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 '18 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 '18 at 2:40
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The standard deviation of this average is$O(N^{-1/2})$.
– herb steinberg
Nov 18 '18 at 22:35
This statement does not help me.
– user1511417
Nov 19 '18 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 '18 at 22:59
And why doesn't this deviation apply to a deterministic numerical integration?
– user1511417
Nov 21 '18 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 '18 at 2:40