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Estimating the value of $pi$ with a Monte Carlo dartboard: $

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5

$begingroup$

I'm trying to figure out which is the proper way to estimate $pi$ using the Monte Carlo method randomly distributing points in a square that also contains an inscribed circle.

Some sources say to use the comparison of $sqrtx^2+y^2le 1$, while others use $sqrtx^2+y^2<1$.

                                          

Here's some example code from a wikipedia article:

def monte_carlo_pi(nsamples):
 acc = 0
 for i in range(nsamples):
 x = random.random()
 y = random.random()
 if (x**2 + y**2) < 1.0:
 acc += 1
 return 4.0 * acc / nsamples

Instead of posting a long list of websites that use $le 1$ or $< 1$, I've made the list and stored it on the following websites:
See either: socrates.io or markdown.press or markdownshare for examples using less than and less than or equal to.

math simulation monte-carlo

share|improve this question

edited May 9 at 10:26

MachineLearner

610112

asked May 9 at 5:17

I_Don't_CodeI_Don't_Code

262

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know you are trying to have a logical, math, reason which one, but that asides your first list of resources referring to <= is way more legit that the other list. I have also found in books, articles, university lectures mostly <=, like in Oak Ridge National Lab: olcf.ornl.gov/tutorials/monte-carlo-pi. End of the way it is method that randomly samples, and there might be that there is no math-proven right or wrong here, but rather you gotta pick the one that converge faster; maybe.
  $endgroup$
  – TwinPenguins
  May 9 at 5:55
  

  
  
  
  

add a comment | 

5

$begingroup$

I'm trying to figure out which is the proper way to estimate $pi$ using the Monte Carlo method randomly distributing points in a square that also contains an inscribed circle.

Some sources say to use the comparison of $sqrtx^2+y^2le 1$, while others use $sqrtx^2+y^2<1$.

                                          

Here's some example code from a wikipedia article:

def monte_carlo_pi(nsamples):
 acc = 0
 for i in range(nsamples):
 x = random.random()
 y = random.random()
 if (x**2 + y**2) < 1.0:
 acc += 1
 return 4.0 * acc / nsamples

Instead of posting a long list of websites that use $le 1$ or $< 1$, I've made the list and stored it on the following websites:
See either: socrates.io or markdown.press or markdownshare for examples using less than and less than or equal to.

math simulation monte-carlo

share|improve this question

edited May 9 at 10:26

MachineLearner

610112

asked May 9 at 5:17

I_Don't_CodeI_Don't_Code

262

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know you are trying to have a logical, math, reason which one, but that asides your first list of resources referring to <= is way more legit that the other list. I have also found in books, articles, university lectures mostly <=, like in Oak Ridge National Lab: olcf.ornl.gov/tutorials/monte-carlo-pi. End of the way it is method that randomly samples, and there might be that there is no math-proven right or wrong here, but rather you gotta pick the one that converge faster; maybe.
  $endgroup$
  – TwinPenguins
  May 9 at 5:55
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to figure out which is the proper way to estimate $pi$ using the Monte Carlo method randomly distributing points in a square that also contains an inscribed circle.

Some sources say to use the comparison of $sqrtx^2+y^2le 1$, while others use $sqrtx^2+y^2<1$.

                                          

Here's some example code from a wikipedia article:

def monte_carlo_pi(nsamples):
 acc = 0
 for i in range(nsamples):
 x = random.random()
 y = random.random()
 if (x**2 + y**2) < 1.0:
 acc += 1
 return 4.0 * acc / nsamples

Instead of posting a long list of websites that use $le 1$ or $< 1$, I've made the list and stored it on the following websites:
See either: socrates.io or markdown.press or markdownshare for examples using less than and less than or equal to.

math simulation monte-carlo

share|improve this question

edited May 9 at 10:26

MachineLearner

610112

asked May 9 at 5:17

I_Don't_CodeI_Don't_Code

262

$endgroup$

def monte_carlo_pi(nsamples):
 acc = 0
 for i in range(nsamples):
 x = random.random()
 y = random.random()
 if (x**2 + y**2) < 1.0:
 acc += 1
 return 4.0 * acc / nsamples

share|improve this question

edited May 9 at 10:26

MachineLearner

610112

asked May 9 at 5:17

I_Don't_CodeI_Don't_Code

262

share|improve this question

edited May 9 at 10:26

MachineLearner

610112

asked May 9 at 5:17

I_Don't_CodeI_Don't_Code

262

edited May 9 at 10:26

MachineLearner

610112

edited May 9 at 10:26

MachineLearner

610112

asked May 9 at 5:17

I_Don't_CodeI_Don't_Code

262

asked May 9 at 5:17

I_Don't_CodeI_Don't_Code

262

1 Answer
1

active

oldest

votes

6

$begingroup$

Short answer: Both formulations lead to the same answer.

---

Mathematical explanation:

In order to understand that let us look at two similar problems. Imagine we want to integrate a function $f(x)=x^2$ on two intervals $I_1=[0,1]$ (including both bounds) and $I_2=(0,1)$ (excluding both bounds).

For $I_1$ we have

$$int_0^1 x^2~dx=1/3.$$

For the second interval, we need to introduce a positive dummy parameter $varepsilon$ then we can calculate the integral as

$$lim_varepsilonto 0int_0+varepsilon^1-varepsilonx^2~dx=1/3.$$

So the line of separation between both intervals does not contribute to the integral (area) because it has an infinitesimal width. The same argument can be applied to the circle area. But including the line of the circle will make the calculations easier because we do not need to introduce a dummy variable.

---

Numerical explanation:

Because of the numerical precision of your computer, it will be very unlikely that the generated random numbers will lead to points that are really on the line of the circle. Numerically it will be impossible to obtain any value in which at least one of the coordinates is irrational. Only if both values are rational and if $x^2+y^2=1$ they can lie on the circle. But this case is also very unlikely.

share|improve this answer

edited May 10 at 5:09

answered May 9 at 6:11

MachineLearnerMachineLearner

610112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say that it is easier without the dummy variable $epsilon$ that approaches 0, does this only apply to the math? There is no dummy variable in the Monte Carlo simulation, after all.
  $endgroup$
  – I_Don't_Code
  May 9 at 6:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @I_Don't_Code Yes, this is only related to the mathematical derivation.
  $endgroup$
  – MachineLearner
  May 9 at 6:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So if I'm reading this correctly, it mathematically doesn't make a difference using $le1$ or $< 1$, so it wouldn't affect the accuracy of the simulation, even if my computer had an infinite decimal precision.
  $endgroup$
  – I_Don't_Code
  May 9 at 6:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @I_Don't_Code: If the number of samples goes to infinity it will not make any difference. Remember that you are trying to approximate the area of the circle with points. It will always make a difference for a finite number of points. But with an increasing number of data points the difference will become smaller.
  $endgroup$
  – MachineLearner
  May 9 at 6:40
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @I_Don't_Code: You could also count the number of points that for which $0.99999999999 leq sqrtx^2+y^2leq 1.00000000001$ (You can use the numerical precision of your programming language for the bounds). To see the number of values which lie very close to the line of the circle. Then you can directly see the bias.
  $endgroup$
  – MachineLearner
  May 9 at 6:47
  

  
  
  
  

add a comment | 

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6

$begingroup$

Short answer: Both formulations lead to the same answer.

---

Mathematical explanation:

In order to understand that let us look at two similar problems. Imagine we want to integrate a function $f(x)=x^2$ on two intervals $I_1=[0,1]$ (including both bounds) and $I_2=(0,1)$ (excluding both bounds).

For $I_1$ we have

$$int_0^1 x^2~dx=1/3.$$

For the second interval, we need to introduce a positive dummy parameter $varepsilon$ then we can calculate the integral as

$$lim_varepsilonto 0int_0+varepsilon^1-varepsilonx^2~dx=1/3.$$

So the line of separation between both intervals does not contribute to the integral (area) because it has an infinitesimal width. The same argument can be applied to the circle area. But including the line of the circle will make the calculations easier because we do not need to introduce a dummy variable.

---

Numerical explanation:

Because of the numerical precision of your computer, it will be very unlikely that the generated random numbers will lead to points that are really on the line of the circle. Numerically it will be impossible to obtain any value in which at least one of the coordinates is irrational. Only if both values are rational and if $x^2+y^2=1$ they can lie on the circle. But this case is also very unlikely.

share|improve this answer

edited May 10 at 5:09

answered May 9 at 6:11

MachineLearnerMachineLearner

610112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say that it is easier without the dummy variable $epsilon$ that approaches 0, does this only apply to the math? There is no dummy variable in the Monte Carlo simulation, after all.
  $endgroup$
  – I_Don't_Code
  May 9 at 6:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @I_Don't_Code Yes, this is only related to the mathematical derivation.
  $endgroup$
  – MachineLearner
  May 9 at 6:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So if I'm reading this correctly, it mathematically doesn't make a difference using $le1$ or $< 1$, so it wouldn't affect the accuracy of the simulation, even if my computer had an infinite decimal precision.
  $endgroup$
  – I_Don't_Code
  May 9 at 6:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @I_Don't_Code: If the number of samples goes to infinity it will not make any difference. Remember that you are trying to approximate the area of the circle with points. It will always make a difference for a finite number of points. But with an increasing number of data points the difference will become smaller.
  $endgroup$
  – MachineLearner
  May 9 at 6:40
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @I_Don't_Code: You could also count the number of points that for which $0.99999999999 leq sqrtx^2+y^2leq 1.00000000001$ (You can use the numerical precision of your programming language for the bounds). To see the number of values which lie very close to the line of the circle. Then you can directly see the bias.
  $endgroup$
  – MachineLearner
  May 9 at 6:47
  

  
  
  
  

add a comment | 

6

$begingroup$

Short answer: Both formulations lead to the same answer.

---

Mathematical explanation:

In order to understand that let us look at two similar problems. Imagine we want to integrate a function $f(x)=x^2$ on two intervals $I_1=[0,1]$ (including both bounds) and $I_2=(0,1)$ (excluding both bounds).

For $I_1$ we have

$$int_0^1 x^2~dx=1/3.$$

For the second interval, we need to introduce a positive dummy parameter $varepsilon$ then we can calculate the integral as

$$lim_varepsilonto 0int_0+varepsilon^1-varepsilonx^2~dx=1/3.$$

So the line of separation between both intervals does not contribute to the integral (area) because it has an infinitesimal width. The same argument can be applied to the circle area. But including the line of the circle will make the calculations easier because we do not need to introduce a dummy variable.

---

Numerical explanation:

Because of the numerical precision of your computer, it will be very unlikely that the generated random numbers will lead to points that are really on the line of the circle. Numerically it will be impossible to obtain any value in which at least one of the coordinates is irrational. Only if both values are rational and if $x^2+y^2=1$ they can lie on the circle. But this case is also very unlikely.

share|improve this answer

edited May 10 at 5:09

answered May 9 at 6:11

MachineLearnerMachineLearner

610112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say that it is easier without the dummy variable $epsilon$ that approaches 0, does this only apply to the math? There is no dummy variable in the Monte Carlo simulation, after all.
  $endgroup$
  – I_Don't_Code
  May 9 at 6:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @I_Don't_Code Yes, this is only related to the mathematical derivation.
  $endgroup$
  – MachineLearner
  May 9 at 6:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So if I'm reading this correctly, it mathematically doesn't make a difference using $le1$ or $< 1$, so it wouldn't affect the accuracy of the simulation, even if my computer had an infinite decimal precision.
  $endgroup$
  – I_Don't_Code
  May 9 at 6:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @I_Don't_Code: If the number of samples goes to infinity it will not make any difference. Remember that you are trying to approximate the area of the circle with points. It will always make a difference for a finite number of points. But with an increasing number of data points the difference will become smaller.
  $endgroup$
  – MachineLearner
  May 9 at 6:40
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @I_Don't_Code: You could also count the number of points that for which $0.99999999999 leq sqrtx^2+y^2leq 1.00000000001$ (You can use the numerical precision of your programming language for the bounds). To see the number of values which lie very close to the line of the circle. Then you can directly see the bias.
  $endgroup$
  – MachineLearner
  May 9 at 6:47
  

  
  
  
  

add a comment | 

$begingroup$

Short answer: Both formulations lead to the same answer.

---

Mathematical explanation:

In order to understand that let us look at two similar problems. Imagine we want to integrate a function $f(x)=x^2$ on two intervals $I_1=[0,1]$ (including both bounds) and $I_2=(0,1)$ (excluding both bounds).

For $I_1$ we have

$$int_0^1 x^2~dx=1/3.$$

For the second interval, we need to introduce a positive dummy parameter $varepsilon$ then we can calculate the integral as

$$lim_varepsilonto 0int_0+varepsilon^1-varepsilonx^2~dx=1/3.$$

So the line of separation between both intervals does not contribute to the integral (area) because it has an infinitesimal width. The same argument can be applied to the circle area. But including the line of the circle will make the calculations easier because we do not need to introduce a dummy variable.

---

Numerical explanation:

Because of the numerical precision of your computer, it will be very unlikely that the generated random numbers will lead to points that are really on the line of the circle. Numerically it will be impossible to obtain any value in which at least one of the coordinates is irrational. Only if both values are rational and if $x^2+y^2=1$ they can lie on the circle. But this case is also very unlikely.

share|improve this answer

edited May 10 at 5:09

answered May 9 at 6:11

MachineLearnerMachineLearner

610112

$endgroup$

share|improve this answer

edited May 10 at 5:09

answered May 9 at 6:11

MachineLearnerMachineLearner

610112

answered May 9 at 6:11

MachineLearnerMachineLearner

610112

answered May 9 at 6:11

MachineLearnerMachineLearner

610112