Mixing
Numerical evidence of law of iterated logarithm (random walk)
$begingroup$
The law of iterated logarithm states that for a random walk $$S_n = X_1 + X_2 + ... X_n$$ with $X_i$ independent random variables such that $P(X_i = 1) = P(X_i = -1) = 1/2$, we have
$$limsup_{n rightarrow infty} S_n / sqrt{2 n log log n} = 1, qquad rm{a.s.}$$
Here is Python code to test it:
import numpy as np
import matplotlib.pyplot as plt
N = 10*1000*1000
B = 2 * np.random.binomial(1, 0.5, N) - 1 # N independent +1/-1 each of them with probability 1/2
B = np.cumsum(B) # random walk
plt.plot(B)
plt.show()
C = B / np.sqrt(2 * np.arange(N) * np.log(np.log(np.arange(N))))
M = np.maximum.accumulate(C[::-1])[::-1] # limsup, see http://stackoverflow.com/questions/35149843/running-max-limsup-in-numpy-what-optimization
plt.plot(M)
plt.show()
Question:
I have done it lots of times, but the ratio is nearly always decreasing to 0, instead of having a limit 1.
Where is the problem?
Here's the kind of plot I have most often for the ratio (which should approach $1$):
probability-theory random-variables random-walk simulation
asked Feb 2 '16 at 10:37
BasjBasj
4121529
$endgroup$
add a comment |
$begingroup$
The law of iterated logarithm states that for a random walk $$S_n = X_1 + X_2 + ... X_n$$ with $X_i$ independent random variables such that $P(X_i = 1) = P(X_i = -1) = 1/2$, we have
$$limsup_{n rightarrow infty} S_n / sqrt{2 n log log n} = 1, qquad rm{a.s.}$$
Here is Python code to test it:
import numpy as np
import matplotlib.pyplot as plt
N = 10*1000*1000
B = 2 * np.random.binomial(1, 0.5, N) - 1 # N independent +1/-1 each of them with probability 1/2
B = np.cumsum(B) # random walk
plt.plot(B)
plt.show()
C = B / np.sqrt(2 * np.arange(N) * np.log(np.log(np.arange(N))))
M = np.maximum.accumulate(C[::-1])[::-1] # limsup, see http://stackoverflow.com/questions/35149843/running-max-limsup-in-numpy-what-optimization
plt.plot(M)
plt.show()
Question:
I have done it lots of times, but the ratio is nearly always decreasing to 0, instead of having a limit 1.
Where is the problem?
Here's the kind of plot I have most often for the ratio (which should approach $1$):
probability-theory random-variables random-walk simulation
asked Feb 2 '16 at 10:37
BasjBasj
4121529
$endgroup$
add a comment |
$begingroup$
The law of iterated logarithm states that for a random walk $$S_n = X_1 + X_2 + ... X_n$$ with $X_i$ independent random variables such that $P(X_i = 1) = P(X_i = -1) = 1/2$, we have
$$limsup_{n rightarrow infty} S_n / sqrt{2 n log log n} = 1, qquad rm{a.s.}$$
Here is Python code to test it:
import numpy as np
import matplotlib.pyplot as plt
N = 10*1000*1000
B = 2 * np.random.binomial(1, 0.5, N) - 1 # N independent +1/-1 each of them with probability 1/2
B = np.cumsum(B) # random walk
plt.plot(B)
plt.show()
C = B / np.sqrt(2 * np.arange(N) * np.log(np.log(np.arange(N))))
M = np.maximum.accumulate(C[::-1])[::-1] # limsup, see http://stackoverflow.com/questions/35149843/running-max-limsup-in-numpy-what-optimization
plt.plot(M)
plt.show()
Question:
I have done it lots of times, but the ratio is nearly always decreasing to 0, instead of having a limit 1.
Where is the problem?
Here's the kind of plot I have most often for the ratio (which should approach $1$):
probability-theory random-variables random-walk simulation
asked Feb 2 '16 at 10:37
BasjBasj
4121529
$endgroup$
The law of iterated logarithm states that for a random walk $$S_n = X_1 + X_2 + ... X_n$$ with $X_i$ independent random variables such that $P(X_i = 1) = P(X_i = -1) = 1/2$, we have
$$limsup_{n rightarrow infty} S_n / sqrt{2 n log log n} = 1, qquad rm{a.s.}$$
Here is Python code to test it:
import numpy as np
import matplotlib.pyplot as plt
N = 10*1000*1000
B = 2 * np.random.binomial(1, 0.5, N) - 1 # N independent +1/-1 each of them with probability 1/2
B = np.cumsum(B) # random walk
plt.plot(B)
plt.show()
C = B / np.sqrt(2 * np.arange(N) * np.log(np.log(np.arange(N))))
M = np.maximum.accumulate(C[::-1])[::-1] # limsup, see http://stackoverflow.com/questions/35149843/running-max-limsup-in-numpy-what-optimization
plt.plot(M)
plt.show()
Question:
I have done it lots of times, but the ratio is nearly always decreasing to 0, instead of having a limit 1.
Where is the problem?
Here's the kind of plot I have most often for the ratio (which should approach $1$):
probability-theory random-variables random-walk simulation
probability-theory random-variables random-walk simulation
asked Feb 2 '16 at 10:37
BasjBasj
4121529
asked Feb 2 '16 at 10:37
BasjBasj
4121529
edited Jan 29 at 11:19
Basj
asked Feb 2 '16 at 10:37
BasjBasj
4121529
asked Feb 2 '16 at 10:37
BasjBasj
4121529
asked Feb 2 '16 at 10:37
BasjBasj
4121529
4121529
add a comment |
add a comment |
1 Answer
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oldest
votes
$begingroup$
I think the problem is that the number of attempts that can be used in a numerical simulation $n$ is finite.
Notice this: if $Y_n=frac{S_n}{sqrt{2nloglog n}}$, by properties of random walk we know $mathbb{E}[Y_n]=frac{mathbb{E}[S_n]}{sqrt{2nloglog n}}=0$ and
$$
Var[Y_n]=frac{Var[S_n]}{2nloglog n}=frac{n}{2nloglog n}=frac{1}{2loglog n}to 0
$$
which implies $Y_n$ converges to 0 in distribution (we can prove it using Chebyshev's inequality).
In particular, if we define $Y_{k,n}=max_{kleq ell leq n}Y_ell$ (which is the variable you are using in your code, instead of the variable $Z_k=sup_{ell geq k}Y_{ell}$ which is the variable one should use), then "$Y_{k,n}searrow_{kto n} Y_n$", which in turn converges to 0. So, in the large majority of cases, in your simulations $Y_{k,n}$ should converge to 0.
answered Feb 2 '16 at 14:36
Nate RiverNate River
2,2201712
$endgroup$
1
$begingroup$
@Basj "in the large majority of cases, in your simulations Yk,n should converge to 0." Indeed, for samples of length $10^n$, one obtains a limit at least $alpha$ with probability roughly $10^{-nalpha^2}$. For example, to observe a limit at least $.9$ in samples of length $10^7$ requires to repeat the simulation a number of times of the order of $5cdot10^5$.
$endgroup$
– Did
Feb 2 '16 at 15:45
$begingroup$
Many thanks @NateRiver. You're right: without noticing it, I was indeed working with $max_{k leq ell leq n} Y_ell rightarrow_{k rightarrow n} Y_n$ and not with $sup_{k leq ell} Y_ell$. That clearly explains why it converges to $0$. Thanks for that. Now do you have an idea how I could do a numerical simulation showing that $(2 n log log n)^{1/2}$ is the right magnitude order, and that $sqrt{n} (log log n)^{2/3}$ is not the right order of magnitude? Because of the double log (very small values), it's difficult to see this in a simulation.
$endgroup$
– Basj
Feb 2 '16 at 16:50
$begingroup$
@Did Could I get something by considering $Y_{n/2, n} = max_{n/2 leq ell leq n} Y_ell$ ?
$endgroup$
– Basj
Feb 2 '16 at 16:53
$begingroup$
Did and @NateRiver, maybe you would have an idea for this really linked question math.stackexchange.com/questions/1637989/… ?
$endgroup$
– Basj
Feb 2 '16 at 22:02
$begingroup$
@Basj one quick and dirty idea to avoid the problem of your simulations going almost certainly to 0, is to consider the same variables $Y_{k,n}$, but only plot them for $kin {1,...,n/2}$. That said, I ran some simulations with this trick and while they didn't converge to 0 as $kto n/2$, they didn't converge to 1, either. I'm kind of afraid that the only way to get results somewhat closer to 1 is to use way larger values of $n$ than the ones being used, which sadly results in memory problems : (.
$endgroup$
– Nate River
Feb 4 '16 at 0:56
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
I think the problem is that the number of attempts that can be used in a numerical simulation $n$ is finite.
Notice this: if $Y_n=frac{S_n}{sqrt{2nloglog n}}$, by properties of random walk we know $mathbb{E}[Y_n]=frac{mathbb{E}[S_n]}{sqrt{2nloglog n}}=0$ and
$$
Var[Y_n]=frac{Var[S_n]}{2nloglog n}=frac{n}{2nloglog n}=frac{1}{2loglog n}to 0
$$
which implies $Y_n$ converges to 0 in distribution (we can prove it using Chebyshev's inequality).
In particular, if we define $Y_{k,n}=max_{kleq ell leq n}Y_ell$ (which is the variable you are using in your code, instead of the variable $Z_k=sup_{ell geq k}Y_{ell}$ which is the variable one should use), then "$Y_{k,n}searrow_{kto n} Y_n$", which in turn converges to 0. So, in the large majority of cases, in your simulations $Y_{k,n}$ should converge to 0.
answered Feb 2 '16 at 14:36
Nate RiverNate River
2,2201712
$endgroup$
1
$begingroup$
@Basj "in the large majority of cases, in your simulations Yk,n should converge to 0." Indeed, for samples of length $10^n$, one obtains a limit at least $alpha$ with probability roughly $10^{-nalpha^2}$. For example, to observe a limit at least $.9$ in samples of length $10^7$ requires to repeat the simulation a number of times of the order of $5cdot10^5$.
$endgroup$
– Did
Feb 2 '16 at 15:45
$begingroup$
Many thanks @NateRiver. You're right: without noticing it, I was indeed working with $max_{k leq ell leq n} Y_ell rightarrow_{k rightarrow n} Y_n$ and not with $sup_{k leq ell} Y_ell$. That clearly explains why it converges to $0$. Thanks for that. Now do you have an idea how I could do a numerical simulation showing that $(2 n log log n)^{1/2}$ is the right magnitude order, and that $sqrt{n} (log log n)^{2/3}$ is not the right order of magnitude? Because of the double log (very small values), it's difficult to see this in a simulation.
$endgroup$
– Basj
Feb 2 '16 at 16:50
$begingroup$
@Did Could I get something by considering $Y_{n/2, n} = max_{n/2 leq ell leq n} Y_ell$ ?
$endgroup$
– Basj
Feb 2 '16 at 16:53
$begingroup$
Did and @NateRiver, maybe you would have an idea for this really linked question math.stackexchange.com/questions/1637989/… ?
$endgroup$
– Basj
Feb 2 '16 at 22:02
$begingroup$
@Basj one quick and dirty idea to avoid the problem of your simulations going almost certainly to 0, is to consider the same variables $Y_{k,n}$, but only plot them for $kin {1,...,n/2}$. That said, I ran some simulations with this trick and while they didn't converge to 0 as $kto n/2$, they didn't converge to 1, either. I'm kind of afraid that the only way to get results somewhat closer to 1 is to use way larger values of $n$ than the ones being used, which sadly results in memory problems : (.
$endgroup$
– Nate River
Feb 4 '16 at 0:56
|
show 1 more comment
$begingroup$
I think the problem is that the number of attempts that can be used in a numerical simulation $n$ is finite.
Notice this: if $Y_n=frac{S_n}{sqrt{2nloglog n}}$, by properties of random walk we know $mathbb{E}[Y_n]=frac{mathbb{E}[S_n]}{sqrt{2nloglog n}}=0$ and
$$
Var[Y_n]=frac{Var[S_n]}{2nloglog n}=frac{n}{2nloglog n}=frac{1}{2loglog n}to 0
$$
which implies $Y_n$ converges to 0 in distribution (we can prove it using Chebyshev's inequality).
In particular, if we define $Y_{k,n}=max_{kleq ell leq n}Y_ell$ (which is the variable you are using in your code, instead of the variable $Z_k=sup_{ell geq k}Y_{ell}$ which is the variable one should use), then "$Y_{k,n}searrow_{kto n} Y_n$", which in turn converges to 0. So, in the large majority of cases, in your simulations $Y_{k,n}$ should converge to 0.
answered Feb 2 '16 at 14:36
Nate RiverNate River
2,2201712
$endgroup$
1
$begingroup$
@Basj "in the large majority of cases, in your simulations Yk,n should converge to 0." Indeed, for samples of length $10^n$, one obtains a limit at least $alpha$ with probability roughly $10^{-nalpha^2}$. For example, to observe a limit at least $.9$ in samples of length $10^7$ requires to repeat the simulation a number of times of the order of $5cdot10^5$.
$endgroup$
– Did
Feb 2 '16 at 15:45
$begingroup$
Many thanks @NateRiver. You're right: without noticing it, I was indeed working with $max_{k leq ell leq n} Y_ell rightarrow_{k rightarrow n} Y_n$ and not with $sup_{k leq ell} Y_ell$. That clearly explains why it converges to $0$. Thanks for that. Now do you have an idea how I could do a numerical simulation showing that $(2 n log log n)^{1/2}$ is the right magnitude order, and that $sqrt{n} (log log n)^{2/3}$ is not the right order of magnitude? Because of the double log (very small values), it's difficult to see this in a simulation.
$endgroup$
– Basj
Feb 2 '16 at 16:50
$begingroup$
@Did Could I get something by considering $Y_{n/2, n} = max_{n/2 leq ell leq n} Y_ell$ ?
$endgroup$
– Basj
Feb 2 '16 at 16:53
$begingroup$
Did and @NateRiver, maybe you would have an idea for this really linked question math.stackexchange.com/questions/1637989/… ?
$endgroup$
– Basj
Feb 2 '16 at 22:02
$begingroup$
@Basj one quick and dirty idea to avoid the problem of your simulations going almost certainly to 0, is to consider the same variables $Y_{k,n}$, but only plot them for $kin {1,...,n/2}$. That said, I ran some simulations with this trick and while they didn't converge to 0 as $kto n/2$, they didn't converge to 1, either. I'm kind of afraid that the only way to get results somewhat closer to 1 is to use way larger values of $n$ than the ones being used, which sadly results in memory problems : (.
$endgroup$
– Nate River
Feb 4 '16 at 0:56
|
show 1 more comment
$begingroup$
I think the problem is that the number of attempts that can be used in a numerical simulation $n$ is finite.
Notice this: if $Y_n=frac{S_n}{sqrt{2nloglog n}}$, by properties of random walk we know $mathbb{E}[Y_n]=frac{mathbb{E}[S_n]}{sqrt{2nloglog n}}=0$ and
$$
Var[Y_n]=frac{Var[S_n]}{2nloglog n}=frac{n}{2nloglog n}=frac{1}{2loglog n}to 0
$$
which implies $Y_n$ converges to 0 in distribution (we can prove it using Chebyshev's inequality).
In particular, if we define $Y_{k,n}=max_{kleq ell leq n}Y_ell$ (which is the variable you are using in your code, instead of the variable $Z_k=sup_{ell geq k}Y_{ell}$ which is the variable one should use), then "$Y_{k,n}searrow_{kto n} Y_n$", which in turn converges to 0. So, in the large majority of cases, in your simulations $Y_{k,n}$ should converge to 0.
answered Feb 2 '16 at 14:36
Nate RiverNate River
2,2201712
$endgroup$
I think the problem is that the number of attempts that can be used in a numerical simulation $n$ is finite.
Notice this: if $Y_n=frac{S_n}{sqrt{2nloglog n}}$, by properties of random walk we know $mathbb{E}[Y_n]=frac{mathbb{E}[S_n]}{sqrt{2nloglog n}}=0$ and
$$
Var[Y_n]=frac{Var[S_n]}{2nloglog n}=frac{n}{2nloglog n}=frac{1}{2loglog n}to 0
$$
which implies $Y_n$ converges to 0 in distribution (we can prove it using Chebyshev's inequality).
In particular, if we define $Y_{k,n}=max_{kleq ell leq n}Y_ell$ (which is the variable you are using in your code, instead of the variable $Z_k=sup_{ell geq k}Y_{ell}$ which is the variable one should use), then "$Y_{k,n}searrow_{kto n} Y_n$", which in turn converges to 0. So, in the large majority of cases, in your simulations $Y_{k,n}$ should converge to 0.
answered Feb 2 '16 at 14:36
Nate RiverNate River
2,2201712
answered Feb 2 '16 at 14:36
Nate RiverNate River
2,2201712
answered Feb 2 '16 at 14:36
Nate RiverNate River
2,2201712
answered Feb 2 '16 at 14:36
Nate RiverNate River
2,2201712
2,2201712
1
$begingroup$
@Basj "in the large majority of cases, in your simulations Yk,n should converge to 0." Indeed, for samples of length $10^n$, one obtains a limit at least $alpha$ with probability roughly $10^{-nalpha^2}$. For example, to observe a limit at least $.9$ in samples of length $10^7$ requires to repeat the simulation a number of times of the order of $5cdot10^5$.
$endgroup$
– Did
Feb 2 '16 at 15:45
$begingroup$
Many thanks @NateRiver. You're right: without noticing it, I was indeed working with $max_{k leq ell leq n} Y_ell rightarrow_{k rightarrow n} Y_n$ and not with $sup_{k leq ell} Y_ell$. That clearly explains why it converges to $0$. Thanks for that. Now do you have an idea how I could do a numerical simulation showing that $(2 n log log n)^{1/2}$ is the right magnitude order, and that $sqrt{n} (log log n)^{2/3}$ is not the right order of magnitude? Because of the double log (very small values), it's difficult to see this in a simulation.
$endgroup$
– Basj
Feb 2 '16 at 16:50
$begingroup$
@Did Could I get something by considering $Y_{n/2, n} = max_{n/2 leq ell leq n} Y_ell$ ?
$endgroup$
– Basj
Feb 2 '16 at 16:53
$begingroup$
Did and @NateRiver, maybe you would have an idea for this really linked question math.stackexchange.com/questions/1637989/… ?
$endgroup$
– Basj
Feb 2 '16 at 22:02
$begingroup$
@Basj one quick and dirty idea to avoid the problem of your simulations going almost certainly to 0, is to consider the same variables $Y_{k,n}$, but only plot them for $kin {1,...,n/2}$. That said, I ran some simulations with this trick and while they didn't converge to 0 as $kto n/2$, they didn't converge to 1, either. I'm kind of afraid that the only way to get results somewhat closer to 1 is to use way larger values of $n$ than the ones being used, which sadly results in memory problems : (.
$endgroup$
– Nate River
Feb 4 '16 at 0:56
|
show 1 more comment
1
$begingroup$
@Basj "in the large majority of cases, in your simulations Yk,n should converge to 0." Indeed, for samples of length $10^n$, one obtains a limit at least $alpha$ with probability roughly $10^{-nalpha^2}$. For example, to observe a limit at least $.9$ in samples of length $10^7$ requires to repeat the simulation a number of times of the order of $5cdot10^5$.
$endgroup$
– Did
Feb 2 '16 at 15:45
$begingroup$
Many thanks @NateRiver. You're right: without noticing it, I was indeed working with $max_{k leq ell leq n} Y_ell rightarrow_{k rightarrow n} Y_n$ and not with $sup_{k leq ell} Y_ell$. That clearly explains why it converges to $0$. Thanks for that. Now do you have an idea how I could do a numerical simulation showing that $(2 n log log n)^{1/2}$ is the right magnitude order, and that $sqrt{n} (log log n)^{2/3}$ is not the right order of magnitude? Because of the double log (very small values), it's difficult to see this in a simulation.
$endgroup$
– Basj
Feb 2 '16 at 16:50
$begingroup$
@Did Could I get something by considering $Y_{n/2, n} = max_{n/2 leq ell leq n} Y_ell$ ?
$endgroup$
– Basj
Feb 2 '16 at 16:53
$begingroup$
Did and @NateRiver, maybe you would have an idea for this really linked question math.stackexchange.com/questions/1637989/… ?
$endgroup$
– Basj
Feb 2 '16 at 22:02
$begingroup$
@Basj one quick and dirty idea to avoid the problem of your simulations going almost certainly to 0, is to consider the same variables $Y_{k,n}$, but only plot them for $kin {1,...,n/2}$. That said, I ran some simulations with this trick and while they didn't converge to 0 as $kto n/2$, they didn't converge to 1, either. I'm kind of afraid that the only way to get results somewhat closer to 1 is to use way larger values of $n$ than the ones being used, which sadly results in memory problems : (.
$endgroup$
– Nate River
Feb 4 '16 at 0:56
1
1
$begingroup$
@Basj "in the large majority of cases, in your simulations Yk,n should converge to 0." Indeed, for samples of length $10^n$, one obtains a limit at least $alpha$ with probability roughly $10^{-nalpha^2}$. For example, to observe a limit at least $.9$ in samples of length $10^7$ requires to repeat the simulation a number of times of the order of $5cdot10^5$.
$endgroup$
– Did
Feb 2 '16 at 15:45
$begingroup$
@Basj "in the large majority of cases, in your simulations Yk,n should converge to 0." Indeed, for samples of length $10^n$, one obtains a limit at least $alpha$ with probability roughly $10^{-nalpha^2}$. For example, to observe a limit at least $.9$ in samples of length $10^7$ requires to repeat the simulation a number of times of the order of $5cdot10^5$.
$endgroup$
– Did
Feb 2 '16 at 15:45
$begingroup$
Many thanks @NateRiver. You're right: without noticing it, I was indeed working with $max_{k leq ell leq n} Y_ell rightarrow_{k rightarrow n} Y_n$ and not with $sup_{k leq ell} Y_ell$. That clearly explains why it converges to $0$. Thanks for that. Now do you have an idea how I could do a numerical simulation showing that $(2 n log log n)^{1/2}$ is the right magnitude order, and that $sqrt{n} (log log n)^{2/3}$ is not the right order of magnitude? Because of the double log (very small values), it's difficult to see this in a simulation.
$endgroup$
– Basj
Feb 2 '16 at 16:50
$begingroup$
Many thanks @NateRiver. You're right: without noticing it, I was indeed working with $max_{k leq ell leq n} Y_ell rightarrow_{k rightarrow n} Y_n$ and not with $sup_{k leq ell} Y_ell$. That clearly explains why it converges to $0$. Thanks for that. Now do you have an idea how I could do a numerical simulation showing that $(2 n log log n)^{1/2}$ is the right magnitude order, and that $sqrt{n} (log log n)^{2/3}$ is not the right order of magnitude? Because of the double log (very small values), it's difficult to see this in a simulation.
$endgroup$
– Basj
Feb 2 '16 at 16:50
$begingroup$
@Did Could I get something by considering $Y_{n/2, n} = max_{n/2 leq ell leq n} Y_ell$ ?
$endgroup$
– Basj
Feb 2 '16 at 16:53
$begingroup$
@Did Could I get something by considering $Y_{n/2, n} = max_{n/2 leq ell leq n} Y_ell$ ?
$endgroup$
– Basj
Feb 2 '16 at 16:53
$begingroup$
Did and @NateRiver, maybe you would have an idea for this really linked question math.stackexchange.com/questions/1637989/… ?
$endgroup$
– Basj
Feb 2 '16 at 22:02
$begingroup$
Did and @NateRiver, maybe you would have an idea for this really linked question math.stackexchange.com/questions/1637989/… ?
$endgroup$
– Basj
Feb 2 '16 at 22:02
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@Basj one quick and dirty idea to avoid the problem of your simulations going almost certainly to 0, is to consider the same variables $Y_{k,n}$, but only plot them for $kin {1,...,n/2}$. That said, I ran some simulations with this trick and while they didn't converge to 0 as $kto n/2$, they didn't converge to 1, either. I'm kind of afraid that the only way to get results somewhat closer to 1 is to use way larger values of $n$ than the ones being used, which sadly results in memory problems : (.
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– Nate River
Feb 4 '16 at 0:56
$begingroup$
@Basj one quick and dirty idea to avoid the problem of your simulations going almost certainly to 0, is to consider the same variables $Y_{k,n}$, but only plot them for $kin {1,...,n/2}$. That said, I ran some simulations with this trick and while they didn't converge to 0 as $kto n/2$, they didn't converge to 1, either. I'm kind of afraid that the only way to get results somewhat closer to 1 is to use way larger values of $n$ than the ones being used, which sadly results in memory problems : (.
$endgroup$
– Nate River
Feb 4 '16 at 0:56
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