How to plot the normal distribution?
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According to 'An Introduction to Probability Theory and Its Applications', Vol. 1 by Feller the number of inversions in a random permutations at large numbers satisfy CLT with dedicated mean and variance.
However, I am practically intrested in how to plot the figure of the normal distribution (what to calculate for it)?
I understand that the figure may depends on the fact of how large the numbers are. Any explanations to clarify the topic are highly welcomed. Thank you in advance.
probability-theory probability-distributions
asked Dec 20 '18 at 19:43
Mikhail GaichenkovMikhail Gaichenkov
6610
$endgroup$
add a comment |
$begingroup$
According to 'An Introduction to Probability Theory and Its Applications', Vol. 1 by Feller the number of inversions in a random permutations at large numbers satisfy CLT with dedicated mean and variance.
However, I am practically intrested in how to plot the figure of the normal distribution (what to calculate for it)?
I understand that the figure may depends on the fact of how large the numbers are. Any explanations to clarify the topic are highly welcomed. Thank you in advance.
probability-theory probability-distributions
asked Dec 20 '18 at 19:43
Mikhail GaichenkovMikhail Gaichenkov
6610
$endgroup$
$begingroup$
Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
$endgroup$
– J.G.
Dec 20 '18 at 19:52
$begingroup$
The PDF one, which depends on the number of elements
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:06
add a comment |
$begingroup$
According to 'An Introduction to Probability Theory and Its Applications', Vol. 1 by Feller the number of inversions in a random permutations at large numbers satisfy CLT with dedicated mean and variance.
However, I am practically intrested in how to plot the figure of the normal distribution (what to calculate for it)?
I understand that the figure may depends on the fact of how large the numbers are. Any explanations to clarify the topic are highly welcomed. Thank you in advance.
probability-theory probability-distributions
asked Dec 20 '18 at 19:43
Mikhail GaichenkovMikhail Gaichenkov
6610
$endgroup$
According to 'An Introduction to Probability Theory and Its Applications', Vol. 1 by Feller the number of inversions in a random permutations at large numbers satisfy CLT with dedicated mean and variance.
However, I am practically intrested in how to plot the figure of the normal distribution (what to calculate for it)?
I understand that the figure may depends on the fact of how large the numbers are. Any explanations to clarify the topic are highly welcomed. Thank you in advance.
probability-theory probability-distributions
probability-theory probability-distributions
asked Dec 20 '18 at 19:43
Mikhail GaichenkovMikhail Gaichenkov
6610
asked Dec 20 '18 at 19:43
Mikhail GaichenkovMikhail Gaichenkov
6610
asked Dec 20 '18 at 19:43
Mikhail GaichenkovMikhail Gaichenkov
6610
asked Dec 20 '18 at 19:43
Mikhail GaichenkovMikhail Gaichenkov
6610
asked Dec 20 '18 at 19:43
Mikhail GaichenkovMikhail Gaichenkov
6610
6610
$begingroup$
Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
$endgroup$
– J.G.
Dec 20 '18 at 19:52
$begingroup$
The PDF one, which depends on the number of elements
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:06
add a comment |
$begingroup$
Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
$endgroup$
– J.G.
Dec 20 '18 at 19:52
$begingroup$
The PDF one, which depends on the number of elements
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:06
$begingroup$
Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
$endgroup$
– J.G.
Dec 20 '18 at 19:52
$begingroup$
Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
$endgroup$
– J.G.
Dec 20 '18 at 19:52
$begingroup$
The PDF one, which depends on the number of elements
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:06
$begingroup$
The PDF one, which depends on the number of elements
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is how to plot the density function of $N(0,1)$:
$$
f(x) = frac{e^{-frac{x^2}{2}}}{sqrt{2 pi}} .
$$
In Mathematica, a one-liner:
Plot[PDF[NormalDistribution, x], {x, -4, 4}]
In Python, slightly more verbose:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-4, 4, 101)
y = np.exp(-x*x/2) / np.sqrt(2*np.pi)
plt.plot(x, y)
plt.show()
answered Dec 20 '18 at 19:49
FedericoFederico
5,124514
$endgroup$
$begingroup$
Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 19:56
$begingroup$
Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
$endgroup$
– Federico
Dec 20 '18 at 19:58
$begingroup$
Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:04
$begingroup$
They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
$endgroup$
– Federico
Dec 20 '18 at 20:08
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is how to plot the density function of $N(0,1)$:
$$
f(x) = frac{e^{-frac{x^2}{2}}}{sqrt{2 pi}} .
$$
In Mathematica, a one-liner:
Plot[PDF[NormalDistribution, x], {x, -4, 4}]
In Python, slightly more verbose:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-4, 4, 101)
y = np.exp(-x*x/2) / np.sqrt(2*np.pi)
plt.plot(x, y)
plt.show()
answered Dec 20 '18 at 19:49
FedericoFederico
5,124514
$endgroup$
$begingroup$
Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 19:56
$begingroup$
Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
$endgroup$
– Federico
Dec 20 '18 at 19:58
$begingroup$
Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:04
$begingroup$
They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
$endgroup$
– Federico
Dec 20 '18 at 20:08
add a comment |
$begingroup$
Here is how to plot the density function of $N(0,1)$:
$$
f(x) = frac{e^{-frac{x^2}{2}}}{sqrt{2 pi}} .
$$
In Mathematica, a one-liner:
Plot[PDF[NormalDistribution, x], {x, -4, 4}]
In Python, slightly more verbose:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-4, 4, 101)
y = np.exp(-x*x/2) / np.sqrt(2*np.pi)
plt.plot(x, y)
plt.show()
answered Dec 20 '18 at 19:49
FedericoFederico
5,124514
$endgroup$
$begingroup$
Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 19:56
$begingroup$
Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
$endgroup$
– Federico
Dec 20 '18 at 19:58
$begingroup$
Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:04
$begingroup$
They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
$endgroup$
– Federico
Dec 20 '18 at 20:08
add a comment |
$begingroup$
Here is how to plot the density function of $N(0,1)$:
$$
f(x) = frac{e^{-frac{x^2}{2}}}{sqrt{2 pi}} .
$$
In Mathematica, a one-liner:
Plot[PDF[NormalDistribution, x], {x, -4, 4}]
In Python, slightly more verbose:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-4, 4, 101)
y = np.exp(-x*x/2) / np.sqrt(2*np.pi)
plt.plot(x, y)
plt.show()
answered Dec 20 '18 at 19:49
FedericoFederico
5,124514
$endgroup$
Here is how to plot the density function of $N(0,1)$:
$$
f(x) = frac{e^{-frac{x^2}{2}}}{sqrt{2 pi}} .
$$
In Mathematica, a one-liner:
Plot[PDF[NormalDistribution, x], {x, -4, 4}]
In Python, slightly more verbose:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-4, 4, 101)
y = np.exp(-x*x/2) / np.sqrt(2*np.pi)
plt.plot(x, y)
plt.show()
answered Dec 20 '18 at 19:49
FedericoFederico
5,124514
answered Dec 20 '18 at 19:49
FedericoFederico
5,124514
answered Dec 20 '18 at 19:49
FedericoFederico
5,124514
answered Dec 20 '18 at 19:49
FedericoFederico
5,124514
5,124514
$begingroup$
Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 19:56
$begingroup$
Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
$endgroup$
– Federico
Dec 20 '18 at 19:58
$begingroup$
Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:04
$begingroup$
They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
$endgroup$
– Federico
Dec 20 '18 at 20:08
add a comment |
$begingroup$
Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 19:56
$begingroup$
Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
$endgroup$
– Federico
Dec 20 '18 at 19:58
$begingroup$
Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:04
$begingroup$
They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
$endgroup$
– Federico
Dec 20 '18 at 20:08
$begingroup$
Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 19:56
$begingroup$
Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 19:56
$begingroup$
Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
$endgroup$
– Federico
Dec 20 '18 at 19:58
$begingroup$
Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
$endgroup$
– Federico
Dec 20 '18 at 19:58
$begingroup$
Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:04
$begingroup$
Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:04
$begingroup$
They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
$endgroup$
– Federico
Dec 20 '18 at 20:08
$begingroup$
They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
$endgroup$
– Federico
Dec 20 '18 at 20:08
add a comment |
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$begingroup$
Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
$endgroup$
– J.G.
Dec 20 '18 at 19:52
$begingroup$
The PDF one, which depends on the number of elements
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:06