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Transformation of random variables and joint distributions
Plotting confidence intervalsWhat is the PDF of a variable where a parameter is itself a random variable?NProbability not reliability analysis?Mathematica function to calculate equivalent NormalDistribution from a WeibullDistributionPDF for square of Rician random variable?Convolve discrete random variables efficientlyDistribution of Function of Random Sum of Random VariablesSketching Normal Probability Distributions GraphsConstruct Distribution Histogram From Random VariableNormal distribution plot construction
$begingroup$
Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation
$y_i$ ~ NormalDistribution[0,$σ_y$ ]
I want to obtain with Mathematica:
The distribution of:
$x = bary = frac sum_i=1^ny_in$The joint distribution of $ (x,y_i )$
Thank you for your helpful comments
probability-or-statistics
New contributor
$endgroup$
add a comment |
$begingroup$
Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation
$y_i$ ~ NormalDistribution[0,$σ_y$ ]
I want to obtain with Mathematica:
The distribution of:
$x = bary = frac sum_i=1^ny_in$The joint distribution of $ (x,y_i )$
Thank you for your helpful comments
probability-or-statistics
New contributor
$endgroup$
4
$begingroup$
What have you tried? For example, have you seen the documentation onTransformedDistribution
andProbabilityDistribution
?
$endgroup$
– JimB
6 hours ago
$begingroup$
@JimB . I tried thisTransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result isNormalDistribution[0, [Sigma]y]
. However, the correct result should beNormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
5 hours ago
$begingroup$
You need to "index" the variabley
or else Mathematica thinks it is a single variable.
$endgroup$
– JimB
1 hour ago
add a comment |
$begingroup$
Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation
$y_i$ ~ NormalDistribution[0,$σ_y$ ]
I want to obtain with Mathematica:
The distribution of:
$x = bary = frac sum_i=1^ny_in$The joint distribution of $ (x,y_i )$
Thank you for your helpful comments
probability-or-statistics
New contributor
$endgroup$
Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation
$y_i$ ~ NormalDistribution[0,$σ_y$ ]
I want to obtain with Mathematica:
The distribution of:
$x = bary = frac sum_i=1^ny_in$The joint distribution of $ (x,y_i )$
Thank you for your helpful comments
probability-or-statistics
probability-or-statistics
New contributor
New contributor
edited 1 hour ago
mjw
9679
9679
New contributor
asked 6 hours ago
Andrea2810Andrea2810
162
162
New contributor
New contributor
4
$begingroup$
What have you tried? For example, have you seen the documentation onTransformedDistribution
andProbabilityDistribution
?
$endgroup$
– JimB
6 hours ago
$begingroup$
@JimB . I tried thisTransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result isNormalDistribution[0, [Sigma]y]
. However, the correct result should beNormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
5 hours ago
$begingroup$
You need to "index" the variabley
or else Mathematica thinks it is a single variable.
$endgroup$
– JimB
1 hour ago
add a comment |
4
$begingroup$
What have you tried? For example, have you seen the documentation onTransformedDistribution
andProbabilityDistribution
?
$endgroup$
– JimB
6 hours ago
$begingroup$
@JimB . I tried thisTransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result isNormalDistribution[0, [Sigma]y]
. However, the correct result should beNormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
5 hours ago
$begingroup$
You need to "index" the variabley
or else Mathematica thinks it is a single variable.
$endgroup$
– JimB
1 hour ago
4
4
$begingroup$
What have you tried? For example, have you seen the documentation on
TransformedDistribution
and ProbabilityDistribution
?$endgroup$
– JimB
6 hours ago
$begingroup$
What have you tried? For example, have you seen the documentation on
TransformedDistribution
and ProbabilityDistribution
?$endgroup$
– JimB
6 hours ago
$begingroup$
@JimB . I tried this
TransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result is NormalDistribution[0, [Sigma]y]
. However, the correct result should be NormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
5 hours ago
$begingroup$
@JimB . I tried this
TransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result is NormalDistribution[0, [Sigma]y]
. However, the correct result should be NormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
5 hours ago
$begingroup$
You need to "index" the variable
y
or else Mathematica thinks it is a single variable.$endgroup$
– JimB
1 hour ago
$begingroup$
You need to "index" the variable
y
or else Mathematica thinks it is a single variable.$endgroup$
– JimB
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.
First the distribution of the mean:
marginalDistribution = TransformedDistribution[Sum[y[i], i, n]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n],
Assumptions -> [Sigma] > 0]
#, marginalDistribution/.n-># &/@Range[2,10]
$$
beginarraycc
2 & textNormalDistributionleft[0,fracsigma sqrt2right] \
3 & textNormalDistributionleft[0,fracsigma sqrt3right] \
4 & textNormalDistributionleft[0,fracsigma 2right] \
5 & textNormalDistributionleft[0,fracsigma sqrt5right] \
6 & textNormalDistributionleft[0,fracsigma sqrt6right] \
7 & textNormalDistributionleft[0,fracsigma sqrt7right] \
8 & textNormalDistributionleft[0,fracsigma 2 sqrt2right] \
9 & textNormalDistributionleft[0,fracsigma 3right] \
10 & textNormalDistributionleft[0,fracsigma sqrt10right] \
endarray
$$
So we see that the marginal distribution of $bary$ is
NormalDistribution[0, σ/Sqrt[n]]
The joint distribution of $bary$ and, say, $y_1$ is given by
jointDistribution = TransformedDistribution[y[1], Sum[y[i], i, n]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n]]
#, jointDistribution /. n -> # & /@ Range[2, 10] // TableForm
$$
beginarraycc
2 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^22 \
fracsigma ^22 & fracsigma ^22 \
endarray
right)right] \
3 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^23 \
fracsigma ^23 & fracsigma ^23 \
endarray
right)right] \
4 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^24 \
fracsigma ^24 & fracsigma ^24 \
endarray
right)right] \
5 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^25 \
fracsigma ^25 & fracsigma ^25 \
endarray
right)right] \
6 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^26 \
fracsigma ^26 & fracsigma ^26 \
endarray
right)right] \
7 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^27 \
fracsigma ^27 & fracsigma ^27 \
endarray
right)right] \
8 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^28 \
fracsigma ^28 & fracsigma ^28 \
endarray
right)right] \
9 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^29 \
fracsigma ^29 & fracsigma ^29 \
endarray
right)right] \
10 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^210 \
fracsigma ^210 & fracsigma ^210 \
endarray
right)right] \
endarray
$$
So the general distribution is a multivariate normal
MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n]
The general form of the joint density function can then be found with
FullSimplify[PDF[MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n], y, ybar],
Assumptions -> σ > 0, n > 1]
$$fracn e^-fracn left(n textybar^2+y^2-2 y textybarright)2 (n-1) sigma ^22 pi sqrtn-1 sigma ^2$$
$endgroup$
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
44 mins ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
42 mins ago
add a comment |
$begingroup$
Here is the distribution of $x=overliney$ (Part I of your question):
n = 5; (* for example *)
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
TransformedDistribution[Sum[y[k]/n, k, 5], a]
The result is
NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
UPDATE
Okay, here is how to do it with $n$ a variable:
a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
p[n_] := TransformedDistribution[Sum[y[k]/n, k, n], a[n]];
Now
x [Distributed] p[5] (* n=5, for example *)
Again, the result is
x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
$endgroup$
$begingroup$
I am not sure, but shouldn't be n instead of 5 hereTransformedDistribution[Sum[y[k]/n, k, 5], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw
$endgroup$
– Andrea2810
2 hours ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
Let's go with five because it is clearer. The result isNormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica putsAbs[]
around $sigma$. Obviously, $sigma>0$.
$endgroup$
– mjw
2 hours ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
1 hour ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
$endgroup$
– mjw
1 hour ago
|
show 5 more comments
$begingroup$
just modified @mjw's answer,
n = 100;(*for example*)ClearAll[y];
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
meanDist = TransformedDistribution[Sum[y[k]/100, k, 100], a]
JointDistribution can be composed by ProductDistribution,
if these random variables are independent.
if not,you have to use Copula
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, 100000];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, 100000];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]
=>
1., -0.00256777, -0.00256777, 1.
I'm not sure about correlation,but it's okay.
$endgroup$
$begingroup$
I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 hours ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 hours ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.
First the distribution of the mean:
marginalDistribution = TransformedDistribution[Sum[y[i], i, n]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n],
Assumptions -> [Sigma] > 0]
#, marginalDistribution/.n-># &/@Range[2,10]
$$
beginarraycc
2 & textNormalDistributionleft[0,fracsigma sqrt2right] \
3 & textNormalDistributionleft[0,fracsigma sqrt3right] \
4 & textNormalDistributionleft[0,fracsigma 2right] \
5 & textNormalDistributionleft[0,fracsigma sqrt5right] \
6 & textNormalDistributionleft[0,fracsigma sqrt6right] \
7 & textNormalDistributionleft[0,fracsigma sqrt7right] \
8 & textNormalDistributionleft[0,fracsigma 2 sqrt2right] \
9 & textNormalDistributionleft[0,fracsigma 3right] \
10 & textNormalDistributionleft[0,fracsigma sqrt10right] \
endarray
$$
So we see that the marginal distribution of $bary$ is
NormalDistribution[0, σ/Sqrt[n]]
The joint distribution of $bary$ and, say, $y_1$ is given by
jointDistribution = TransformedDistribution[y[1], Sum[y[i], i, n]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n]]
#, jointDistribution /. n -> # & /@ Range[2, 10] // TableForm
$$
beginarraycc
2 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^22 \
fracsigma ^22 & fracsigma ^22 \
endarray
right)right] \
3 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^23 \
fracsigma ^23 & fracsigma ^23 \
endarray
right)right] \
4 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^24 \
fracsigma ^24 & fracsigma ^24 \
endarray
right)right] \
5 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^25 \
fracsigma ^25 & fracsigma ^25 \
endarray
right)right] \
6 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^26 \
fracsigma ^26 & fracsigma ^26 \
endarray
right)right] \
7 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^27 \
fracsigma ^27 & fracsigma ^27 \
endarray
right)right] \
8 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^28 \
fracsigma ^28 & fracsigma ^28 \
endarray
right)right] \
9 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^29 \
fracsigma ^29 & fracsigma ^29 \
endarray
right)right] \
10 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^210 \
fracsigma ^210 & fracsigma ^210 \
endarray
right)right] \
endarray
$$
So the general distribution is a multivariate normal
MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n]
The general form of the joint density function can then be found with
FullSimplify[PDF[MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n], y, ybar],
Assumptions -> σ > 0, n > 1]
$$fracn e^-fracn left(n textybar^2+y^2-2 y textybarright)2 (n-1) sigma ^22 pi sqrtn-1 sigma ^2$$
$endgroup$
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
44 mins ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
42 mins ago
add a comment |
$begingroup$
I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.
First the distribution of the mean:
marginalDistribution = TransformedDistribution[Sum[y[i], i, n]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n],
Assumptions -> [Sigma] > 0]
#, marginalDistribution/.n-># &/@Range[2,10]
$$
beginarraycc
2 & textNormalDistributionleft[0,fracsigma sqrt2right] \
3 & textNormalDistributionleft[0,fracsigma sqrt3right] \
4 & textNormalDistributionleft[0,fracsigma 2right] \
5 & textNormalDistributionleft[0,fracsigma sqrt5right] \
6 & textNormalDistributionleft[0,fracsigma sqrt6right] \
7 & textNormalDistributionleft[0,fracsigma sqrt7right] \
8 & textNormalDistributionleft[0,fracsigma 2 sqrt2right] \
9 & textNormalDistributionleft[0,fracsigma 3right] \
10 & textNormalDistributionleft[0,fracsigma sqrt10right] \
endarray
$$
So we see that the marginal distribution of $bary$ is
NormalDistribution[0, σ/Sqrt[n]]
The joint distribution of $bary$ and, say, $y_1$ is given by
jointDistribution = TransformedDistribution[y[1], Sum[y[i], i, n]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n]]
#, jointDistribution /. n -> # & /@ Range[2, 10] // TableForm
$$
beginarraycc
2 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^22 \
fracsigma ^22 & fracsigma ^22 \
endarray
right)right] \
3 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^23 \
fracsigma ^23 & fracsigma ^23 \
endarray
right)right] \
4 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^24 \
fracsigma ^24 & fracsigma ^24 \
endarray
right)right] \
5 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^25 \
fracsigma ^25 & fracsigma ^25 \
endarray
right)right] \
6 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^26 \
fracsigma ^26 & fracsigma ^26 \
endarray
right)right] \
7 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^27 \
fracsigma ^27 & fracsigma ^27 \
endarray
right)right] \
8 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^28 \
fracsigma ^28 & fracsigma ^28 \
endarray
right)right] \
9 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^29 \
fracsigma ^29 & fracsigma ^29 \
endarray
right)right] \
10 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^210 \
fracsigma ^210 & fracsigma ^210 \
endarray
right)right] \
endarray
$$
So the general distribution is a multivariate normal
MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n]
The general form of the joint density function can then be found with
FullSimplify[PDF[MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n], y, ybar],
Assumptions -> σ > 0, n > 1]
$$fracn e^-fracn left(n textybar^2+y^2-2 y textybarright)2 (n-1) sigma ^22 pi sqrtn-1 sigma ^2$$
$endgroup$
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
44 mins ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
42 mins ago
add a comment |
$begingroup$
I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.
First the distribution of the mean:
marginalDistribution = TransformedDistribution[Sum[y[i], i, n]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n],
Assumptions -> [Sigma] > 0]
#, marginalDistribution/.n-># &/@Range[2,10]
$$
beginarraycc
2 & textNormalDistributionleft[0,fracsigma sqrt2right] \
3 & textNormalDistributionleft[0,fracsigma sqrt3right] \
4 & textNormalDistributionleft[0,fracsigma 2right] \
5 & textNormalDistributionleft[0,fracsigma sqrt5right] \
6 & textNormalDistributionleft[0,fracsigma sqrt6right] \
7 & textNormalDistributionleft[0,fracsigma sqrt7right] \
8 & textNormalDistributionleft[0,fracsigma 2 sqrt2right] \
9 & textNormalDistributionleft[0,fracsigma 3right] \
10 & textNormalDistributionleft[0,fracsigma sqrt10right] \
endarray
$$
So we see that the marginal distribution of $bary$ is
NormalDistribution[0, σ/Sqrt[n]]
The joint distribution of $bary$ and, say, $y_1$ is given by
jointDistribution = TransformedDistribution[y[1], Sum[y[i], i, n]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n]]
#, jointDistribution /. n -> # & /@ Range[2, 10] // TableForm
$$
beginarraycc
2 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^22 \
fracsigma ^22 & fracsigma ^22 \
endarray
right)right] \
3 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^23 \
fracsigma ^23 & fracsigma ^23 \
endarray
right)right] \
4 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^24 \
fracsigma ^24 & fracsigma ^24 \
endarray
right)right] \
5 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^25 \
fracsigma ^25 & fracsigma ^25 \
endarray
right)right] \
6 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^26 \
fracsigma ^26 & fracsigma ^26 \
endarray
right)right] \
7 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^27 \
fracsigma ^27 & fracsigma ^27 \
endarray
right)right] \
8 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^28 \
fracsigma ^28 & fracsigma ^28 \
endarray
right)right] \
9 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^29 \
fracsigma ^29 & fracsigma ^29 \
endarray
right)right] \
10 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^210 \
fracsigma ^210 & fracsigma ^210 \
endarray
right)right] \
endarray
$$
So the general distribution is a multivariate normal
MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n]
The general form of the joint density function can then be found with
FullSimplify[PDF[MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n], y, ybar],
Assumptions -> σ > 0, n > 1]
$$fracn e^-fracn left(n textybar^2+y^2-2 y textybarright)2 (n-1) sigma ^22 pi sqrtn-1 sigma ^2$$
$endgroup$
I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.
First the distribution of the mean:
marginalDistribution = TransformedDistribution[Sum[y[i], i, n]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n],
Assumptions -> [Sigma] > 0]
#, marginalDistribution/.n-># &/@Range[2,10]
$$
beginarraycc
2 & textNormalDistributionleft[0,fracsigma sqrt2right] \
3 & textNormalDistributionleft[0,fracsigma sqrt3right] \
4 & textNormalDistributionleft[0,fracsigma 2right] \
5 & textNormalDistributionleft[0,fracsigma sqrt5right] \
6 & textNormalDistributionleft[0,fracsigma sqrt6right] \
7 & textNormalDistributionleft[0,fracsigma sqrt7right] \
8 & textNormalDistributionleft[0,fracsigma 2 sqrt2right] \
9 & textNormalDistributionleft[0,fracsigma 3right] \
10 & textNormalDistributionleft[0,fracsigma sqrt10right] \
endarray
$$
So we see that the marginal distribution of $bary$ is
NormalDistribution[0, σ/Sqrt[n]]
The joint distribution of $bary$ and, say, $y_1$ is given by
jointDistribution = TransformedDistribution[y[1], Sum[y[i], i, n]/n,
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n]]
#, jointDistribution /. n -> # & /@ Range[2, 10] // TableForm
$$
beginarraycc
2 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^22 \
fracsigma ^22 & fracsigma ^22 \
endarray
right)right] \
3 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^23 \
fracsigma ^23 & fracsigma ^23 \
endarray
right)right] \
4 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^24 \
fracsigma ^24 & fracsigma ^24 \
endarray
right)right] \
5 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^25 \
fracsigma ^25 & fracsigma ^25 \
endarray
right)right] \
6 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^26 \
fracsigma ^26 & fracsigma ^26 \
endarray
right)right] \
7 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^27 \
fracsigma ^27 & fracsigma ^27 \
endarray
right)right] \
8 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^28 \
fracsigma ^28 & fracsigma ^28 \
endarray
right)right] \
9 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^29 \
fracsigma ^29 & fracsigma ^29 \
endarray
right)right] \
10 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^210 \
fracsigma ^210 & fracsigma ^210 \
endarray
right)right] \
endarray
$$
So the general distribution is a multivariate normal
MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n]
The general form of the joint density function can then be found with
FullSimplify[PDF[MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n], y, ybar],
Assumptions -> σ > 0, n > 1]
$$fracn e^-fracn left(n textybar^2+y^2-2 y textybarright)2 (n-1) sigma ^22 pi sqrtn-1 sigma ^2$$
edited 54 mins ago
answered 1 hour ago
JimBJimB
18k12863
18k12863
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
44 mins ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
42 mins ago
add a comment |
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
44 mins ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
42 mins ago
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
44 mins ago
$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
44 mins ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
42 mins ago
$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
42 mins ago
add a comment |
$begingroup$
Here is the distribution of $x=overliney$ (Part I of your question):
n = 5; (* for example *)
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
TransformedDistribution[Sum[y[k]/n, k, 5], a]
The result is
NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
UPDATE
Okay, here is how to do it with $n$ a variable:
a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
p[n_] := TransformedDistribution[Sum[y[k]/n, k, n], a[n]];
Now
x [Distributed] p[5] (* n=5, for example *)
Again, the result is
x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
$endgroup$
$begingroup$
I am not sure, but shouldn't be n instead of 5 hereTransformedDistribution[Sum[y[k]/n, k, 5], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw
$endgroup$
– Andrea2810
2 hours ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
Let's go with five because it is clearer. The result isNormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica putsAbs[]
around $sigma$. Obviously, $sigma>0$.
$endgroup$
– mjw
2 hours ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
1 hour ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
$endgroup$
– mjw
1 hour ago
|
show 5 more comments
$begingroup$
Here is the distribution of $x=overliney$ (Part I of your question):
n = 5; (* for example *)
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
TransformedDistribution[Sum[y[k]/n, k, 5], a]
The result is
NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
UPDATE
Okay, here is how to do it with $n$ a variable:
a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
p[n_] := TransformedDistribution[Sum[y[k]/n, k, n], a[n]];
Now
x [Distributed] p[5] (* n=5, for example *)
Again, the result is
x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
$endgroup$
$begingroup$
I am not sure, but shouldn't be n instead of 5 hereTransformedDistribution[Sum[y[k]/n, k, 5], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw
$endgroup$
– Andrea2810
2 hours ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
Let's go with five because it is clearer. The result isNormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica putsAbs[]
around $sigma$. Obviously, $sigma>0$.
$endgroup$
– mjw
2 hours ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
1 hour ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
$endgroup$
– mjw
1 hour ago
|
show 5 more comments
$begingroup$
Here is the distribution of $x=overliney$ (Part I of your question):
n = 5; (* for example *)
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
TransformedDistribution[Sum[y[k]/n, k, 5], a]
The result is
NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
UPDATE
Okay, here is how to do it with $n$ a variable:
a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
p[n_] := TransformedDistribution[Sum[y[k]/n, k, n], a[n]];
Now
x [Distributed] p[5] (* n=5, for example *)
Again, the result is
x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
$endgroup$
Here is the distribution of $x=overliney$ (Part I of your question):
n = 5; (* for example *)
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
TransformedDistribution[Sum[y[k]/n, k, 5], a]
The result is
NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
UPDATE
Okay, here is how to do it with $n$ a variable:
a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
p[n_] := TransformedDistribution[Sum[y[k]/n, k, n], a[n]];
Now
x [Distributed] p[5] (* n=5, for example *)
Again, the result is
x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]
edited 1 hour ago
answered 3 hours ago
mjwmjw
9679
9679
$begingroup$
I am not sure, but shouldn't be n instead of 5 hereTransformedDistribution[Sum[y[k]/n, k, 5], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw
$endgroup$
– Andrea2810
2 hours ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
Let's go with five because it is clearer. The result isNormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica putsAbs[]
around $sigma$. Obviously, $sigma>0$.
$endgroup$
– mjw
2 hours ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
1 hour ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
$endgroup$
– mjw
1 hour ago
|
show 5 more comments
$begingroup$
I am not sure, but shouldn't be n instead of 5 hereTransformedDistribution[Sum[y[k]/n, k, 5], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw
$endgroup$
– Andrea2810
2 hours ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
Let's go with five because it is clearer. The result isNormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica putsAbs[]
around $sigma$. Obviously, $sigma>0$.
$endgroup$
– mjw
2 hours ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
1 hour ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
$endgroup$
– mjw
1 hour ago
$begingroup$
I am not sure, but shouldn't be n instead of 5 here
TransformedDistribution[Sum[y[k]/n, k, 5], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw$endgroup$
– Andrea2810
2 hours ago
$begingroup$
I am not sure, but shouldn't be n instead of 5 here
TransformedDistribution[Sum[y[k]/n, k, 5], a]
? And what if I want to leave n, without assigning a value to n? Thanks @mjw$endgroup$
– Andrea2810
2 hours ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
Let's go with five because it is clearer. The result is
NormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica puts Abs[]
around $sigma$. Obviously, $sigma>0$.$endgroup$
– mjw
2 hours ago
$begingroup$
Let's go with five because it is clearer. The result is
NormalDistribution[0,[Sigma]/Sqrt[5]]
. Not sure why Mathematica puts Abs[]
around $sigma$. Obviously, $sigma>0$.$endgroup$
– mjw
2 hours ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
1 hour ago
$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
1 hour ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
$endgroup$
– mjw
1 hour ago
$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
$endgroup$
– mjw
1 hour ago
|
show 5 more comments
$begingroup$
just modified @mjw's answer,
n = 100;(*for example*)ClearAll[y];
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
meanDist = TransformedDistribution[Sum[y[k]/100, k, 100], a]
JointDistribution can be composed by ProductDistribution,
if these random variables are independent.
if not,you have to use Copula
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, 100000];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, 100000];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]
=>
1., -0.00256777, -0.00256777, 1.
I'm not sure about correlation,but it's okay.
$endgroup$
$begingroup$
I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 hours ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 hours ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 hours ago
add a comment |
$begingroup$
just modified @mjw's answer,
n = 100;(*for example*)ClearAll[y];
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
meanDist = TransformedDistribution[Sum[y[k]/100, k, 100], a]
JointDistribution can be composed by ProductDistribution,
if these random variables are independent.
if not,you have to use Copula
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, 100000];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, 100000];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]
=>
1., -0.00256777, -0.00256777, 1.
I'm not sure about correlation,but it's okay.
$endgroup$
$begingroup$
I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 hours ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 hours ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 hours ago
add a comment |
$begingroup$
just modified @mjw's answer,
n = 100;(*for example*)ClearAll[y];
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
meanDist = TransformedDistribution[Sum[y[k]/100, k, 100], a]
JointDistribution can be composed by ProductDistribution,
if these random variables are independent.
if not,you have to use Copula
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, 100000];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, 100000];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]
=>
1., -0.00256777, -0.00256777, 1.
I'm not sure about correlation,but it's okay.
$endgroup$
just modified @mjw's answer,
n = 100;(*for example*)ClearAll[y];
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
meanDist = TransformedDistribution[Sum[y[k]/100, k, 100], a]
JointDistribution can be composed by ProductDistribution,
if these random variables are independent.
if not,you have to use Copula
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D
joint = ProductDistribution[meanDist,
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, 100000];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, 100000];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]
=>
1., -0.00256777, -0.00256777, 1.
I'm not sure about correlation,but it's okay.
edited 1 hour ago
answered 2 hours ago
XminerXminer
19918
19918
$begingroup$
I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 hours ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 hours ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 hours ago
add a comment |
$begingroup$
I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 hours ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 hours ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 hours ago
$begingroup$
I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 hours ago
$begingroup$
I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 hours ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 hours ago
$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 hours ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 hours ago
$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 hours ago
add a comment |
Andrea2810 is a new contributor. Be nice, and check out our Code of Conduct.
Andrea2810 is a new contributor. Be nice, and check out our Code of Conduct.
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4
$begingroup$
What have you tried? For example, have you seen the documentation on
TransformedDistribution
andProbabilityDistribution
?$endgroup$
– JimB
6 hours ago
$begingroup$
@JimB . I tried this
TransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]
. The result isNormalDistribution[0, [Sigma]y]
. However, the correct result should beNormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
5 hours ago
$begingroup$
You need to "index" the variable
y
or else Mathematica thinks it is a single variable.$endgroup$
– JimB
1 hour ago