Special case of higher moments of multivariate normal distribution

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Consider $X=(X_1,ldots, X_n)$, $nin mathbb N$, such that $X_i sim mathcal N (x_i, t)$ independently for $(x_1,ldots, x_n) in mathbb R^n$ and $tgeq 0$. Let $p = 2 q$ for $ qin mathbb N$. I am intersted in an expression for
$$mathbb E[| X | ^p] = mathbb E [(X_1^2 +ldots + X_n^2)^q]$$
which should be a polynomial in $t$, but I do not know how to figure it out as clear expression.

probability probability-distributions normal-distribution

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edited Aug 6 at 8:57

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asked Aug 6 at 8:43

Falrach

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Consider $X=(X_1,ldots, X_n)$, $nin mathbb N$, such that $X_i sim mathcal N (x_i, t)$ independently for $(x_1,ldots, x_n) in mathbb R^n$ and $tgeq 0$. Let $p = 2 q$ for $ qin mathbb N$. I am intersted in an expression for
$$mathbb E[| X | ^p] = mathbb E [(X_1^2 +ldots + X_n^2)^q]$$
which should be a polynomial in $t$, but I do not know how to figure it out as clear expression.

probability probability-distributions normal-distribution

share|cite|improve this question

edited Aug 6 at 8:57

pointguard0

1,136818

asked Aug 6 at 8:43

Falrach

1,222123

add a comment | 

up vote
1
down vote

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up vote
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down vote

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1

Consider $X=(X_1,ldots, X_n)$, $nin mathbb N$, such that $X_i sim mathcal N (x_i, t)$ independently for $(x_1,ldots, x_n) in mathbb R^n$ and $tgeq 0$. Let $p = 2 q$ for $ qin mathbb N$. I am intersted in an expression for
$$mathbb E[| X | ^p] = mathbb E [(X_1^2 +ldots + X_n^2)^q]$$
which should be a polynomial in $t$, but I do not know how to figure it out as clear expression.

probability probability-distributions normal-distribution

share|cite|improve this question

edited Aug 6 at 8:57

pointguard0

1,136818

asked Aug 6 at 8:43

Falrach

1,222123

Consider $X=(X_1,ldots, X_n)$, $nin mathbb N$, such that $X_i sim mathcal N (x_i, t)$ independently for $(x_1,ldots, x_n) in mathbb R^n$ and $tgeq 0$. Let $p = 2 q$ for $ qin mathbb N$. I am intersted in an expression for
$$mathbb E[| X | ^p] = mathbb E [(X_1^2 +ldots + X_n^2)^q]$$
which should be a polynomial in $t$, but I do not know how to figure it out as clear expression.

probability probability-distributions normal-distribution

share|cite|improve this question

edited Aug 6 at 8:57

pointguard0

1,136818

asked Aug 6 at 8:43

Falrach

1,222123

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 8:57

pointguard0

1,136818

edited Aug 6 at 8:57

pointguard0

1,136818

edited Aug 6 at 8:57

pointguard0

1,136818

1,136818

asked Aug 6 at 8:43

Falrach

1,222123

asked Aug 6 at 8:43

Falrach

1,222123

asked Aug 6 at 8:43

Falrach

1,222123

1,222123

add a comment | 

add a comment | 

1 Answer
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HINT:

All you need to do is to compute the even moments of normal distribution and carefully expand the brackets. Since, $X_i$ are not standard normal then the expressions become very messy even for small $q$. To illustrate what I mean I computed for the first two values of $q$ so that you could get the idea.

---

For $q = 1$ you clearly get $$mathbbE | X |^2 = nt + sum_i=1^n x_i^2.$$

For $q = 2$ similarly $$mathbbE | X |^4 = sum_i=1^n mathbbE X_i^4 + sum_ineq j mathbbE X_i^2 cdot mathbbE X_j^2 = 3nt^2 + sum_i=1^n (x_i^4 + 6x_i^2 t) + sum_i neq j(t + x_i^2)(t + x_j^2).$$

The latter could be written in a canonical form of polynomial as follows
$$
mathbbE | X |^4 = t^2(n^2 + 2n) + tleft(sum_i=1^n 6x_i^2 + sum_i neq j (x_i^2 + x_j^2)right) + sum_i=1^n x_i^4 + sum_i neq j x_i^2 x_j^2.
$$

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answered Aug 6 at 9:02

pointguard0

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1 Answer
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1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
1
down vote

HINT:

All you need to do is to compute the even moments of normal distribution and carefully expand the brackets. Since, $X_i$ are not standard normal then the expressions become very messy even for small $q$. To illustrate what I mean I computed for the first two values of $q$ so that you could get the idea.

---

For $q = 1$ you clearly get $$mathbbE | X |^2 = nt + sum_i=1^n x_i^2.$$

For $q = 2$ similarly $$mathbbE | X |^4 = sum_i=1^n mathbbE X_i^4 + sum_ineq j mathbbE X_i^2 cdot mathbbE X_j^2 = 3nt^2 + sum_i=1^n (x_i^4 + 6x_i^2 t) + sum_i neq j(t + x_i^2)(t + x_j^2).$$

The latter could be written in a canonical form of polynomial as follows
$$
mathbbE | X |^4 = t^2(n^2 + 2n) + tleft(sum_i=1^n 6x_i^2 + sum_i neq j (x_i^2 + x_j^2)right) + sum_i=1^n x_i^4 + sum_i neq j x_i^2 x_j^2.
$$

share|cite|improve this answer

answered Aug 6 at 9:02

pointguard0

1,136818

add a comment | 

up vote
1
down vote

HINT:

All you need to do is to compute the even moments of normal distribution and carefully expand the brackets. Since, $X_i$ are not standard normal then the expressions become very messy even for small $q$. To illustrate what I mean I computed for the first two values of $q$ so that you could get the idea.

---

For $q = 1$ you clearly get $$mathbbE | X |^2 = nt + sum_i=1^n x_i^2.$$

For $q = 2$ similarly $$mathbbE | X |^4 = sum_i=1^n mathbbE X_i^4 + sum_ineq j mathbbE X_i^2 cdot mathbbE X_j^2 = 3nt^2 + sum_i=1^n (x_i^4 + 6x_i^2 t) + sum_i neq j(t + x_i^2)(t + x_j^2).$$

The latter could be written in a canonical form of polynomial as follows
$$
mathbbE | X |^4 = t^2(n^2 + 2n) + tleft(sum_i=1^n 6x_i^2 + sum_i neq j (x_i^2 + x_j^2)right) + sum_i=1^n x_i^4 + sum_i neq j x_i^2 x_j^2.
$$

share|cite|improve this answer

answered Aug 6 at 9:02

pointguard0

1,136818

add a comment | 

up vote
1
down vote

up vote
1
down vote

HINT:

All you need to do is to compute the even moments of normal distribution and carefully expand the brackets. Since, $X_i$ are not standard normal then the expressions become very messy even for small $q$. To illustrate what I mean I computed for the first two values of $q$ so that you could get the idea.

---

For $q = 1$ you clearly get $$mathbbE | X |^2 = nt + sum_i=1^n x_i^2.$$

For $q = 2$ similarly $$mathbbE | X |^4 = sum_i=1^n mathbbE X_i^4 + sum_ineq j mathbbE X_i^2 cdot mathbbE X_j^2 = 3nt^2 + sum_i=1^n (x_i^4 + 6x_i^2 t) + sum_i neq j(t + x_i^2)(t + x_j^2).$$

The latter could be written in a canonical form of polynomial as follows
$$
mathbbE | X |^4 = t^2(n^2 + 2n) + tleft(sum_i=1^n 6x_i^2 + sum_i neq j (x_i^2 + x_j^2)right) + sum_i=1^n x_i^4 + sum_i neq j x_i^2 x_j^2.
$$

share|cite|improve this answer

answered Aug 6 at 9:02

pointguard0

1,136818

HINT:

All you need to do is to compute the even moments of normal distribution and carefully expand the brackets. Since, $X_i$ are not standard normal then the expressions become very messy even for small $q$. To illustrate what I mean I computed for the first two values of $q$ so that you could get the idea.

---

For $q = 1$ you clearly get $$mathbbE | X |^2 = nt + sum_i=1^n x_i^2.$$

For $q = 2$ similarly $$mathbbE | X |^4 = sum_i=1^n mathbbE X_i^4 + sum_ineq j mathbbE X_i^2 cdot mathbbE X_j^2 = 3nt^2 + sum_i=1^n (x_i^4 + 6x_i^2 t) + sum_i neq j(t + x_i^2)(t + x_j^2).$$

The latter could be written in a canonical form of polynomial as follows
$$
mathbbE | X |^4 = t^2(n^2 + 2n) + tleft(sum_i=1^n 6x_i^2 + sum_i neq j (x_i^2 + x_j^2)right) + sum_i=1^n x_i^4 + sum_i neq j x_i^2 x_j^2.
$$

share|cite|improve this answer

answered Aug 6 at 9:02

pointguard0

1,136818

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 9:02

pointguard0

1,136818

answered Aug 6 at 9:02

pointguard0

1,136818

answered Aug 6 at 9:02

pointguard0

1,136818

1,136818

add a comment | 

add a comment | 

 

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