Gaussian expectation for a rational function
up vote
1
down vote
favorite
Suppose $X_1, X_2, ..., X_N$ are i.i.d with Gaussian distribution with unit mean and variance $sigma^2$, can we find the following expectation ?
begin{equation}
mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]
end{equation}
Actually I wanted to calculate following expectation, then I realized I need that :
begin{align}
&mathbb{E} left[frac{X_1^2+X_2^2+...+X_N^2}{1+X_1^2+X_2^2+...+X_N^2}right]\
&qquad=mathbb{E} left[frac{1+X_1^2+X_2^2+...+X_N^2}{1+X_1^2+X_2^2+...+X_N^2}right]-mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]\
&qquad=1-mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]
end{align}
definite-integrals normal-distribution expectation gaussian-integral
edited Aug 20 at 2:47
6005
35.5k751125
asked Aug 12 '17 at 1:20
Alireza
1939
add a comment |
up vote
1
down vote
favorite
Suppose $X_1, X_2, ..., X_N$ are i.i.d with Gaussian distribution with unit mean and variance $sigma^2$, can we find the following expectation ?
begin{equation}
mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]
end{equation}
Actually I wanted to calculate following expectation, then I realized I need that :
begin{align}
&mathbb{E} left[frac{X_1^2+X_2^2+...+X_N^2}{1+X_1^2+X_2^2+...+X_N^2}right]\
&qquad=mathbb{E} left[frac{1+X_1^2+X_2^2+...+X_N^2}{1+X_1^2+X_2^2+...+X_N^2}right]-mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]\
&qquad=1-mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]
end{align}
definite-integrals normal-distribution expectation gaussian-integral
edited Aug 20 at 2:47
6005
35.5k751125
asked Aug 12 '17 at 1:20
Alireza
1939
Do you know about the non-central chi-squared distribution?
– kimchi lover
Aug 12 '17 at 1:25
@kimchilover I know, but I am actually looking into expectation of inverse of (Chi-Square +1)
– Alireza
Aug 12 '17 at 1:27
So you have a 1 dimensional integral: a rational function integrated against a gamma-ish density thing. I guess I'm asking, have you written out such an integral and checked conditions for convergence, and checked for its presence in handbooks of definite integrals? Can you do the same for the central chi-squared distribution, as a warm up exercise?
– kimchi lover
Aug 12 '17 at 1:31
@kimchilover I believ I have looked at that before, I 'll try it again with more devotion, I thought I may be missing something obvious and somebody can help me.
– Alireza
Aug 12 '17 at 1:36
Just for reference, we have $$mathbb{E}left[frac{1}{1+X_1^2+cdots+X_N^2}right]=int_{0}^{infty}frac{e^{-t}}{(1+2sigma^2t)^{N/2}},dt.$$ This integral can be represented in terms of incomplete gamma function.
– Sangchul Lee
Aug 20 at 4:43
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $X_1, X_2, ..., X_N$ are i.i.d with Gaussian distribution with unit mean and variance $sigma^2$, can we find the following expectation ?
begin{equation}
mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]
end{equation}
Actually I wanted to calculate following expectation, then I realized I need that :
begin{align}
&mathbb{E} left[frac{X_1^2+X_2^2+...+X_N^2}{1+X_1^2+X_2^2+...+X_N^2}right]\
&qquad=mathbb{E} left[frac{1+X_1^2+X_2^2+...+X_N^2}{1+X_1^2+X_2^2+...+X_N^2}right]-mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]\
&qquad=1-mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]
end{align}
definite-integrals normal-distribution expectation gaussian-integral
edited Aug 20 at 2:47
6005
35.5k751125
asked Aug 12 '17 at 1:20
Alireza
1939
Suppose $X_1, X_2, ..., X_N$ are i.i.d with Gaussian distribution with unit mean and variance $sigma^2$, can we find the following expectation ?
begin{equation}
mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]
end{equation}
Actually I wanted to calculate following expectation, then I realized I need that :
begin{align}
&mathbb{E} left[frac{X_1^2+X_2^2+...+X_N^2}{1+X_1^2+X_2^2+...+X_N^2}right]\
&qquad=mathbb{E} left[frac{1+X_1^2+X_2^2+...+X_N^2}{1+X_1^2+X_2^2+...+X_N^2}right]-mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]\
&qquad=1-mathbb{E} left[frac{1}{1+X_1^2+X_2^2+...+X_N^2}right]
end{align}
definite-integrals normal-distribution expectation gaussian-integral
definite-integrals normal-distribution expectation gaussian-integral
edited Aug 20 at 2:47
6005
35.5k751125
asked Aug 12 '17 at 1:20
Alireza
1939
edited Aug 20 at 2:47
6005
35.5k751125
asked Aug 12 '17 at 1:20
Alireza
1939
edited Aug 20 at 2:47
6005
35.5k751125
edited Aug 20 at 2:47
6005
35.5k751125
edited Aug 20 at 2:47
6005
35.5k751125
35.5k751125
asked Aug 12 '17 at 1:20
Alireza
1939
asked Aug 12 '17 at 1:20
Alireza
1939
asked Aug 12 '17 at 1:20
Alireza
1939
1939
Do you know about the non-central chi-squared distribution?
– kimchi lover
Aug 12 '17 at 1:25
@kimchilover I know, but I am actually looking into expectation of inverse of (Chi-Square +1)
– Alireza
Aug 12 '17 at 1:27
So you have a 1 dimensional integral: a rational function integrated against a gamma-ish density thing. I guess I'm asking, have you written out such an integral and checked conditions for convergence, and checked for its presence in handbooks of definite integrals? Can you do the same for the central chi-squared distribution, as a warm up exercise?
– kimchi lover
Aug 12 '17 at 1:31
@kimchilover I believ I have looked at that before, I 'll try it again with more devotion, I thought I may be missing something obvious and somebody can help me.
– Alireza
Aug 12 '17 at 1:36
Just for reference, we have $$mathbb{E}left[frac{1}{1+X_1^2+cdots+X_N^2}right]=int_{0}^{infty}frac{e^{-t}}{(1+2sigma^2t)^{N/2}},dt.$$ This integral can be represented in terms of incomplete gamma function.
– Sangchul Lee
Aug 20 at 4:43
add a comment |
Do you know about the non-central chi-squared distribution?
– kimchi lover
Aug 12 '17 at 1:25
@kimchilover I know, but I am actually looking into expectation of inverse of (Chi-Square +1)
– Alireza
Aug 12 '17 at 1:27
So you have a 1 dimensional integral: a rational function integrated against a gamma-ish density thing. I guess I'm asking, have you written out such an integral and checked conditions for convergence, and checked for its presence in handbooks of definite integrals? Can you do the same for the central chi-squared distribution, as a warm up exercise?
– kimchi lover
Aug 12 '17 at 1:31
@kimchilover I believ I have looked at that before, I 'll try it again with more devotion, I thought I may be missing something obvious and somebody can help me.
– Alireza
Aug 12 '17 at 1:36
Just for reference, we have $$mathbb{E}left[frac{1}{1+X_1^2+cdots+X_N^2}right]=int_{0}^{infty}frac{e^{-t}}{(1+2sigma^2t)^{N/2}},dt.$$ This integral can be represented in terms of incomplete gamma function.
– Sangchul Lee
Aug 20 at 4:43
Do you know about the non-central chi-squared distribution?
– kimchi lover
Aug 12 '17 at 1:25
Do you know about the non-central chi-squared distribution?
– kimchi lover
Aug 12 '17 at 1:25
@kimchilover I know, but I am actually looking into expectation of inverse of (Chi-Square +1)
– Alireza
Aug 12 '17 at 1:27
@kimchilover I know, but I am actually looking into expectation of inverse of (Chi-Square +1)
– Alireza
Aug 12 '17 at 1:27
So you have a 1 dimensional integral: a rational function integrated against a gamma-ish density thing. I guess I'm asking, have you written out such an integral and checked conditions for convergence, and checked for its presence in handbooks of definite integrals? Can you do the same for the central chi-squared distribution, as a warm up exercise?
– kimchi lover
Aug 12 '17 at 1:31
So you have a 1 dimensional integral: a rational function integrated against a gamma-ish density thing. I guess I'm asking, have you written out such an integral and checked conditions for convergence, and checked for its presence in handbooks of definite integrals? Can you do the same for the central chi-squared distribution, as a warm up exercise?
– kimchi lover
Aug 12 '17 at 1:31
@kimchilover I believ I have looked at that before, I 'll try it again with more devotion, I thought I may be missing something obvious and somebody can help me.
– Alireza
Aug 12 '17 at 1:36
@kimchilover I believ I have looked at that before, I 'll try it again with more devotion, I thought I may be missing something obvious and somebody can help me.
– Alireza
Aug 12 '17 at 1:36
Just for reference, we have $$mathbb{E}left[frac{1}{1+X_1^2+cdots+X_N^2}right]=int_{0}^{infty}frac{e^{-t}}{(1+2sigma^2t)^{N/2}},dt.$$ This integral can be represented in terms of incomplete gamma function.
– Sangchul Lee
Aug 20 at 4:43
Just for reference, we have $$mathbb{E}left[frac{1}{1+X_1^2+cdots+X_N^2}right]=int_{0}^{infty}frac{e^{-t}}{(1+2sigma^2t)^{N/2}},dt.$$ This integral can be represented in terms of incomplete gamma function.
– Sangchul Lee
Aug 20 at 4:43
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Using my favorite trick:
$$F(a)=int dX_1 cdots dX_N {e^{-asum X^2_i}over{1+sum X^2_i}}$$
$$F(a)={1over 1-{dover da}}int dX_1 cdots dX_N e^{-asum X^2_i}$$
$$(1-{dover da})F(a)=int dX_1 cdots dX_N e^{-asum X^2_i} $$
define
$$F(a)=u(a)e^{a}$$
and we get
$$u'(a)e^{a}=Cint dX_1 cdots dX_N e^{-asum X^2_i}=C a^{-N/2}$$
$$u'(a)=e^{-a}a^{-N/2}$$
$$F(a)=C e^a int_a da' e^{-a'}a'^{-N/2}$$
setting $a=1$ will give the required integral
answered Nov 22 at 8:11
user617446
261
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user617446 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Using my favorite trick:
$$F(a)=int dX_1 cdots dX_N {e^{-asum X^2_i}over{1+sum X^2_i}}$$
$$F(a)={1over 1-{dover da}}int dX_1 cdots dX_N e^{-asum X^2_i}$$
$$(1-{dover da})F(a)=int dX_1 cdots dX_N e^{-asum X^2_i} $$
define
$$F(a)=u(a)e^{a}$$
and we get
$$u'(a)e^{a}=Cint dX_1 cdots dX_N e^{-asum X^2_i}=C a^{-N/2}$$
$$u'(a)=e^{-a}a^{-N/2}$$
$$F(a)=C e^a int_a da' e^{-a'}a'^{-N/2}$$
setting $a=1$ will give the required integral
answered Nov 22 at 8:11
user617446
261
New contributor
user617446 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
Using my favorite trick:
$$F(a)=int dX_1 cdots dX_N {e^{-asum X^2_i}over{1+sum X^2_i}}$$
$$F(a)={1over 1-{dover da}}int dX_1 cdots dX_N e^{-asum X^2_i}$$
$$(1-{dover da})F(a)=int dX_1 cdots dX_N e^{-asum X^2_i} $$
define
$$F(a)=u(a)e^{a}$$
and we get
$$u'(a)e^{a}=Cint dX_1 cdots dX_N e^{-asum X^2_i}=C a^{-N/2}$$
$$u'(a)=e^{-a}a^{-N/2}$$
$$F(a)=C e^a int_a da' e^{-a'}a'^{-N/2}$$
setting $a=1$ will give the required integral
answered Nov 22 at 8:11
user617446
261
New contributor
user617446 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
Using my favorite trick:
$$F(a)=int dX_1 cdots dX_N {e^{-asum X^2_i}over{1+sum X^2_i}}$$
$$F(a)={1over 1-{dover da}}int dX_1 cdots dX_N e^{-asum X^2_i}$$
$$(1-{dover da})F(a)=int dX_1 cdots dX_N e^{-asum X^2_i} $$
define
$$F(a)=u(a)e^{a}$$
and we get
$$u'(a)e^{a}=Cint dX_1 cdots dX_N e^{-asum X^2_i}=C a^{-N/2}$$
$$u'(a)=e^{-a}a^{-N/2}$$
$$F(a)=C e^a int_a da' e^{-a'}a'^{-N/2}$$
setting $a=1$ will give the required integral
answered Nov 22 at 8:11
user617446
261
New contributor
user617446 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Using my favorite trick:
$$F(a)=int dX_1 cdots dX_N {e^{-asum X^2_i}over{1+sum X^2_i}}$$
$$F(a)={1over 1-{dover da}}int dX_1 cdots dX_N e^{-asum X^2_i}$$
$$(1-{dover da})F(a)=int dX_1 cdots dX_N e^{-asum X^2_i} $$
define
$$F(a)=u(a)e^{a}$$
and we get
$$u'(a)e^{a}=Cint dX_1 cdots dX_N e^{-asum X^2_i}=C a^{-N/2}$$
$$u'(a)=e^{-a}a^{-N/2}$$
$$F(a)=C e^a int_a da' e^{-a'}a'^{-N/2}$$
setting $a=1$ will give the required integral
answered Nov 22 at 8:11
user617446
261
New contributor
user617446 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 22 at 8:11
user617446
261
New contributor
user617446 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 22 at 8:11
user617446
261
answered Nov 22 at 8:11
user617446
261
261
New contributor
user617446 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user617446 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user617446 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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Do you know about the non-central chi-squared distribution?
– kimchi lover
Aug 12 '17 at 1:25
@kimchilover I know, but I am actually looking into expectation of inverse of (Chi-Square +1)
– Alireza
Aug 12 '17 at 1:27
So you have a 1 dimensional integral: a rational function integrated against a gamma-ish density thing. I guess I'm asking, have you written out such an integral and checked conditions for convergence, and checked for its presence in handbooks of definite integrals? Can you do the same for the central chi-squared distribution, as a warm up exercise?
– kimchi lover
Aug 12 '17 at 1:31
@kimchilover I believ I have looked at that before, I 'll try it again with more devotion, I thought I may be missing something obvious and somebody can help me.
– Alireza
Aug 12 '17 at 1:36
Just for reference, we have $$mathbb{E}left[frac{1}{1+X_1^2+cdots+X_N^2}right]=int_{0}^{infty}frac{e^{-t}}{(1+2sigma^2t)^{N/2}},dt.$$ This integral can be represented in terms of incomplete gamma function.
– Sangchul Lee
Aug 20 at 4:43