Mixing
The moments of the sum of variables following standard normal distribution divided by the sum of the squares of them
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Let i.i.d. $t_isim N(0,1), i = 1,2,dots,n$, and
beginequation
X = fracsum_i t_isum_i t_i^2.
endequation
How to calculate the first two moments of $X$, i.e., $mathrmE(X)$ and $mathrmE(X^2)$?
I did some simulation studies and am almost sure that $mathrmE(X) = 0$ and $mathrmE(X^2) = frac1n-2$. However, I failed to get the derivation in detail.
Thanks in advance.
expectation normal-distribution moment-generating-functions ratio chi-squared
asked Jul 25 at 12:50
Frank Li
62
add a comment |Â
up vote
1
down vote
favorite
Let i.i.d. $t_isim N(0,1), i = 1,2,dots,n$, and
beginequation
X = fracsum_i t_isum_i t_i^2.
endequation
How to calculate the first two moments of $X$, i.e., $mathrmE(X)$ and $mathrmE(X^2)$?
I did some simulation studies and am almost sure that $mathrmE(X) = 0$ and $mathrmE(X^2) = frac1n-2$. However, I failed to get the derivation in detail.
Thanks in advance.
expectation normal-distribution moment-generating-functions ratio chi-squared
asked Jul 25 at 12:50
Frank Li
62
Welcome to MSE. What have you tried? You will find you'll get a better reception if you show your work in progress, no matter how small. See math.meta.stackexchange.com/questions/9959/…
– Daniel Buck
Jul 25 at 13:00
1
Thanks for reminding me. Some progress added.
– Frank Li
Jul 25 at 13:09
Well, $E(X)=0$ is trivial because each $t_i$ has a distribution symmetric about $0$. (You have to verify the tails don't make for an $infty-infty$ situation, but that's easy in this case.)
– J.G.
Jul 25 at 13:15
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let i.i.d. $t_isim N(0,1), i = 1,2,dots,n$, and
beginequation
X = fracsum_i t_isum_i t_i^2.
endequation
How to calculate the first two moments of $X$, i.e., $mathrmE(X)$ and $mathrmE(X^2)$?
I did some simulation studies and am almost sure that $mathrmE(X) = 0$ and $mathrmE(X^2) = frac1n-2$. However, I failed to get the derivation in detail.
Thanks in advance.
expectation normal-distribution moment-generating-functions ratio chi-squared
asked Jul 25 at 12:50
Frank Li
62
Let i.i.d. $t_isim N(0,1), i = 1,2,dots,n$, and
beginequation
X = fracsum_i t_isum_i t_i^2.
endequation
How to calculate the first two moments of $X$, i.e., $mathrmE(X)$ and $mathrmE(X^2)$?
I did some simulation studies and am almost sure that $mathrmE(X) = 0$ and $mathrmE(X^2) = frac1n-2$. However, I failed to get the derivation in detail.
Thanks in advance.
expectation normal-distribution moment-generating-functions ratio chi-squared
asked Jul 25 at 12:50
Frank Li
62
edited Jul 25 at 13:34
asked Jul 25 at 12:50
Frank Li
62
asked Jul 25 at 12:50
Frank Li
62
asked Jul 25 at 12:50
Frank Li
62
62
Welcome to MSE. What have you tried? You will find you'll get a better reception if you show your work in progress, no matter how small. See math.meta.stackexchange.com/questions/9959/…
– Daniel Buck
Jul 25 at 13:00
1
Thanks for reminding me. Some progress added.
– Frank Li
Jul 25 at 13:09
Well, $E(X)=0$ is trivial because each $t_i$ has a distribution symmetric about $0$. (You have to verify the tails don't make for an $infty-infty$ situation, but that's easy in this case.)
– J.G.
Jul 25 at 13:15
add a comment |Â
Welcome to MSE. What have you tried? You will find you'll get a better reception if you show your work in progress, no matter how small. See math.meta.stackexchange.com/questions/9959/…
– Daniel Buck
Jul 25 at 13:00
1
Thanks for reminding me. Some progress added.
– Frank Li
Jul 25 at 13:09
Well, $E(X)=0$ is trivial because each $t_i$ has a distribution symmetric about $0$. (You have to verify the tails don't make for an $infty-infty$ situation, but that's easy in this case.)
– J.G.
Jul 25 at 13:15
Welcome to MSE. What have you tried? You will find you'll get a better reception if you show your work in progress, no matter how small. See math.meta.stackexchange.com/questions/9959/…
– Daniel Buck
Jul 25 at 13:00
Welcome to MSE. What have you tried? You will find you'll get a better reception if you show your work in progress, no matter how small. See math.meta.stackexchange.com/questions/9959/…
– Daniel Buck
Jul 25 at 13:00
1
1
Thanks for reminding me. Some progress added.
– Frank Li
Jul 25 at 13:09
Thanks for reminding me. Some progress added.
– Frank Li
Jul 25 at 13:09
Well, $E(X)=0$ is trivial because each $t_i$ has a distribution symmetric about $0$. (You have to verify the tails don't make for an $infty-infty$ situation, but that's easy in this case.)
– J.G.
Jul 25 at 13:15
Well, $E(X)=0$ is trivial because each $t_i$ has a distribution symmetric about $0$. (You have to verify the tails don't make for an $infty-infty$ situation, but that's easy in this case.)
– J.G.
Jul 25 at 13:15
add a comment |Â
1 Answer
1
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up vote
1
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I'll assume that you forgot to state that the $t_i$ are independent.
The expectation of any function that's antisymmetric in one of the $t_i$ is zero. Thus, $E(X)=0$ without further calculation.
For $Eleft(X^2right)$, the mixed pairs in the sum have zero expectation because they're odd in both factors, so what remains is
$$
Eleft(X^2right)=fracsum_it_i^2left(sum_it_i^2right)^2=frac1sum_it_i^2=r^-2;,
$$
where $r$ is the radial coordinate in $n$-dimensional spherical coordinates for the $t_i$. The density for $r$ is proportional to $r^n-1mathrm e^-frac12r^2$, so the expectation of $r^-2$ is
$$
Eleft(X^2right)=fracint_0^infty r^n-3mathrm e^-frac12r^2mathrm drint_0^infty r^n-1mathrm e^-frac12r^2mathrm dr=frac2^frac n2-2Gammaleft(frac n2-1right)2^frac n2-1Gammaleft(frac n2right)=frac1n-2;.
$$
answered Jul 25 at 13:32
joriki
164k10180328
Yes I forgot. $t_i$ are independent. And thank you very much.
– Frank Li
Jul 25 at 13:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I'll assume that you forgot to state that the $t_i$ are independent.
The expectation of any function that's antisymmetric in one of the $t_i$ is zero. Thus, $E(X)=0$ without further calculation.
For $Eleft(X^2right)$, the mixed pairs in the sum have zero expectation because they're odd in both factors, so what remains is
$$
Eleft(X^2right)=fracsum_it_i^2left(sum_it_i^2right)^2=frac1sum_it_i^2=r^-2;,
$$
where $r$ is the radial coordinate in $n$-dimensional spherical coordinates for the $t_i$. The density for $r$ is proportional to $r^n-1mathrm e^-frac12r^2$, so the expectation of $r^-2$ is
$$
Eleft(X^2right)=fracint_0^infty r^n-3mathrm e^-frac12r^2mathrm drint_0^infty r^n-1mathrm e^-frac12r^2mathrm dr=frac2^frac n2-2Gammaleft(frac n2-1right)2^frac n2-1Gammaleft(frac n2right)=frac1n-2;.
$$
answered Jul 25 at 13:32
joriki
164k10180328
Yes I forgot. $t_i$ are independent. And thank you very much.
– Frank Li
Jul 25 at 13:34
add a comment |Â
up vote
1
down vote
I'll assume that you forgot to state that the $t_i$ are independent.
The expectation of any function that's antisymmetric in one of the $t_i$ is zero. Thus, $E(X)=0$ without further calculation.
For $Eleft(X^2right)$, the mixed pairs in the sum have zero expectation because they're odd in both factors, so what remains is
$$
Eleft(X^2right)=fracsum_it_i^2left(sum_it_i^2right)^2=frac1sum_it_i^2=r^-2;,
$$
where $r$ is the radial coordinate in $n$-dimensional spherical coordinates for the $t_i$. The density for $r$ is proportional to $r^n-1mathrm e^-frac12r^2$, so the expectation of $r^-2$ is
$$
Eleft(X^2right)=fracint_0^infty r^n-3mathrm e^-frac12r^2mathrm drint_0^infty r^n-1mathrm e^-frac12r^2mathrm dr=frac2^frac n2-2Gammaleft(frac n2-1right)2^frac n2-1Gammaleft(frac n2right)=frac1n-2;.
$$
answered Jul 25 at 13:32
joriki
164k10180328
Yes I forgot. $t_i$ are independent. And thank you very much.
– Frank Li
Jul 25 at 13:34
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I'll assume that you forgot to state that the $t_i$ are independent.
The expectation of any function that's antisymmetric in one of the $t_i$ is zero. Thus, $E(X)=0$ without further calculation.
For $Eleft(X^2right)$, the mixed pairs in the sum have zero expectation because they're odd in both factors, so what remains is
$$
Eleft(X^2right)=fracsum_it_i^2left(sum_it_i^2right)^2=frac1sum_it_i^2=r^-2;,
$$
where $r$ is the radial coordinate in $n$-dimensional spherical coordinates for the $t_i$. The density for $r$ is proportional to $r^n-1mathrm e^-frac12r^2$, so the expectation of $r^-2$ is
$$
Eleft(X^2right)=fracint_0^infty r^n-3mathrm e^-frac12r^2mathrm drint_0^infty r^n-1mathrm e^-frac12r^2mathrm dr=frac2^frac n2-2Gammaleft(frac n2-1right)2^frac n2-1Gammaleft(frac n2right)=frac1n-2;.
$$
answered Jul 25 at 13:32
joriki
164k10180328
I'll assume that you forgot to state that the $t_i$ are independent.
The expectation of any function that's antisymmetric in one of the $t_i$ is zero. Thus, $E(X)=0$ without further calculation.
For $Eleft(X^2right)$, the mixed pairs in the sum have zero expectation because they're odd in both factors, so what remains is
$$
Eleft(X^2right)=fracsum_it_i^2left(sum_it_i^2right)^2=frac1sum_it_i^2=r^-2;,
$$
where $r$ is the radial coordinate in $n$-dimensional spherical coordinates for the $t_i$. The density for $r$ is proportional to $r^n-1mathrm e^-frac12r^2$, so the expectation of $r^-2$ is
$$
Eleft(X^2right)=fracint_0^infty r^n-3mathrm e^-frac12r^2mathrm drint_0^infty r^n-1mathrm e^-frac12r^2mathrm dr=frac2^frac n2-2Gammaleft(frac n2-1right)2^frac n2-1Gammaleft(frac n2right)=frac1n-2;.
$$
answered Jul 25 at 13:32
joriki
164k10180328
answered Jul 25 at 13:32
joriki
164k10180328
answered Jul 25 at 13:32
joriki
164k10180328
answered Jul 25 at 13:32
joriki
164k10180328
164k10180328
Yes I forgot. $t_i$ are independent. And thank you very much.
– Frank Li
Jul 25 at 13:34
add a comment |Â
Yes I forgot. $t_i$ are independent. And thank you very much.
– Frank Li
Jul 25 at 13:34
Yes I forgot. $t_i$ are independent. And thank you very much.
– Frank Li
Jul 25 at 13:34
Yes I forgot. $t_i$ are independent. And thank you very much.
– Frank Li
Jul 25 at 13:34
add a comment |Â
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Welcome to MSE. What have you tried? You will find you'll get a better reception if you show your work in progress, no matter how small. See math.meta.stackexchange.com/questions/9959/…
– Daniel Buck
Jul 25 at 13:00
1
Thanks for reminding me. Some progress added.
– Frank Li
Jul 25 at 13:09
Well, $E(X)=0$ is trivial because each $t_i$ has a distribution symmetric about $0$. (You have to verify the tails don't make for an $infty-infty$ situation, but that's easy in this case.)
– J.G.
Jul 25 at 13:15