Gaussian expectation for a rational function











up vote
1
down vote

favorite
1












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







share|cite|improve this question
























  • 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

















up vote
1
down vote

favorite
1












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







share|cite|improve this question
























  • 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















up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 20 at 2:47









6005

35.6k751125




35.6k751125










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




















  • 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












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






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2390834%2fgaussian-expectation-for-a-rational-function%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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






    share|cite|improve this answer

























      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






      share|cite|improve this answer























        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






        share|cite|improve this answer












        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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 8:11









        user617446

        261




        261






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2390834%2fgaussian-expectation-for-a-rational-function%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more