What is $E|langle Arangle|$?












10












$begingroup$


Suppose $A$ is a random subset of $S_n$, such that each element of $S_n$ independently belongs to $A$ with probability p. What is the expectation of $|langle Arangle|$?



The case with $p = 1$ ($E|langle Arangle| = n!$) is quite obvious, however, I do not know, how to deal with the situation when $0 < p < 1$.



Any help will be appreciated.






probability abstract-algebra group-theory permutations expectation







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $p=0$ then $E=1$, as the identity is always an element of any subgroup.
    $endgroup$
    – Berci
    Jun 9 '18 at 15:33






  • 3




    $begingroup$
    The expected size of $A$ is $n!/p$. If this is at least $2$, then the probability that $langle A rangle = A_n$ or $S_n$ approaches $1$ as $n to infty$.
    $endgroup$
    – Derek Holt
    Jun 9 '18 at 17:05










  • $begingroup$
    Sorry I mean the expected size of $A$ is $n!p$.
    $endgroup$
    – Derek Holt
    Jun 9 '18 at 17:43






  • 1




    $begingroup$
    What's $langle Arangle$ and $|langle Arangle|$, btw?
    $endgroup$
    – Saad
    Jul 16 '18 at 12:13






  • 1




    $begingroup$
    @Alex $langle Arangle$ is the subgroup generated by the set $A$, and $|langle Arangle|$ is the order of this subgroup.
    $endgroup$
    – user1729
    Jul 16 '18 at 12:16
















10












$begingroup$


Suppose $A$ is a random subset of $S_n$, such that each element of $S_n$ independently belongs to $A$ with probability p. What is the expectation of $|langle Arangle|$?



The case with $p = 1$ ($E|langle Arangle| = n!$) is quite obvious, however, I do not know, how to deal with the situation when $0 < p < 1$.



Any help will be appreciated.






probability abstract-algebra group-theory permutations expectation







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $p=0$ then $E=1$, as the identity is always an element of any subgroup.
    $endgroup$
    – Berci
    Jun 9 '18 at 15:33






  • 3




    $begingroup$
    The expected size of $A$ is $n!/p$. If this is at least $2$, then the probability that $langle A rangle = A_n$ or $S_n$ approaches $1$ as $n to infty$.
    $endgroup$
    – Derek Holt
    Jun 9 '18 at 17:05










  • $begingroup$
    Sorry I mean the expected size of $A$ is $n!p$.
    $endgroup$
    – Derek Holt
    Jun 9 '18 at 17:43






  • 1




    $begingroup$
    What's $langle Arangle$ and $|langle Arangle|$, btw?
    $endgroup$
    – Saad
    Jul 16 '18 at 12:13






  • 1




    $begingroup$
    @Alex $langle Arangle$ is the subgroup generated by the set $A$, and $|langle Arangle|$ is the order of this subgroup.
    $endgroup$
    – user1729
    Jul 16 '18 at 12:16














10












10








10


4



$begingroup$


Suppose $A$ is a random subset of $S_n$, such that each element of $S_n$ independently belongs to $A$ with probability p. What is the expectation of $|langle Arangle|$?



The case with $p = 1$ ($E|langle Arangle| = n!$) is quite obvious, however, I do not know, how to deal with the situation when $0 < p < 1$.



Any help will be appreciated.






probability abstract-algebra group-theory permutations expectation







share|cite|improve this question











$endgroup$




Suppose $A$ is a random subset of $S_n$, such that each element of $S_n$ independently belongs to $A$ with probability p. What is the expectation of $|langle Arangle|$?



The case with $p = 1$ ($E|langle Arangle| = n!$) is quite obvious, however, I do not know, how to deal with the situation when $0 < p < 1$.



Any help will be appreciated.






probability abstract-algebra group-theory permutations expectation




probability abstract-algebra group-theory permutations expectation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 16 '18 at 11:23







Yanior Weg

















asked Jun 9 '18 at 15:15









Yanior WegYanior Weg

2,54511246




2,54511246












  • $begingroup$
    If $p=0$ then $E=1$, as the identity is always an element of any subgroup.
    $endgroup$
    – Berci
    Jun 9 '18 at 15:33






  • 3




    $begingroup$
    The expected size of $A$ is $n!/p$. If this is at least $2$, then the probability that $langle A rangle = A_n$ or $S_n$ approaches $1$ as $n to infty$.
    $endgroup$
    – Derek Holt
    Jun 9 '18 at 17:05










  • $begingroup$
    Sorry I mean the expected size of $A$ is $n!p$.
    $endgroup$
    – Derek Holt
    Jun 9 '18 at 17:43






  • 1




    $begingroup$
    What's $langle Arangle$ and $|langle Arangle|$, btw?
    $endgroup$
    – Saad
    Jul 16 '18 at 12:13






  • 1




    $begingroup$
    @Alex $langle Arangle$ is the subgroup generated by the set $A$, and $|langle Arangle|$ is the order of this subgroup.
    $endgroup$
    – user1729
    Jul 16 '18 at 12:16


















  • $begingroup$
    If $p=0$ then $E=1$, as the identity is always an element of any subgroup.
    $endgroup$
    – Berci
    Jun 9 '18 at 15:33






  • 3




    $begingroup$
    The expected size of $A$ is $n!/p$. If this is at least $2$, then the probability that $langle A rangle = A_n$ or $S_n$ approaches $1$ as $n to infty$.
    $endgroup$
    – Derek Holt
    Jun 9 '18 at 17:05










  • $begingroup$
    Sorry I mean the expected size of $A$ is $n!p$.
    $endgroup$
    – Derek Holt
    Jun 9 '18 at 17:43






  • 1




    $begingroup$
    What's $langle Arangle$ and $|langle Arangle|$, btw?
    $endgroup$
    – Saad
    Jul 16 '18 at 12:13






  • 1




    $begingroup$
    @Alex $langle Arangle$ is the subgroup generated by the set $A$, and $|langle Arangle|$ is the order of this subgroup.
    $endgroup$
    – user1729
    Jul 16 '18 at 12:16
















$begingroup$
If $p=0$ then $E=1$, as the identity is always an element of any subgroup.
$endgroup$
– Berci
Jun 9 '18 at 15:33




$begingroup$
If $p=0$ then $E=1$, as the identity is always an element of any subgroup.
$endgroup$
– Berci
Jun 9 '18 at 15:33




3




3




$begingroup$
The expected size of $A$ is $n!/p$. If this is at least $2$, then the probability that $langle A rangle = A_n$ or $S_n$ approaches $1$ as $n to infty$.
$endgroup$
– Derek Holt
Jun 9 '18 at 17:05




$begingroup$
The expected size of $A$ is $n!/p$. If this is at least $2$, then the probability that $langle A rangle = A_n$ or $S_n$ approaches $1$ as $n to infty$.
$endgroup$
– Derek Holt
Jun 9 '18 at 17:05












$begingroup$
Sorry I mean the expected size of $A$ is $n!p$.
$endgroup$
– Derek Holt
Jun 9 '18 at 17:43




$begingroup$
Sorry I mean the expected size of $A$ is $n!p$.
$endgroup$
– Derek Holt
Jun 9 '18 at 17:43




1




1




$begingroup$
What's $langle Arangle$ and $|langle Arangle|$, btw?
$endgroup$
– Saad
Jul 16 '18 at 12:13




$begingroup$
What's $langle Arangle$ and $|langle Arangle|$, btw?
$endgroup$
– Saad
Jul 16 '18 at 12:13




1




1




$begingroup$
@Alex $langle Arangle$ is the subgroup generated by the set $A$, and $|langle Arangle|$ is the order of this subgroup.
$endgroup$
– user1729
Jul 16 '18 at 12:16




$begingroup$
@Alex $langle Arangle$ is the subgroup generated by the set $A$, and $|langle Arangle|$ is the order of this subgroup.
$endgroup$
– user1729
Jul 16 '18 at 12:16










1 Answer
1






active

oldest

votes


















1












$begingroup$

Suppose $A$ is a random subset of a finite group $G$, such that each element of $A$ independently belongs to $G$ with probability $p$ (in your case $G = S_n$). One can see that $forall H leq G (P(A subset H) = (1 - p)^{|G| - |H|}$. It is also true, that $P(A subset H) = Sigma_{K leq H} P(langle A rangle = K)$. Thus, $P(langle A rangle = H) = Sigma_{K leq H} mu(H, K)P(A subset K) = Sigma_{K leq H} mu(H, K)(1 - p)^{|G| - |K|}$, where $mu$ is the Moebius function for subgroup lattice of $G$. Thus we have the formula:
$$E|langle A rangle| = Sigma_{H leq G} Sigma_{K leq H} |H|mu(H, K)(1 - p)^{|G| - |K|}$$



That is one of the possible forms of answer, despite it is quite hard to anyhow simplify it, as one must know the structure of subgroup lattice of $G$ to calculate the Moebius function (In your case it is especially hard as the subgroup lattice of $S_n$ is not well described).



However, one can prove an interesting fact about the asymptotic behavior of $E|langle Arangle|$ (that also works not only for your case, but for my generalization too). That is $lim_{|G| to infty} frac{E|langle Arangle|}{|G|} = 1$. That is due to the fact, that $|G| geq E|langle Arangle| geq |G| - |G|^3 (1 - p)^{frac{|G|}{2}}$ (that follows from the inequalities $|H| < frac{|G|}{2}$ for any proper subgroup of $G$, $|{H leq G}| leq |G|$ and $|mu(H, K)| leq 1$). Now that means, that $1 geq frac{E|langle Arangle|}{|G|} geq 1 - |G|^2(1 - p)^{frac{|G|}{2}}$. This, combined with the fact, that $lim_{n to infty} n^2(1 - p)^{frac{n}{2}} = 0$, gives us the statement we need.






share|cite|improve this answer











$endgroup$













    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',
    autoActivateHeartbeat: false,
    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%2f2813655%2fwhat-is-e-langle-a-rangle%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









    1












    $begingroup$

    Suppose $A$ is a random subset of a finite group $G$, such that each element of $A$ independently belongs to $G$ with probability $p$ (in your case $G = S_n$). One can see that $forall H leq G (P(A subset H) = (1 - p)^{|G| - |H|}$. It is also true, that $P(A subset H) = Sigma_{K leq H} P(langle A rangle = K)$. Thus, $P(langle A rangle = H) = Sigma_{K leq H} mu(H, K)P(A subset K) = Sigma_{K leq H} mu(H, K)(1 - p)^{|G| - |K|}$, where $mu$ is the Moebius function for subgroup lattice of $G$. Thus we have the formula:
    $$E|langle A rangle| = Sigma_{H leq G} Sigma_{K leq H} |H|mu(H, K)(1 - p)^{|G| - |K|}$$



    That is one of the possible forms of answer, despite it is quite hard to anyhow simplify it, as one must know the structure of subgroup lattice of $G$ to calculate the Moebius function (In your case it is especially hard as the subgroup lattice of $S_n$ is not well described).



    However, one can prove an interesting fact about the asymptotic behavior of $E|langle Arangle|$ (that also works not only for your case, but for my generalization too). That is $lim_{|G| to infty} frac{E|langle Arangle|}{|G|} = 1$. That is due to the fact, that $|G| geq E|langle Arangle| geq |G| - |G|^3 (1 - p)^{frac{|G|}{2}}$ (that follows from the inequalities $|H| < frac{|G|}{2}$ for any proper subgroup of $G$, $|{H leq G}| leq |G|$ and $|mu(H, K)| leq 1$). Now that means, that $1 geq frac{E|langle Arangle|}{|G|} geq 1 - |G|^2(1 - p)^{frac{|G|}{2}}$. This, combined with the fact, that $lim_{n to infty} n^2(1 - p)^{frac{n}{2}} = 0$, gives us the statement we need.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Suppose $A$ is a random subset of a finite group $G$, such that each element of $A$ independently belongs to $G$ with probability $p$ (in your case $G = S_n$). One can see that $forall H leq G (P(A subset H) = (1 - p)^{|G| - |H|}$. It is also true, that $P(A subset H) = Sigma_{K leq H} P(langle A rangle = K)$. Thus, $P(langle A rangle = H) = Sigma_{K leq H} mu(H, K)P(A subset K) = Sigma_{K leq H} mu(H, K)(1 - p)^{|G| - |K|}$, where $mu$ is the Moebius function for subgroup lattice of $G$. Thus we have the formula:
      $$E|langle A rangle| = Sigma_{H leq G} Sigma_{K leq H} |H|mu(H, K)(1 - p)^{|G| - |K|}$$



      That is one of the possible forms of answer, despite it is quite hard to anyhow simplify it, as one must know the structure of subgroup lattice of $G$ to calculate the Moebius function (In your case it is especially hard as the subgroup lattice of $S_n$ is not well described).



      However, one can prove an interesting fact about the asymptotic behavior of $E|langle Arangle|$ (that also works not only for your case, but for my generalization too). That is $lim_{|G| to infty} frac{E|langle Arangle|}{|G|} = 1$. That is due to the fact, that $|G| geq E|langle Arangle| geq |G| - |G|^3 (1 - p)^{frac{|G|}{2}}$ (that follows from the inequalities $|H| < frac{|G|}{2}$ for any proper subgroup of $G$, $|{H leq G}| leq |G|$ and $|mu(H, K)| leq 1$). Now that means, that $1 geq frac{E|langle Arangle|}{|G|} geq 1 - |G|^2(1 - p)^{frac{|G|}{2}}$. This, combined with the fact, that $lim_{n to infty} n^2(1 - p)^{frac{n}{2}} = 0$, gives us the statement we need.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Suppose $A$ is a random subset of a finite group $G$, such that each element of $A$ independently belongs to $G$ with probability $p$ (in your case $G = S_n$). One can see that $forall H leq G (P(A subset H) = (1 - p)^{|G| - |H|}$. It is also true, that $P(A subset H) = Sigma_{K leq H} P(langle A rangle = K)$. Thus, $P(langle A rangle = H) = Sigma_{K leq H} mu(H, K)P(A subset K) = Sigma_{K leq H} mu(H, K)(1 - p)^{|G| - |K|}$, where $mu$ is the Moebius function for subgroup lattice of $G$. Thus we have the formula:
        $$E|langle A rangle| = Sigma_{H leq G} Sigma_{K leq H} |H|mu(H, K)(1 - p)^{|G| - |K|}$$



        That is one of the possible forms of answer, despite it is quite hard to anyhow simplify it, as one must know the structure of subgroup lattice of $G$ to calculate the Moebius function (In your case it is especially hard as the subgroup lattice of $S_n$ is not well described).



        However, one can prove an interesting fact about the asymptotic behavior of $E|langle Arangle|$ (that also works not only for your case, but for my generalization too). That is $lim_{|G| to infty} frac{E|langle Arangle|}{|G|} = 1$. That is due to the fact, that $|G| geq E|langle Arangle| geq |G| - |G|^3 (1 - p)^{frac{|G|}{2}}$ (that follows from the inequalities $|H| < frac{|G|}{2}$ for any proper subgroup of $G$, $|{H leq G}| leq |G|$ and $|mu(H, K)| leq 1$). Now that means, that $1 geq frac{E|langle Arangle|}{|G|} geq 1 - |G|^2(1 - p)^{frac{|G|}{2}}$. This, combined with the fact, that $lim_{n to infty} n^2(1 - p)^{frac{n}{2}} = 0$, gives us the statement we need.






        share|cite|improve this answer











        $endgroup$



        Suppose $A$ is a random subset of a finite group $G$, such that each element of $A$ independently belongs to $G$ with probability $p$ (in your case $G = S_n$). One can see that $forall H leq G (P(A subset H) = (1 - p)^{|G| - |H|}$. It is also true, that $P(A subset H) = Sigma_{K leq H} P(langle A rangle = K)$. Thus, $P(langle A rangle = H) = Sigma_{K leq H} mu(H, K)P(A subset K) = Sigma_{K leq H} mu(H, K)(1 - p)^{|G| - |K|}$, where $mu$ is the Moebius function for subgroup lattice of $G$. Thus we have the formula:
        $$E|langle A rangle| = Sigma_{H leq G} Sigma_{K leq H} |H|mu(H, K)(1 - p)^{|G| - |K|}$$



        That is one of the possible forms of answer, despite it is quite hard to anyhow simplify it, as one must know the structure of subgroup lattice of $G$ to calculate the Moebius function (In your case it is especially hard as the subgroup lattice of $S_n$ is not well described).



        However, one can prove an interesting fact about the asymptotic behavior of $E|langle Arangle|$ (that also works not only for your case, but for my generalization too). That is $lim_{|G| to infty} frac{E|langle Arangle|}{|G|} = 1$. That is due to the fact, that $|G| geq E|langle Arangle| geq |G| - |G|^3 (1 - p)^{frac{|G|}{2}}$ (that follows from the inequalities $|H| < frac{|G|}{2}$ for any proper subgroup of $G$, $|{H leq G}| leq |G|$ and $|mu(H, K)| leq 1$). Now that means, that $1 geq frac{E|langle Arangle|}{|G|} geq 1 - |G|^2(1 - p)^{frac{|G|}{2}}$. This, combined with the fact, that $lim_{n to infty} n^2(1 - p)^{frac{n}{2}} = 0$, gives us the statement we need.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 25 at 22:34

























        answered Jan 25 at 21:05









        Yanior WegYanior Weg

        2,54511246




        2,54511246






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2813655%2fwhat-is-e-langle-a-rangle%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

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
            Read more

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
            Read more