How to prove that the following estimators are biased and consistent?












-1












$begingroup$


Given a random variable $X$ following a geometric distribution with parameter $p.$ Then one estimator that can be obtained by considering the second moment $E[X^{2}]=frac{2-p}{p^2}$, which is
$$hat{p}_1=frac{-1+sqrt{1+frac{8}{n}sum_{i=1}^{n}X_i^2}}{frac{2}{n}sum_{i=1}^{n}X_i^2}.$$
Another family of estimators can be obtained by observing that $E[mathbf{1}_{[k,infty)}X_1] = P[X_1>k]=(1-p)^{k}$ and so
$$hat{p}_2 = 1- frac{logleft(frac{1}{n}sum_{i=1}^{n}mathbb{1}_{[k,+infty)}(X_i)right)}{k}.$$
I want to determine whether $hat{p}_1$ and $hat{p}_2$ are biased and consistent, but this seems difficult since I am not able to figure the distribution of these estimators. Perhaps I have to use inequalities, but I am not sure how to proceed. Any hints will be much appreciated.



Edit:
Let $Y=sum_{i=1}^{n}X_i^2/n$ then
$$E[hat{p}_1]=E[f(Y)]$$
where
$$f(y) = frac{-1+sqrt{1+8y}}{2y}$$
then since $f''(y)>0$ using the Jensen Inequality we have that:
$$E[f(Y)]>f(E[Y])implies E[hat{p}_1]>p.$$
For the second estimator, we try Jensen with $Y=frac{1}{n}sum_{i=1}^{n}mathbb{1}_{[k,+infty)}(X_i)$ and $$f(y)=1-frac{log(y)}{k}.$$Then $$f''(y)=-frac{1}{yk}<0$$ and so we have that
$$E[hat{p}_2]=E[f(Y)]<f(E[Y])=p.$$
I think this argument shows that $hat{p}_1$ and $hat{p}_2$ are biased. I am not very sure as to how the law of large numbers will work with the random variable $Y=sum_{i=1}^{n}X_i^2/n$ since it is not exactly an average. Perhaps someone could explain this to me.






probability-theory statistics statistical-inference probability-limit-theorems







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Both are consistent by the law of large numbers. The second one is biased by Jensen inequality. Probably the first one as well, also by convexity...
    $endgroup$
    – Did
    Dec 31 '18 at 22:33












  • $begingroup$
    Hmmm... All these arguments were already mentioned to you à propos your previous recent question. Why are you not trying to apply them in the present context?
    $endgroup$
    – Did
    Dec 31 '18 at 22:39










  • $begingroup$
    Hey, I made some edits. I hope my question makes more sense now.
    $endgroup$
    – model_checker
    Jan 1 at 0:33










  • $begingroup$
    1. In the log case, your $f''$ is wrong. 2. The random variable $Y=frac1nsum X_k^2$ is exactly an average hence the LLN applies perfectly.
    $endgroup$
    – Did
    Jan 1 at 11:35


















-1












$begingroup$


Given a random variable $X$ following a geometric distribution with parameter $p.$ Then one estimator that can be obtained by considering the second moment $E[X^{2}]=frac{2-p}{p^2}$, which is
$$hat{p}_1=frac{-1+sqrt{1+frac{8}{n}sum_{i=1}^{n}X_i^2}}{frac{2}{n}sum_{i=1}^{n}X_i^2}.$$
Another family of estimators can be obtained by observing that $E[mathbf{1}_{[k,infty)}X_1] = P[X_1>k]=(1-p)^{k}$ and so
$$hat{p}_2 = 1- frac{logleft(frac{1}{n}sum_{i=1}^{n}mathbb{1}_{[k,+infty)}(X_i)right)}{k}.$$
I want to determine whether $hat{p}_1$ and $hat{p}_2$ are biased and consistent, but this seems difficult since I am not able to figure the distribution of these estimators. Perhaps I have to use inequalities, but I am not sure how to proceed. Any hints will be much appreciated.



Edit:
Let $Y=sum_{i=1}^{n}X_i^2/n$ then
$$E[hat{p}_1]=E[f(Y)]$$
where
$$f(y) = frac{-1+sqrt{1+8y}}{2y}$$
then since $f''(y)>0$ using the Jensen Inequality we have that:
$$E[f(Y)]>f(E[Y])implies E[hat{p}_1]>p.$$
For the second estimator, we try Jensen with $Y=frac{1}{n}sum_{i=1}^{n}mathbb{1}_{[k,+infty)}(X_i)$ and $$f(y)=1-frac{log(y)}{k}.$$Then $$f''(y)=-frac{1}{yk}<0$$ and so we have that
$$E[hat{p}_2]=E[f(Y)]<f(E[Y])=p.$$
I think this argument shows that $hat{p}_1$ and $hat{p}_2$ are biased. I am not very sure as to how the law of large numbers will work with the random variable $Y=sum_{i=1}^{n}X_i^2/n$ since it is not exactly an average. Perhaps someone could explain this to me.






probability-theory statistics statistical-inference probability-limit-theorems







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Both are consistent by the law of large numbers. The second one is biased by Jensen inequality. Probably the first one as well, also by convexity...
    $endgroup$
    – Did
    Dec 31 '18 at 22:33












  • $begingroup$
    Hmmm... All these arguments were already mentioned to you à propos your previous recent question. Why are you not trying to apply them in the present context?
    $endgroup$
    – Did
    Dec 31 '18 at 22:39










  • $begingroup$
    Hey, I made some edits. I hope my question makes more sense now.
    $endgroup$
    – model_checker
    Jan 1 at 0:33










  • $begingroup$
    1. In the log case, your $f''$ is wrong. 2. The random variable $Y=frac1nsum X_k^2$ is exactly an average hence the LLN applies perfectly.
    $endgroup$
    – Did
    Jan 1 at 11:35
















-1












-1








-1





$begingroup$


Given a random variable $X$ following a geometric distribution with parameter $p.$ Then one estimator that can be obtained by considering the second moment $E[X^{2}]=frac{2-p}{p^2}$, which is
$$hat{p}_1=frac{-1+sqrt{1+frac{8}{n}sum_{i=1}^{n}X_i^2}}{frac{2}{n}sum_{i=1}^{n}X_i^2}.$$
Another family of estimators can be obtained by observing that $E[mathbf{1}_{[k,infty)}X_1] = P[X_1>k]=(1-p)^{k}$ and so
$$hat{p}_2 = 1- frac{logleft(frac{1}{n}sum_{i=1}^{n}mathbb{1}_{[k,+infty)}(X_i)right)}{k}.$$
I want to determine whether $hat{p}_1$ and $hat{p}_2$ are biased and consistent, but this seems difficult since I am not able to figure the distribution of these estimators. Perhaps I have to use inequalities, but I am not sure how to proceed. Any hints will be much appreciated.



Edit:
Let $Y=sum_{i=1}^{n}X_i^2/n$ then
$$E[hat{p}_1]=E[f(Y)]$$
where
$$f(y) = frac{-1+sqrt{1+8y}}{2y}$$
then since $f''(y)>0$ using the Jensen Inequality we have that:
$$E[f(Y)]>f(E[Y])implies E[hat{p}_1]>p.$$
For the second estimator, we try Jensen with $Y=frac{1}{n}sum_{i=1}^{n}mathbb{1}_{[k,+infty)}(X_i)$ and $$f(y)=1-frac{log(y)}{k}.$$Then $$f''(y)=-frac{1}{yk}<0$$ and so we have that
$$E[hat{p}_2]=E[f(Y)]<f(E[Y])=p.$$
I think this argument shows that $hat{p}_1$ and $hat{p}_2$ are biased. I am not very sure as to how the law of large numbers will work with the random variable $Y=sum_{i=1}^{n}X_i^2/n$ since it is not exactly an average. Perhaps someone could explain this to me.






probability-theory statistics statistical-inference probability-limit-theorems







share|cite|improve this question











$endgroup$




Given a random variable $X$ following a geometric distribution with parameter $p.$ Then one estimator that can be obtained by considering the second moment $E[X^{2}]=frac{2-p}{p^2}$, which is
$$hat{p}_1=frac{-1+sqrt{1+frac{8}{n}sum_{i=1}^{n}X_i^2}}{frac{2}{n}sum_{i=1}^{n}X_i^2}.$$
Another family of estimators can be obtained by observing that $E[mathbf{1}_{[k,infty)}X_1] = P[X_1>k]=(1-p)^{k}$ and so
$$hat{p}_2 = 1- frac{logleft(frac{1}{n}sum_{i=1}^{n}mathbb{1}_{[k,+infty)}(X_i)right)}{k}.$$
I want to determine whether $hat{p}_1$ and $hat{p}_2$ are biased and consistent, but this seems difficult since I am not able to figure the distribution of these estimators. Perhaps I have to use inequalities, but I am not sure how to proceed. Any hints will be much appreciated.



Edit:
Let $Y=sum_{i=1}^{n}X_i^2/n$ then
$$E[hat{p}_1]=E[f(Y)]$$
where
$$f(y) = frac{-1+sqrt{1+8y}}{2y}$$
then since $f''(y)>0$ using the Jensen Inequality we have that:
$$E[f(Y)]>f(E[Y])implies E[hat{p}_1]>p.$$
For the second estimator, we try Jensen with $Y=frac{1}{n}sum_{i=1}^{n}mathbb{1}_{[k,+infty)}(X_i)$ and $$f(y)=1-frac{log(y)}{k}.$$Then $$f''(y)=-frac{1}{yk}<0$$ and so we have that
$$E[hat{p}_2]=E[f(Y)]<f(E[Y])=p.$$
I think this argument shows that $hat{p}_1$ and $hat{p}_2$ are biased. I am not very sure as to how the law of large numbers will work with the random variable $Y=sum_{i=1}^{n}X_i^2/n$ since it is not exactly an average. Perhaps someone could explain this to me.






probability-theory statistics statistical-inference probability-limit-theorems




probability-theory statistics statistical-inference probability-limit-theorems






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 0:33







model_checker

















asked Dec 31 '18 at 22:26









model_checkermodel_checker

4,45021931




4,45021931








  • 1




    $begingroup$
    Both are consistent by the law of large numbers. The second one is biased by Jensen inequality. Probably the first one as well, also by convexity...
    $endgroup$
    – Did
    Dec 31 '18 at 22:33












  • $begingroup$
    Hmmm... All these arguments were already mentioned to you à propos your previous recent question. Why are you not trying to apply them in the present context?
    $endgroup$
    – Did
    Dec 31 '18 at 22:39










  • $begingroup$
    Hey, I made some edits. I hope my question makes more sense now.
    $endgroup$
    – model_checker
    Jan 1 at 0:33










  • $begingroup$
    1. In the log case, your $f''$ is wrong. 2. The random variable $Y=frac1nsum X_k^2$ is exactly an average hence the LLN applies perfectly.
    $endgroup$
    – Did
    Jan 1 at 11:35
















  • 1




    $begingroup$
    Both are consistent by the law of large numbers. The second one is biased by Jensen inequality. Probably the first one as well, also by convexity...
    $endgroup$
    – Did
    Dec 31 '18 at 22:33












  • $begingroup$
    Hmmm... All these arguments were already mentioned to you à propos your previous recent question. Why are you not trying to apply them in the present context?
    $endgroup$
    – Did
    Dec 31 '18 at 22:39










  • $begingroup$
    Hey, I made some edits. I hope my question makes more sense now.
    $endgroup$
    – model_checker
    Jan 1 at 0:33










  • $begingroup$
    1. In the log case, your $f''$ is wrong. 2. The random variable $Y=frac1nsum X_k^2$ is exactly an average hence the LLN applies perfectly.
    $endgroup$
    – Did
    Jan 1 at 11:35










1




1




$begingroup$
Both are consistent by the law of large numbers. The second one is biased by Jensen inequality. Probably the first one as well, also by convexity...
$endgroup$
– Did
Dec 31 '18 at 22:33






$begingroup$
Both are consistent by the law of large numbers. The second one is biased by Jensen inequality. Probably the first one as well, also by convexity...
$endgroup$
– Did
Dec 31 '18 at 22:33














$begingroup$
Hmmm... All these arguments were already mentioned to you à propos your previous recent question. Why are you not trying to apply them in the present context?
$endgroup$
– Did
Dec 31 '18 at 22:39




$begingroup$
Hmmm... All these arguments were already mentioned to you à propos your previous recent question. Why are you not trying to apply them in the present context?
$endgroup$
– Did
Dec 31 '18 at 22:39












$begingroup$
Hey, I made some edits. I hope my question makes more sense now.
$endgroup$
– model_checker
Jan 1 at 0:33




$begingroup$
Hey, I made some edits. I hope my question makes more sense now.
$endgroup$
– model_checker
Jan 1 at 0:33












$begingroup$
1. In the log case, your $f''$ is wrong. 2. The random variable $Y=frac1nsum X_k^2$ is exactly an average hence the LLN applies perfectly.
$endgroup$
– Did
Jan 1 at 11:35






$begingroup$
1. In the log case, your $f''$ is wrong. 2. The random variable $Y=frac1nsum X_k^2$ is exactly an average hence the LLN applies perfectly.
$endgroup$
– Did
Jan 1 at 11:35












1 Answer
1






active

oldest

votes


















1












$begingroup$

For the first estimator we can use the strong law of large numbers to deduce that since $(X_i)_{igeq 1}$ are i.i.d, it follows that $sum_1^n X_i/nstackrel{text{a.s}}{to} EX=1/p$ whence
$$
hat{p}=frac{1}{sum_1^n X_i/n}stackrel{text{a.s}}{to} frac{1}{1/p}=p
$$
as $nto infty$. So the first estimator is strongly consistent.






share|cite|improve this answer









$endgroup$














    Your Answer








    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%2f3058088%2fhow-to-prove-that-the-following-estimators-are-biased-and-consistent%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












    $begingroup$

    For the first estimator we can use the strong law of large numbers to deduce that since $(X_i)_{igeq 1}$ are i.i.d, it follows that $sum_1^n X_i/nstackrel{text{a.s}}{to} EX=1/p$ whence
    $$
    hat{p}=frac{1}{sum_1^n X_i/n}stackrel{text{a.s}}{to} frac{1}{1/p}=p
    $$
    as $nto infty$. So the first estimator is strongly consistent.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      For the first estimator we can use the strong law of large numbers to deduce that since $(X_i)_{igeq 1}$ are i.i.d, it follows that $sum_1^n X_i/nstackrel{text{a.s}}{to} EX=1/p$ whence
      $$
      hat{p}=frac{1}{sum_1^n X_i/n}stackrel{text{a.s}}{to} frac{1}{1/p}=p
      $$
      as $nto infty$. So the first estimator is strongly consistent.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        For the first estimator we can use the strong law of large numbers to deduce that since $(X_i)_{igeq 1}$ are i.i.d, it follows that $sum_1^n X_i/nstackrel{text{a.s}}{to} EX=1/p$ whence
        $$
        hat{p}=frac{1}{sum_1^n X_i/n}stackrel{text{a.s}}{to} frac{1}{1/p}=p
        $$
        as $nto infty$. So the first estimator is strongly consistent.






        share|cite|improve this answer









        $endgroup$



        For the first estimator we can use the strong law of large numbers to deduce that since $(X_i)_{igeq 1}$ are i.i.d, it follows that $sum_1^n X_i/nstackrel{text{a.s}}{to} EX=1/p$ whence
        $$
        hat{p}=frac{1}{sum_1^n X_i/n}stackrel{text{a.s}}{to} frac{1}{1/p}=p
        $$
        as $nto infty$. So the first estimator is strongly consistent.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 22:35









        Foobaz JohnFoobaz John

        23k41552




        23k41552






























            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%2f3058088%2fhow-to-prove-that-the-following-estimators-are-biased-and-consistent%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

            Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

            Image
            2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
            Read more

            ComboBox Display Member on multiple fields

            Image
            2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
            Read more

            Is it possible to collect Nectar points via Trainline?

            Image
            2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
            Read more