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
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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.






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    $begingroup$

    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

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