How to prove that the following estimators are biased and consistent?
$begingroup$
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
asked Dec 31 '18 at 22:26
model_checkermodel_checker
4,25621931
$endgroup$
add a comment |
$begingroup$
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
asked Dec 31 '18 at 22:26
model_checkermodel_checker
4,25621931
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 31 '18 at 22:26
model_checkermodel_checker
4,25621931
$endgroup$
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
asked Dec 31 '18 at 22:26
model_checkermodel_checker
4,25621931
asked Dec 31 '18 at 22:26
model_checkermodel_checker
4,25621931
edited Jan 1 at 0:33
model_checker
asked Dec 31 '18 at 22:26
model_checkermodel_checker
4,25621931
asked Dec 31 '18 at 22:26
model_checkermodel_checker
4,25621931
asked Dec 31 '18 at 22:26
model_checkermodel_checker
4,25621931
4,25621931
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
add a comment |
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
add a comment |
1 Answer
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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.
answered Dec 31 '18 at 22:35
Foobaz JohnFoobaz John
22.8k41452
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add a comment |
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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.
answered Dec 31 '18 at 22:35
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 31 '18 at 22:35
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 31 '18 at 22:35
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
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.
answered Dec 31 '18 at 22:35
Foobaz JohnFoobaz John
22.8k41452
answered Dec 31 '18 at 22:35
Foobaz JohnFoobaz John
22.8k41452
answered Dec 31 '18 at 22:35
Foobaz JohnFoobaz John
22.8k41452
answered Dec 31 '18 at 22:35
Foobaz JohnFoobaz John
22.8k41452
22.8k41452
add a comment |
add a comment |
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$begingroup$
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