Mixing
Is this estimator biased [Geometric Distribution]?
Using the method of moments we get that
$$hat{theta} = frac{n}{sum_{i=1}^{n}X_i},$$
where $X_isim text{Geo}(p)$ for $i=1,2,3,cdots,n$ and pmf $P(X_i=k) = (1-p)^{k-1}p$ for $k=1,2,3,cdots.$
Then $bar{X}=sum_{i=1}^{n}X_isim text{NB}(n,p)$ where NB stands for negative binomial distribution with pmf
$$P(bar{X}=k)=binom{k+n-1}{k}(1-p)^{k-1}p^{n}$$
where $k=1,2,3,cdots.$ Thus
$$mathbb{E}left[frac{n}{bar{X}}right] = nsum_{k=1}^{infty}frac{1}{k}binom{k+n-1}{k}(1-p)^{k-1}p^{n}.$$
I am not sure how to compute this sum analytically. Maybe I can say that
$$E[n/bar{X}]>np^n $$
and so $hat{theta}$ is not a biased estimator. Furthermore, I am also interested in the consistency of the estimator, but I am not sure how to do this. Any hints in this regard will also be much appreciated.
probability-theory statistical-inference
asked Dec 23 '18 at 17:54
Hello_WorldHello_World
4,14021630
add a comment |
Using the method of moments we get that
$$hat{theta} = frac{n}{sum_{i=1}^{n}X_i},$$
where $X_isim text{Geo}(p)$ for $i=1,2,3,cdots,n$ and pmf $P(X_i=k) = (1-p)^{k-1}p$ for $k=1,2,3,cdots.$
Then $bar{X}=sum_{i=1}^{n}X_isim text{NB}(n,p)$ where NB stands for negative binomial distribution with pmf
$$P(bar{X}=k)=binom{k+n-1}{k}(1-p)^{k-1}p^{n}$$
where $k=1,2,3,cdots.$ Thus
$$mathbb{E}left[frac{n}{bar{X}}right] = nsum_{k=1}^{infty}frac{1}{k}binom{k+n-1}{k}(1-p)^{k-1}p^{n}.$$
I am not sure how to compute this sum analytically. Maybe I can say that
$$E[n/bar{X}]>np^n $$
and so $hat{theta}$ is not a biased estimator. Furthermore, I am also interested in the consistency of the estimator, but I am not sure how to do this. Any hints in this regard will also be much appreciated.
probability-theory statistical-inference
asked Dec 23 '18 at 17:54
Hello_WorldHello_World
4,14021630
add a comment |
Using the method of moments we get that
$$hat{theta} = frac{n}{sum_{i=1}^{n}X_i},$$
where $X_isim text{Geo}(p)$ for $i=1,2,3,cdots,n$ and pmf $P(X_i=k) = (1-p)^{k-1}p$ for $k=1,2,3,cdots.$
Then $bar{X}=sum_{i=1}^{n}X_isim text{NB}(n,p)$ where NB stands for negative binomial distribution with pmf
$$P(bar{X}=k)=binom{k+n-1}{k}(1-p)^{k-1}p^{n}$$
where $k=1,2,3,cdots.$ Thus
$$mathbb{E}left[frac{n}{bar{X}}right] = nsum_{k=1}^{infty}frac{1}{k}binom{k+n-1}{k}(1-p)^{k-1}p^{n}.$$
I am not sure how to compute this sum analytically. Maybe I can say that
$$E[n/bar{X}]>np^n $$
and so $hat{theta}$ is not a biased estimator. Furthermore, I am also interested in the consistency of the estimator, but I am not sure how to do this. Any hints in this regard will also be much appreciated.
probability-theory statistical-inference
asked Dec 23 '18 at 17:54
Hello_WorldHello_World
4,14021630
Using the method of moments we get that
$$hat{theta} = frac{n}{sum_{i=1}^{n}X_i},$$
where $X_isim text{Geo}(p)$ for $i=1,2,3,cdots,n$ and pmf $P(X_i=k) = (1-p)^{k-1}p$ for $k=1,2,3,cdots.$
Then $bar{X}=sum_{i=1}^{n}X_isim text{NB}(n,p)$ where NB stands for negative binomial distribution with pmf
$$P(bar{X}=k)=binom{k+n-1}{k}(1-p)^{k-1}p^{n}$$
where $k=1,2,3,cdots.$ Thus
$$mathbb{E}left[frac{n}{bar{X}}right] = nsum_{k=1}^{infty}frac{1}{k}binom{k+n-1}{k}(1-p)^{k-1}p^{n}.$$
I am not sure how to compute this sum analytically. Maybe I can say that
$$E[n/bar{X}]>np^n $$
and so $hat{theta}$ is not a biased estimator. Furthermore, I am also interested in the consistency of the estimator, but I am not sure how to do this. Any hints in this regard will also be much appreciated.
probability-theory statistical-inference
probability-theory statistical-inference
asked Dec 23 '18 at 17:54
Hello_WorldHello_World
4,14021630
asked Dec 23 '18 at 17:54
Hello_WorldHello_World
4,14021630
asked Dec 23 '18 at 17:54
Hello_WorldHello_World
4,14021630
asked Dec 23 '18 at 17:54
Hello_WorldHello_World
4,14021630
asked Dec 23 '18 at 17:54
Hello_WorldHello_World
4,14021630
4,14021630
add a comment |
add a comment |
2 Answers
2
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oldest
votes
Let $f$ be a strictly convex function and let $X$ be not constant a.s. Then by Jensen's inequality,
$$mathbb{E}f(X) > f(mathbb{E}X).$$
Taking $f(x) = 1/x$ ($x>0$) and $X=bar{X}$, you get:
$$mathbb{E}left(frac{1}{bar{X}}right)>frac{1}{mathbb{E}bar{X}},$$
so indeed:
$$mathbb{E}left(frac{n}{sum_i X_i}right)>p.$$
answered Dec 31 '18 at 21:30
LinAlgLinAlg
8,7911521
add a comment |
For consistency of the estimator, note that since the $X_i$ are i.i.d, we have by the SLLN, that
$$
hat {p}=frac{1}{sum_{i=1}^nX_i/n}stackrel{text{a.s}}{to}frac{1}{EX_1}=frac{1}{1/p}=p
$$
whence the estimator is (strongly) consistent.
answered Dec 23 '18 at 18:17
Foobaz JohnFoobaz John
21.5k41351
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $f$ be a strictly convex function and let $X$ be not constant a.s. Then by Jensen's inequality,
$$mathbb{E}f(X) > f(mathbb{E}X).$$
Taking $f(x) = 1/x$ ($x>0$) and $X=bar{X}$, you get:
$$mathbb{E}left(frac{1}{bar{X}}right)>frac{1}{mathbb{E}bar{X}},$$
so indeed:
$$mathbb{E}left(frac{n}{sum_i X_i}right)>p.$$
answered Dec 31 '18 at 21:30
LinAlgLinAlg
8,7911521
add a comment |
Let $f$ be a strictly convex function and let $X$ be not constant a.s. Then by Jensen's inequality,
$$mathbb{E}f(X) > f(mathbb{E}X).$$
Taking $f(x) = 1/x$ ($x>0$) and $X=bar{X}$, you get:
$$mathbb{E}left(frac{1}{bar{X}}right)>frac{1}{mathbb{E}bar{X}},$$
so indeed:
$$mathbb{E}left(frac{n}{sum_i X_i}right)>p.$$
answered Dec 31 '18 at 21:30
LinAlgLinAlg
8,7911521
add a comment |
Let $f$ be a strictly convex function and let $X$ be not constant a.s. Then by Jensen's inequality,
$$mathbb{E}f(X) > f(mathbb{E}X).$$
Taking $f(x) = 1/x$ ($x>0$) and $X=bar{X}$, you get:
$$mathbb{E}left(frac{1}{bar{X}}right)>frac{1}{mathbb{E}bar{X}},$$
so indeed:
$$mathbb{E}left(frac{n}{sum_i X_i}right)>p.$$
answered Dec 31 '18 at 21:30
LinAlgLinAlg
8,7911521
Let $f$ be a strictly convex function and let $X$ be not constant a.s. Then by Jensen's inequality,
$$mathbb{E}f(X) > f(mathbb{E}X).$$
Taking $f(x) = 1/x$ ($x>0$) and $X=bar{X}$, you get:
$$mathbb{E}left(frac{1}{bar{X}}right)>frac{1}{mathbb{E}bar{X}},$$
so indeed:
$$mathbb{E}left(frac{n}{sum_i X_i}right)>p.$$
answered Dec 31 '18 at 21:30
LinAlgLinAlg
8,7911521
answered Dec 31 '18 at 21:30
LinAlgLinAlg
8,7911521
answered Dec 31 '18 at 21:30
LinAlgLinAlg
8,7911521
answered Dec 31 '18 at 21:30
LinAlgLinAlg
8,7911521
8,7911521
add a comment |
add a comment |
For consistency of the estimator, note that since the $X_i$ are i.i.d, we have by the SLLN, that
$$
hat {p}=frac{1}{sum_{i=1}^nX_i/n}stackrel{text{a.s}}{to}frac{1}{EX_1}=frac{1}{1/p}=p
$$
whence the estimator is (strongly) consistent.
answered Dec 23 '18 at 18:17
Foobaz JohnFoobaz John
21.5k41351
add a comment |
For consistency of the estimator, note that since the $X_i$ are i.i.d, we have by the SLLN, that
$$
hat {p}=frac{1}{sum_{i=1}^nX_i/n}stackrel{text{a.s}}{to}frac{1}{EX_1}=frac{1}{1/p}=p
$$
whence the estimator is (strongly) consistent.
answered Dec 23 '18 at 18:17
Foobaz JohnFoobaz John
21.5k41351
add a comment |
For consistency of the estimator, note that since the $X_i$ are i.i.d, we have by the SLLN, that
$$
hat {p}=frac{1}{sum_{i=1}^nX_i/n}stackrel{text{a.s}}{to}frac{1}{EX_1}=frac{1}{1/p}=p
$$
whence the estimator is (strongly) consistent.
answered Dec 23 '18 at 18:17
Foobaz JohnFoobaz John
21.5k41351
For consistency of the estimator, note that since the $X_i$ are i.i.d, we have by the SLLN, that
$$
hat {p}=frac{1}{sum_{i=1}^nX_i/n}stackrel{text{a.s}}{to}frac{1}{EX_1}=frac{1}{1/p}=p
$$
whence the estimator is (strongly) consistent.
answered Dec 23 '18 at 18:17
Foobaz JohnFoobaz John
21.5k41351
answered Dec 23 '18 at 18:17
Foobaz JohnFoobaz John
21.5k41351
answered Dec 23 '18 at 18:17
Foobaz JohnFoobaz John
21.5k41351
answered Dec 23 '18 at 18:17
Foobaz JohnFoobaz John
21.5k41351
21.5k41351
add a comment |
add a comment |
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