Mixing
Is estimator $dfrac{bar X}{1+bar X}$ of $theta$ is consistent?
$begingroup$
Let $X_1,X_2,X_3.....X_n$ be a random sample from a population X
having the probability density function
$$ f(x;theta) = begin{cases} theta x^{theta -1} & text{if $0 le
xle$ 1} \0 &text{otherwise} end{cases}$$
Is the estimator $hattheta = $ $dfrac{bar X}{1-bar X}$ of $theta$ a Consistent estimator of $theta?$
I am trying to find $E(X) $ here to see if it is equal to $theta$ asymptotically but I am not sure how to find expectation here I am not familiar with finding $E(X)$ of fractions.
statistical-inference parameter-estimation
edited Jan 11 at 7:47
Ahmad Bazzi
8,0812824
asked Jan 11 at 5:22
Daman deepDaman deep
756418
$endgroup$
add a comment |
$begingroup$
Let $X_1,X_2,X_3.....X_n$ be a random sample from a population X
having the probability density function
$$ f(x;theta) = begin{cases} theta x^{theta -1} & text{if $0 le
xle$ 1} \0 &text{otherwise} end{cases}$$
Is the estimator $hattheta = $ $dfrac{bar X}{1-bar X}$ of $theta$ a Consistent estimator of $theta?$
I am trying to find $E(X) $ here to see if it is equal to $theta$ asymptotically but I am not sure how to find expectation here I am not familiar with finding $E(X)$ of fractions.
statistical-inference parameter-estimation
edited Jan 11 at 7:47
Ahmad Bazzi
8,0812824
asked Jan 11 at 5:22
Daman deepDaman deep
756418
$endgroup$
add a comment |
$begingroup$
Let $X_1,X_2,X_3.....X_n$ be a random sample from a population X
having the probability density function
$$ f(x;theta) = begin{cases} theta x^{theta -1} & text{if $0 le
xle$ 1} \0 &text{otherwise} end{cases}$$
Is the estimator $hattheta = $ $dfrac{bar X}{1-bar X}$ of $theta$ a Consistent estimator of $theta?$
I am trying to find $E(X) $ here to see if it is equal to $theta$ asymptotically but I am not sure how to find expectation here I am not familiar with finding $E(X)$ of fractions.
statistical-inference parameter-estimation
edited Jan 11 at 7:47
Ahmad Bazzi
8,0812824
asked Jan 11 at 5:22
Daman deepDaman deep
756418
$endgroup$
Let $X_1,X_2,X_3.....X_n$ be a random sample from a population X
having the probability density function
$$ f(x;theta) = begin{cases} theta x^{theta -1} & text{if $0 le
xle$ 1} \0 &text{otherwise} end{cases}$$
Is the estimator $hattheta = $ $dfrac{bar X}{1-bar X}$ of $theta$ a Consistent estimator of $theta?$
I am trying to find $E(X) $ here to see if it is equal to $theta$ asymptotically but I am not sure how to find expectation here I am not familiar with finding $E(X)$ of fractions.
statistical-inference parameter-estimation
statistical-inference parameter-estimation
edited Jan 11 at 7:47
Ahmad Bazzi
8,0812824
asked Jan 11 at 5:22
Daman deepDaman deep
756418
edited Jan 11 at 7:47
Ahmad Bazzi
8,0812824
asked Jan 11 at 5:22
Daman deepDaman deep
756418
edited Jan 11 at 7:47
Ahmad Bazzi
8,0812824
edited Jan 11 at 7:47
Ahmad Bazzi
8,0812824
edited Jan 11 at 7:47
Ahmad Bazzi
8,0812824
8,0812824
asked Jan 11 at 5:22
Daman deepDaman deep
756418
asked Jan 11 at 5:22
Daman deepDaman deep
756418
asked Jan 11 at 5:22
Daman deepDaman deep
756418
756418
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I am trying to find $E(X) $ here to see if it is equal to $theta$ asymptotically
It's not $E(X)$ that should be equal to $theta$ asymptotically, it's $hat{theta}$.
Let's find $E(X)$ as you suggest $$E(X) = int_0^1 xf(x;theta) dx = int_0^1 theta x^{theta} = frac{theta}{theta+1} $$
Now let's see in the asymptotic regime how $hat{theta}$ behaves,
$$hat{theta} = dfrac{bar X}{1-bar X} rightarrow frac{E(X)}{1-E(X)} = frac{frac{theta}{theta+1} }{1-frac{theta}{theta+1} } = frac{theta}{theta+1} frac{theta+1}{1} = frac{theta}{1} = theta $$
So, what can you say about $hat{theta}$ ?
answered Jan 11 at 7:23
Ahmad BazziAhmad Bazzi
8,0812824
$endgroup$
1
$begingroup$
Hey @Damandeep, yes it is not asymptotically consistent .. what does your textbook say?
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:38
3
$begingroup$
Yeah because they got another estimator which is $hat{theta} = frac{bar{X}}{1- bar{X}}$ .. in your question you have a plus sign instead of a minus in the denominator .. makes a WHOLEEE lot of difference.
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:45
2
$begingroup$
no worries, i have edited your question and my answer .. please see now :) @Damandeep
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:47
2
$begingroup$
no need @Damandeep .. question is still the same ;) .. btw if you found the answer useful you can mark it as correct as well
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:51
1
$begingroup$
An estimator is consistent if, as the sample size increases, the estimates "converge" to the true value of the parameter being estimated. That's it. What you mention ($V(hat{theta}) = 0$) is a consequence of certain types of consistent estimators.
$endgroup$
– Ahmad Bazzi
Jan 11 at 8:44
|
show 9 more comments
$begingroup$
I suppose the answer is no. Perhaps we may do as follows.
Denote
begin{align}
S_n&=frac{1}{n}sum_{j=1}^nX_j,\
T_n&=frac{S_n}{1+S_n}.
end{align}
Let us use the strong law of large number (SLLN) and the dominated convergence theorem (DCT) to go for the consistence.
For one thing, each $X_jinleft[0,1right]$ implies that $S_ninleft[0,1right]$. Consequently, $T_nleft[0,1/2right]$, which implies that $left|T_nright|le 1/2$ for all $ninmathbb{N}$. This upper bound $1/2$ is obviously integrable under the probability measure, i.e., $mathbb{E}(1/2)=1/2<infty$.
For another, by SLLN,
$$
lim_{ntoinfty}S_n=mathbb{E}X_1=frac{theta}{1+theta},
$$
which thus leads to
$$
lim_{ntoinfty}T_n=frac{lim_{ntoinfty}S_n}{1+lim_{ntoinfty}S_n}=frac{theta}{1+2theta}.
$$
Combine the above two facts, and DCT applies. This gives
$$
lim_{ntoinfty}mathbb{E}T_n=mathbb{E}left(lim_{ntoinfty}T_nright)=mathbb{E}frac{theta}{1+2theta}=frac{theta}{1+2theta}netheta.
$$
Therefore, $T_n$ is not a consistent estimator of $theta$.
However, I suppose there might be a typo in $T_n$. In fact, if we consider another estimator
$$
U_n=frac{S_n}{1-S_n},
$$
the answer would be yes.
The scope of its proof is essentially the same as above, although some tricks are necessary. This is because DCT would fail as $U_n$ is no longer bounded from above (more precisely, it is not trivial to figure out some $V$ such that $mathbb{E}V<infty$ and that $left|U_nright|le V$).
To help facilitate our proof, define
$$
U_{mn}=U_ncdot 1_{left{U_nle mright}}.
$$
Note that $U_nge 1$ is guaranteed because $S_ninleft[0,1right]$. Besides, note that $U_{mn}$ is monotone in $m$, i.e.,
$$
U_{mn}le U_{m+1,n}.
$$
Thanks to these two facts, the monotone convergence theorem (MCT) applies. It gives
$$
mathbb{E}left(lim_{mtoinfty}U_{mn}right)=lim_{mtoinfty}mathbb{E}U_{mn}.
$$
Consequently,
$$
lim_{ntoinfty}mathbb{E}U_n=lim_{ntoinfty}mathbb{E}left(lim_{mtoinfty}U_{mn}right)=lim_{ntoinfty}lim_{mtoinfty}mathbb{E}U_{mn}.
$$
Further, for each fixed $m$, $mathbb{E}U_{mn}$ converges as $ntoinfty$ by DCT, and for each fixed $n$, $mathbb{E}U_{mn}$ converges uniformly as $mtoinfty$ due to the uniform cutoff. These facts inspires that
$$
lim_{ntoinfty}lim_{mtoinfty}mathbb{E}U_{mn}=lim_{mtoinfty}lim_{ntoinfty}mathbb{E}U_{mn}.
$$
Thanks to all these arguments, we may safely conclude that
$$
lim_{ntoinfty}mathbb{E}U_n=lim_{mtoinfty}lim_{ntoinfty}mathbb{E}U_{mn}=lim_{mtoinfty}mathbb{E}left(lim_{ntoinfty}U_{mn}right)=cdots=theta.
$$
answered Jan 11 at 7:36
hypernovahypernova
4,779414
$endgroup$
$begingroup$
Thanks for your effort
$endgroup$
– Daman deep
Jan 11 at 7:49
$begingroup$
@Damandeep: No problem :-)
$endgroup$
– hypernova
Jan 11 at 8:06
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
I am trying to find $E(X) $ here to see if it is equal to $theta$ asymptotically
It's not $E(X)$ that should be equal to $theta$ asymptotically, it's $hat{theta}$.
Let's find $E(X)$ as you suggest $$E(X) = int_0^1 xf(x;theta) dx = int_0^1 theta x^{theta} = frac{theta}{theta+1} $$
Now let's see in the asymptotic regime how $hat{theta}$ behaves,
$$hat{theta} = dfrac{bar X}{1-bar X} rightarrow frac{E(X)}{1-E(X)} = frac{frac{theta}{theta+1} }{1-frac{theta}{theta+1} } = frac{theta}{theta+1} frac{theta+1}{1} = frac{theta}{1} = theta $$
So, what can you say about $hat{theta}$ ?
answered Jan 11 at 7:23
Ahmad BazziAhmad Bazzi
8,0812824
$endgroup$
1
$begingroup$
Hey @Damandeep, yes it is not asymptotically consistent .. what does your textbook say?
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:38
3
$begingroup$
Yeah because they got another estimator which is $hat{theta} = frac{bar{X}}{1- bar{X}}$ .. in your question you have a plus sign instead of a minus in the denominator .. makes a WHOLEEE lot of difference.
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:45
2
$begingroup$
no worries, i have edited your question and my answer .. please see now :) @Damandeep
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:47
2
$begingroup$
no need @Damandeep .. question is still the same ;) .. btw if you found the answer useful you can mark it as correct as well
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:51
1
$begingroup$
An estimator is consistent if, as the sample size increases, the estimates "converge" to the true value of the parameter being estimated. That's it. What you mention ($V(hat{theta}) = 0$) is a consequence of certain types of consistent estimators.
$endgroup$
– Ahmad Bazzi
Jan 11 at 8:44
|
show 9 more comments
$begingroup$
I am trying to find $E(X) $ here to see if it is equal to $theta$ asymptotically
It's not $E(X)$ that should be equal to $theta$ asymptotically, it's $hat{theta}$.
Let's find $E(X)$ as you suggest $$E(X) = int_0^1 xf(x;theta) dx = int_0^1 theta x^{theta} = frac{theta}{theta+1} $$
Now let's see in the asymptotic regime how $hat{theta}$ behaves,
$$hat{theta} = dfrac{bar X}{1-bar X} rightarrow frac{E(X)}{1-E(X)} = frac{frac{theta}{theta+1} }{1-frac{theta}{theta+1} } = frac{theta}{theta+1} frac{theta+1}{1} = frac{theta}{1} = theta $$
So, what can you say about $hat{theta}$ ?
answered Jan 11 at 7:23
Ahmad BazziAhmad Bazzi
8,0812824
$endgroup$
1
$begingroup$
Hey @Damandeep, yes it is not asymptotically consistent .. what does your textbook say?
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:38
3
$begingroup$
Yeah because they got another estimator which is $hat{theta} = frac{bar{X}}{1- bar{X}}$ .. in your question you have a plus sign instead of a minus in the denominator .. makes a WHOLEEE lot of difference.
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:45
2
$begingroup$
no worries, i have edited your question and my answer .. please see now :) @Damandeep
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:47
2
$begingroup$
no need @Damandeep .. question is still the same ;) .. btw if you found the answer useful you can mark it as correct as well
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:51
1
$begingroup$
An estimator is consistent if, as the sample size increases, the estimates "converge" to the true value of the parameter being estimated. That's it. What you mention ($V(hat{theta}) = 0$) is a consequence of certain types of consistent estimators.
$endgroup$
– Ahmad Bazzi
Jan 11 at 8:44
|
show 9 more comments
$begingroup$
I am trying to find $E(X) $ here to see if it is equal to $theta$ asymptotically
It's not $E(X)$ that should be equal to $theta$ asymptotically, it's $hat{theta}$.
Let's find $E(X)$ as you suggest $$E(X) = int_0^1 xf(x;theta) dx = int_0^1 theta x^{theta} = frac{theta}{theta+1} $$
Now let's see in the asymptotic regime how $hat{theta}$ behaves,
$$hat{theta} = dfrac{bar X}{1-bar X} rightarrow frac{E(X)}{1-E(X)} = frac{frac{theta}{theta+1} }{1-frac{theta}{theta+1} } = frac{theta}{theta+1} frac{theta+1}{1} = frac{theta}{1} = theta $$
So, what can you say about $hat{theta}$ ?
answered Jan 11 at 7:23
Ahmad BazziAhmad Bazzi
8,0812824
$endgroup$
I am trying to find $E(X) $ here to see if it is equal to $theta$ asymptotically
It's not $E(X)$ that should be equal to $theta$ asymptotically, it's $hat{theta}$.
Let's find $E(X)$ as you suggest $$E(X) = int_0^1 xf(x;theta) dx = int_0^1 theta x^{theta} = frac{theta}{theta+1} $$
Now let's see in the asymptotic regime how $hat{theta}$ behaves,
$$hat{theta} = dfrac{bar X}{1-bar X} rightarrow frac{E(X)}{1-E(X)} = frac{frac{theta}{theta+1} }{1-frac{theta}{theta+1} } = frac{theta}{theta+1} frac{theta+1}{1} = frac{theta}{1} = theta $$
So, what can you say about $hat{theta}$ ?
answered Jan 11 at 7:23
Ahmad BazziAhmad Bazzi
8,0812824
edited Jan 11 at 7:47
answered Jan 11 at 7:23
Ahmad BazziAhmad Bazzi
8,0812824
answered Jan 11 at 7:23
Ahmad BazziAhmad Bazzi
8,0812824
answered Jan 11 at 7:23
Ahmad BazziAhmad Bazzi
8,0812824
8,0812824
1
$begingroup$
Hey @Damandeep, yes it is not asymptotically consistent .. what does your textbook say?
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:38
3
$begingroup$
Yeah because they got another estimator which is $hat{theta} = frac{bar{X}}{1- bar{X}}$ .. in your question you have a plus sign instead of a minus in the denominator .. makes a WHOLEEE lot of difference.
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:45
2
$begingroup$
no worries, i have edited your question and my answer .. please see now :) @Damandeep
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:47
2
$begingroup$
no need @Damandeep .. question is still the same ;) .. btw if you found the answer useful you can mark it as correct as well
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:51
1
$begingroup$
An estimator is consistent if, as the sample size increases, the estimates "converge" to the true value of the parameter being estimated. That's it. What you mention ($V(hat{theta}) = 0$) is a consequence of certain types of consistent estimators.
$endgroup$
– Ahmad Bazzi
Jan 11 at 8:44
|
show 9 more comments
1
$begingroup$
Hey @Damandeep, yes it is not asymptotically consistent .. what does your textbook say?
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:38
3
$begingroup$
Yeah because they got another estimator which is $hat{theta} = frac{bar{X}}{1- bar{X}}$ .. in your question you have a plus sign instead of a minus in the denominator .. makes a WHOLEEE lot of difference.
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:45
2
$begingroup$
no worries, i have edited your question and my answer .. please see now :) @Damandeep
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:47
2
$begingroup$
no need @Damandeep .. question is still the same ;) .. btw if you found the answer useful you can mark it as correct as well
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:51
1
$begingroup$
An estimator is consistent if, as the sample size increases, the estimates "converge" to the true value of the parameter being estimated. That's it. What you mention ($V(hat{theta}) = 0$) is a consequence of certain types of consistent estimators.
$endgroup$
– Ahmad Bazzi
Jan 11 at 8:44
1
1
$begingroup$
Hey @Damandeep, yes it is not asymptotically consistent .. what does your textbook say?
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:38
$begingroup$
Hey @Damandeep, yes it is not asymptotically consistent .. what does your textbook say?
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:38
3
3
$begingroup$
Yeah because they got another estimator which is $hat{theta} = frac{bar{X}}{1- bar{X}}$ .. in your question you have a plus sign instead of a minus in the denominator .. makes a WHOLEEE lot of difference.
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:45
$begingroup$
Yeah because they got another estimator which is $hat{theta} = frac{bar{X}}{1- bar{X}}$ .. in your question you have a plus sign instead of a minus in the denominator .. makes a WHOLEEE lot of difference.
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:45
2
2
$begingroup$
no worries, i have edited your question and my answer .. please see now :) @Damandeep
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:47
$begingroup$
no worries, i have edited your question and my answer .. please see now :) @Damandeep
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:47
2
2
$begingroup$
no need @Damandeep .. question is still the same ;) .. btw if you found the answer useful you can mark it as correct as well
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:51
$begingroup$
no need @Damandeep .. question is still the same ;) .. btw if you found the answer useful you can mark it as correct as well
$endgroup$
– Ahmad Bazzi
Jan 11 at 7:51
1
1
$begingroup$
An estimator is consistent if, as the sample size increases, the estimates "converge" to the true value of the parameter being estimated. That's it. What you mention ($V(hat{theta}) = 0$) is a consequence of certain types of consistent estimators.
$endgroup$
– Ahmad Bazzi
Jan 11 at 8:44
$begingroup$
An estimator is consistent if, as the sample size increases, the estimates "converge" to the true value of the parameter being estimated. That's it. What you mention ($V(hat{theta}) = 0$) is a consequence of certain types of consistent estimators.
$endgroup$
– Ahmad Bazzi
Jan 11 at 8:44
|
show 9 more comments
$begingroup$
I suppose the answer is no. Perhaps we may do as follows.
Denote
begin{align}
S_n&=frac{1}{n}sum_{j=1}^nX_j,\
T_n&=frac{S_n}{1+S_n}.
end{align}
Let us use the strong law of large number (SLLN) and the dominated convergence theorem (DCT) to go for the consistence.
For one thing, each $X_jinleft[0,1right]$ implies that $S_ninleft[0,1right]$. Consequently, $T_nleft[0,1/2right]$, which implies that $left|T_nright|le 1/2$ for all $ninmathbb{N}$. This upper bound $1/2$ is obviously integrable under the probability measure, i.e., $mathbb{E}(1/2)=1/2<infty$.
For another, by SLLN,
$$
lim_{ntoinfty}S_n=mathbb{E}X_1=frac{theta}{1+theta},
$$
which thus leads to
$$
lim_{ntoinfty}T_n=frac{lim_{ntoinfty}S_n}{1+lim_{ntoinfty}S_n}=frac{theta}{1+2theta}.
$$
Combine the above two facts, and DCT applies. This gives
$$
lim_{ntoinfty}mathbb{E}T_n=mathbb{E}left(lim_{ntoinfty}T_nright)=mathbb{E}frac{theta}{1+2theta}=frac{theta}{1+2theta}netheta.
$$
Therefore, $T_n$ is not a consistent estimator of $theta$.
However, I suppose there might be a typo in $T_n$. In fact, if we consider another estimator
$$
U_n=frac{S_n}{1-S_n},
$$
the answer would be yes.
The scope of its proof is essentially the same as above, although some tricks are necessary. This is because DCT would fail as $U_n$ is no longer bounded from above (more precisely, it is not trivial to figure out some $V$ such that $mathbb{E}V<infty$ and that $left|U_nright|le V$).
To help facilitate our proof, define
$$
U_{mn}=U_ncdot 1_{left{U_nle mright}}.
$$
Note that $U_nge 1$ is guaranteed because $S_ninleft[0,1right]$. Besides, note that $U_{mn}$ is monotone in $m$, i.e.,
$$
U_{mn}le U_{m+1,n}.
$$
Thanks to these two facts, the monotone convergence theorem (MCT) applies. It gives
$$
mathbb{E}left(lim_{mtoinfty}U_{mn}right)=lim_{mtoinfty}mathbb{E}U_{mn}.
$$
Consequently,
$$
lim_{ntoinfty}mathbb{E}U_n=lim_{ntoinfty}mathbb{E}left(lim_{mtoinfty}U_{mn}right)=lim_{ntoinfty}lim_{mtoinfty}mathbb{E}U_{mn}.
$$
Further, for each fixed $m$, $mathbb{E}U_{mn}$ converges as $ntoinfty$ by DCT, and for each fixed $n$, $mathbb{E}U_{mn}$ converges uniformly as $mtoinfty$ due to the uniform cutoff. These facts inspires that
$$
lim_{ntoinfty}lim_{mtoinfty}mathbb{E}U_{mn}=lim_{mtoinfty}lim_{ntoinfty}mathbb{E}U_{mn}.
$$
Thanks to all these arguments, we may safely conclude that
$$
lim_{ntoinfty}mathbb{E}U_n=lim_{mtoinfty}lim_{ntoinfty}mathbb{E}U_{mn}=lim_{mtoinfty}mathbb{E}left(lim_{ntoinfty}U_{mn}right)=cdots=theta.
$$
answered Jan 11 at 7:36
hypernovahypernova
4,779414
$endgroup$
$begingroup$
Thanks for your effort
$endgroup$
– Daman deep
Jan 11 at 7:49
$begingroup$
@Damandeep: No problem :-)
$endgroup$
– hypernova
Jan 11 at 8:06
add a comment |
$begingroup$
I suppose the answer is no. Perhaps we may do as follows.
Denote
begin{align}
S_n&=frac{1}{n}sum_{j=1}^nX_j,\
T_n&=frac{S_n}{1+S_n}.
end{align}
Let us use the strong law of large number (SLLN) and the dominated convergence theorem (DCT) to go for the consistence.
For one thing, each $X_jinleft[0,1right]$ implies that $S_ninleft[0,1right]$. Consequently, $T_nleft[0,1/2right]$, which implies that $left|T_nright|le 1/2$ for all $ninmathbb{N}$. This upper bound $1/2$ is obviously integrable under the probability measure, i.e., $mathbb{E}(1/2)=1/2<infty$.
For another, by SLLN,
$$
lim_{ntoinfty}S_n=mathbb{E}X_1=frac{theta}{1+theta},
$$
which thus leads to
$$
lim_{ntoinfty}T_n=frac{lim_{ntoinfty}S_n}{1+lim_{ntoinfty}S_n}=frac{theta}{1+2theta}.
$$
Combine the above two facts, and DCT applies. This gives
$$
lim_{ntoinfty}mathbb{E}T_n=mathbb{E}left(lim_{ntoinfty}T_nright)=mathbb{E}frac{theta}{1+2theta}=frac{theta}{1+2theta}netheta.
$$
Therefore, $T_n$ is not a consistent estimator of $theta$.
However, I suppose there might be a typo in $T_n$. In fact, if we consider another estimator
$$
U_n=frac{S_n}{1-S_n},
$$
the answer would be yes.
The scope of its proof is essentially the same as above, although some tricks are necessary. This is because DCT would fail as $U_n$ is no longer bounded from above (more precisely, it is not trivial to figure out some $V$ such that $mathbb{E}V<infty$ and that $left|U_nright|le V$).
To help facilitate our proof, define
$$
U_{mn}=U_ncdot 1_{left{U_nle mright}}.
$$
Note that $U_nge 1$ is guaranteed because $S_ninleft[0,1right]$. Besides, note that $U_{mn}$ is monotone in $m$, i.e.,
$$
U_{mn}le U_{m+1,n}.
$$
Thanks to these two facts, the monotone convergence theorem (MCT) applies. It gives
$$
mathbb{E}left(lim_{mtoinfty}U_{mn}right)=lim_{mtoinfty}mathbb{E}U_{mn}.
$$
Consequently,
$$
lim_{ntoinfty}mathbb{E}U_n=lim_{ntoinfty}mathbb{E}left(lim_{mtoinfty}U_{mn}right)=lim_{ntoinfty}lim_{mtoinfty}mathbb{E}U_{mn}.
$$
Further, for each fixed $m$, $mathbb{E}U_{mn}$ converges as $ntoinfty$ by DCT, and for each fixed $n$, $mathbb{E}U_{mn}$ converges uniformly as $mtoinfty$ due to the uniform cutoff. These facts inspires that
$$
lim_{ntoinfty}lim_{mtoinfty}mathbb{E}U_{mn}=lim_{mtoinfty}lim_{ntoinfty}mathbb{E}U_{mn}.
$$
Thanks to all these arguments, we may safely conclude that
$$
lim_{ntoinfty}mathbb{E}U_n=lim_{mtoinfty}lim_{ntoinfty}mathbb{E}U_{mn}=lim_{mtoinfty}mathbb{E}left(lim_{ntoinfty}U_{mn}right)=cdots=theta.
$$
answered Jan 11 at 7:36
hypernovahypernova
4,779414
$endgroup$
$begingroup$
Thanks for your effort
$endgroup$
– Daman deep
Jan 11 at 7:49
$begingroup$
@Damandeep: No problem :-)
$endgroup$
– hypernova
Jan 11 at 8:06
add a comment |
$begingroup$
I suppose the answer is no. Perhaps we may do as follows.
Denote
begin{align}
S_n&=frac{1}{n}sum_{j=1}^nX_j,\
T_n&=frac{S_n}{1+S_n}.
end{align}
Let us use the strong law of large number (SLLN) and the dominated convergence theorem (DCT) to go for the consistence.
For one thing, each $X_jinleft[0,1right]$ implies that $S_ninleft[0,1right]$. Consequently, $T_nleft[0,1/2right]$, which implies that $left|T_nright|le 1/2$ for all $ninmathbb{N}$. This upper bound $1/2$ is obviously integrable under the probability measure, i.e., $mathbb{E}(1/2)=1/2<infty$.
For another, by SLLN,
$$
lim_{ntoinfty}S_n=mathbb{E}X_1=frac{theta}{1+theta},
$$
which thus leads to
$$
lim_{ntoinfty}T_n=frac{lim_{ntoinfty}S_n}{1+lim_{ntoinfty}S_n}=frac{theta}{1+2theta}.
$$
Combine the above two facts, and DCT applies. This gives
$$
lim_{ntoinfty}mathbb{E}T_n=mathbb{E}left(lim_{ntoinfty}T_nright)=mathbb{E}frac{theta}{1+2theta}=frac{theta}{1+2theta}netheta.
$$
Therefore, $T_n$ is not a consistent estimator of $theta$.
However, I suppose there might be a typo in $T_n$. In fact, if we consider another estimator
$$
U_n=frac{S_n}{1-S_n},
$$
the answer would be yes.
The scope of its proof is essentially the same as above, although some tricks are necessary. This is because DCT would fail as $U_n$ is no longer bounded from above (more precisely, it is not trivial to figure out some $V$ such that $mathbb{E}V<infty$ and that $left|U_nright|le V$).
To help facilitate our proof, define
$$
U_{mn}=U_ncdot 1_{left{U_nle mright}}.
$$
Note that $U_nge 1$ is guaranteed because $S_ninleft[0,1right]$. Besides, note that $U_{mn}$ is monotone in $m$, i.e.,
$$
U_{mn}le U_{m+1,n}.
$$
Thanks to these two facts, the monotone convergence theorem (MCT) applies. It gives
$$
mathbb{E}left(lim_{mtoinfty}U_{mn}right)=lim_{mtoinfty}mathbb{E}U_{mn}.
$$
Consequently,
$$
lim_{ntoinfty}mathbb{E}U_n=lim_{ntoinfty}mathbb{E}left(lim_{mtoinfty}U_{mn}right)=lim_{ntoinfty}lim_{mtoinfty}mathbb{E}U_{mn}.
$$
Further, for each fixed $m$, $mathbb{E}U_{mn}$ converges as $ntoinfty$ by DCT, and for each fixed $n$, $mathbb{E}U_{mn}$ converges uniformly as $mtoinfty$ due to the uniform cutoff. These facts inspires that
$$
lim_{ntoinfty}lim_{mtoinfty}mathbb{E}U_{mn}=lim_{mtoinfty}lim_{ntoinfty}mathbb{E}U_{mn}.
$$
Thanks to all these arguments, we may safely conclude that
$$
lim_{ntoinfty}mathbb{E}U_n=lim_{mtoinfty}lim_{ntoinfty}mathbb{E}U_{mn}=lim_{mtoinfty}mathbb{E}left(lim_{ntoinfty}U_{mn}right)=cdots=theta.
$$
answered Jan 11 at 7:36
hypernovahypernova
4,779414
$endgroup$
I suppose the answer is no. Perhaps we may do as follows.
Denote
begin{align}
S_n&=frac{1}{n}sum_{j=1}^nX_j,\
T_n&=frac{S_n}{1+S_n}.
end{align}
Let us use the strong law of large number (SLLN) and the dominated convergence theorem (DCT) to go for the consistence.
For one thing, each $X_jinleft[0,1right]$ implies that $S_ninleft[0,1right]$. Consequently, $T_nleft[0,1/2right]$, which implies that $left|T_nright|le 1/2$ for all $ninmathbb{N}$. This upper bound $1/2$ is obviously integrable under the probability measure, i.e., $mathbb{E}(1/2)=1/2<infty$.
For another, by SLLN,
$$
lim_{ntoinfty}S_n=mathbb{E}X_1=frac{theta}{1+theta},
$$
which thus leads to
$$
lim_{ntoinfty}T_n=frac{lim_{ntoinfty}S_n}{1+lim_{ntoinfty}S_n}=frac{theta}{1+2theta}.
$$
Combine the above two facts, and DCT applies. This gives
$$
lim_{ntoinfty}mathbb{E}T_n=mathbb{E}left(lim_{ntoinfty}T_nright)=mathbb{E}frac{theta}{1+2theta}=frac{theta}{1+2theta}netheta.
$$
Therefore, $T_n$ is not a consistent estimator of $theta$.
However, I suppose there might be a typo in $T_n$. In fact, if we consider another estimator
$$
U_n=frac{S_n}{1-S_n},
$$
the answer would be yes.
The scope of its proof is essentially the same as above, although some tricks are necessary. This is because DCT would fail as $U_n$ is no longer bounded from above (more precisely, it is not trivial to figure out some $V$ such that $mathbb{E}V<infty$ and that $left|U_nright|le V$).
To help facilitate our proof, define
$$
U_{mn}=U_ncdot 1_{left{U_nle mright}}.
$$
Note that $U_nge 1$ is guaranteed because $S_ninleft[0,1right]$. Besides, note that $U_{mn}$ is monotone in $m$, i.e.,
$$
U_{mn}le U_{m+1,n}.
$$
Thanks to these two facts, the monotone convergence theorem (MCT) applies. It gives
$$
mathbb{E}left(lim_{mtoinfty}U_{mn}right)=lim_{mtoinfty}mathbb{E}U_{mn}.
$$
Consequently,
$$
lim_{ntoinfty}mathbb{E}U_n=lim_{ntoinfty}mathbb{E}left(lim_{mtoinfty}U_{mn}right)=lim_{ntoinfty}lim_{mtoinfty}mathbb{E}U_{mn}.
$$
Further, for each fixed $m$, $mathbb{E}U_{mn}$ converges as $ntoinfty$ by DCT, and for each fixed $n$, $mathbb{E}U_{mn}$ converges uniformly as $mtoinfty$ due to the uniform cutoff. These facts inspires that
$$
lim_{ntoinfty}lim_{mtoinfty}mathbb{E}U_{mn}=lim_{mtoinfty}lim_{ntoinfty}mathbb{E}U_{mn}.
$$
Thanks to all these arguments, we may safely conclude that
$$
lim_{ntoinfty}mathbb{E}U_n=lim_{mtoinfty}lim_{ntoinfty}mathbb{E}U_{mn}=lim_{mtoinfty}mathbb{E}left(lim_{ntoinfty}U_{mn}right)=cdots=theta.
$$
answered Jan 11 at 7:36
hypernovahypernova
4,779414
answered Jan 11 at 7:36
hypernovahypernova
4,779414
answered Jan 11 at 7:36
hypernovahypernova
4,779414
answered Jan 11 at 7:36
hypernovahypernova
4,779414
4,779414
$begingroup$
Thanks for your effort
$endgroup$
– Daman deep
Jan 11 at 7:49
$begingroup$
@Damandeep: No problem :-)
$endgroup$
– hypernova
Jan 11 at 8:06
add a comment |
$begingroup$
Thanks for your effort
$endgroup$
– Daman deep
Jan 11 at 7:49
$begingroup$
@Damandeep: No problem :-)
$endgroup$
– hypernova
Jan 11 at 8:06
$begingroup$
Thanks for your effort
$endgroup$
– Daman deep
Jan 11 at 7:49
$begingroup$
Thanks for your effort
$endgroup$
– Daman deep
Jan 11 at 7:49
$begingroup$
@Damandeep: No problem :-)
$endgroup$
– hypernova
Jan 11 at 8:06
$begingroup$
@Damandeep: No problem :-)
$endgroup$
– hypernova
Jan 11 at 8:06
add a comment |
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