Find UMVU estimator for $e^{-3 theta}$ given a complete sufficient statistic $X sim Pois(theta)$ with...
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My attempt: We know, since $Xsim Pois(theta)$ that $mathbb{P}_{theta}(X=x)=e^{-theta}theta^{x}/x!$. A given tip is that we must recall that $e^{x}=sum^{infty}_{k=0}frac{x^{k}}{k!}$. I know that once we find an unbiased estimator that is a function of our complete sufficient statistic $X$, that this estimator must then automatically be UMVU. However, I'm not sure how to approach this question by even finding an expression for an unbiased estimator.
Question: How to approach/solve this exercise?
Thanks!
probability probability-theory statistics statistical-inference parameter-estimation
asked Dec 2 at 20:22
S. Crim
10110
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up vote
1
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My attempt: We know, since $Xsim Pois(theta)$ that $mathbb{P}_{theta}(X=x)=e^{-theta}theta^{x}/x!$. A given tip is that we must recall that $e^{x}=sum^{infty}_{k=0}frac{x^{k}}{k!}$. I know that once we find an unbiased estimator that is a function of our complete sufficient statistic $X$, that this estimator must then automatically be UMVU. However, I'm not sure how to approach this question by even finding an expression for an unbiased estimator.
Question: How to approach/solve this exercise?
Thanks!
probability probability-theory statistics statistical-inference parameter-estimation
asked Dec 2 at 20:22
S. Crim
10110
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My attempt: We know, since $Xsim Pois(theta)$ that $mathbb{P}_{theta}(X=x)=e^{-theta}theta^{x}/x!$. A given tip is that we must recall that $e^{x}=sum^{infty}_{k=0}frac{x^{k}}{k!}$. I know that once we find an unbiased estimator that is a function of our complete sufficient statistic $X$, that this estimator must then automatically be UMVU. However, I'm not sure how to approach this question by even finding an expression for an unbiased estimator.
Question: How to approach/solve this exercise?
Thanks!
probability probability-theory statistics statistical-inference parameter-estimation
asked Dec 2 at 20:22
S. Crim
10110
My attempt: We know, since $Xsim Pois(theta)$ that $mathbb{P}_{theta}(X=x)=e^{-theta}theta^{x}/x!$. A given tip is that we must recall that $e^{x}=sum^{infty}_{k=0}frac{x^{k}}{k!}$. I know that once we find an unbiased estimator that is a function of our complete sufficient statistic $X$, that this estimator must then automatically be UMVU. However, I'm not sure how to approach this question by even finding an expression for an unbiased estimator.
Question: How to approach/solve this exercise?
Thanks!
probability probability-theory statistics statistical-inference parameter-estimation
probability probability-theory statistics statistical-inference parameter-estimation
asked Dec 2 at 20:22
S. Crim
10110
asked Dec 2 at 20:22
S. Crim
10110
asked Dec 2 at 20:22
S. Crim
10110
asked Dec 2 at 20:22
S. Crim
10110
asked Dec 2 at 20:22
S. Crim
10110
10110
add a comment |
add a comment |
1 Answer
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We need to find some function $g(k)$ s.t. $mathbb E[g(X)]=e^{-3theta}$ for all $theta>0$. Write this expectation:
$$
mathbb E[g(X)]=sum_{k=0}^infty g(k)dfrac{theta^k}{k!}e^{-theta}=e^{-3theta}.
$$
Multiply both parts by $e^theta$ and get
$$
sum_{k=0}^infty g(k)dfrac{theta^k}{k!}=e^{-2theta} = sum_{k=0}^infty frac{(-2)^ktheta^k}{k!}.
$$
Since these sums are equal for each $theta>0$, we get $g(k)=(-2)^k$ and $$g(X)=(-2)^X$$
is unique unbiased estimator for $e^{-3theta}$.
answered Dec 3 at 3:58
NCh
6,2232723
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We need to find some function $g(k)$ s.t. $mathbb E[g(X)]=e^{-3theta}$ for all $theta>0$. Write this expectation:
$$
mathbb E[g(X)]=sum_{k=0}^infty g(k)dfrac{theta^k}{k!}e^{-theta}=e^{-3theta}.
$$
Multiply both parts by $e^theta$ and get
$$
sum_{k=0}^infty g(k)dfrac{theta^k}{k!}=e^{-2theta} = sum_{k=0}^infty frac{(-2)^ktheta^k}{k!}.
$$
Since these sums are equal for each $theta>0$, we get $g(k)=(-2)^k$ and $$g(X)=(-2)^X$$
is unique unbiased estimator for $e^{-3theta}$.
answered Dec 3 at 3:58
NCh
6,2232723
add a comment |
up vote
1
down vote
accepted
We need to find some function $g(k)$ s.t. $mathbb E[g(X)]=e^{-3theta}$ for all $theta>0$. Write this expectation:
$$
mathbb E[g(X)]=sum_{k=0}^infty g(k)dfrac{theta^k}{k!}e^{-theta}=e^{-3theta}.
$$
Multiply both parts by $e^theta$ and get
$$
sum_{k=0}^infty g(k)dfrac{theta^k}{k!}=e^{-2theta} = sum_{k=0}^infty frac{(-2)^ktheta^k}{k!}.
$$
Since these sums are equal for each $theta>0$, we get $g(k)=(-2)^k$ and $$g(X)=(-2)^X$$
is unique unbiased estimator for $e^{-3theta}$.
answered Dec 3 at 3:58
NCh
6,2232723
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We need to find some function $g(k)$ s.t. $mathbb E[g(X)]=e^{-3theta}$ for all $theta>0$. Write this expectation:
$$
mathbb E[g(X)]=sum_{k=0}^infty g(k)dfrac{theta^k}{k!}e^{-theta}=e^{-3theta}.
$$
Multiply both parts by $e^theta$ and get
$$
sum_{k=0}^infty g(k)dfrac{theta^k}{k!}=e^{-2theta} = sum_{k=0}^infty frac{(-2)^ktheta^k}{k!}.
$$
Since these sums are equal for each $theta>0$, we get $g(k)=(-2)^k$ and $$g(X)=(-2)^X$$
is unique unbiased estimator for $e^{-3theta}$.
answered Dec 3 at 3:58
NCh
6,2232723
We need to find some function $g(k)$ s.t. $mathbb E[g(X)]=e^{-3theta}$ for all $theta>0$. Write this expectation:
$$
mathbb E[g(X)]=sum_{k=0}^infty g(k)dfrac{theta^k}{k!}e^{-theta}=e^{-3theta}.
$$
Multiply both parts by $e^theta$ and get
$$
sum_{k=0}^infty g(k)dfrac{theta^k}{k!}=e^{-2theta} = sum_{k=0}^infty frac{(-2)^ktheta^k}{k!}.
$$
Since these sums are equal for each $theta>0$, we get $g(k)=(-2)^k$ and $$g(X)=(-2)^X$$
is unique unbiased estimator for $e^{-3theta}$.
answered Dec 3 at 3:58
NCh
6,2232723
answered Dec 3 at 3:58
NCh
6,2232723
answered Dec 3 at 3:58
NCh
6,2232723
answered Dec 3 at 3:58
NCh
6,2232723
6,2232723
add a comment |
add a comment |
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