Calculating UMVUE for Poisson distribution
$begingroup$
I've started to learn methods of finding UMVUE distribution. I found some nice examples on this site.
I got stuck while evaluating UMVUE for Poisson distribution for a parametric function $g(theta)$.
How did the author get
$$sum_{t=o}^infty frac{h(t) n^t}{t!} theta^t = e^{n theta}g(theta)?$$
I thought
$$mathbb{E}(h(x)) = g(theta)$$
should be considered.
statistics parameter-estimation
edited Jan 12 at 11:47
Saad
19.7k92352
asked Jan 9 at 17:27
HendrraHendrra
1,163516
$endgroup$
add a comment |
$begingroup$
I've started to learn methods of finding UMVUE distribution. I found some nice examples on this site.
I got stuck while evaluating UMVUE for Poisson distribution for a parametric function $g(theta)$.
How did the author get
$$sum_{t=o}^infty frac{h(t) n^t}{t!} theta^t = e^{n theta}g(theta)?$$
I thought
$$mathbb{E}(h(x)) = g(theta)$$
should be considered.
statistics parameter-estimation
edited Jan 12 at 11:47
Saad
19.7k92352
asked Jan 9 at 17:27
HendrraHendrra
1,163516
$endgroup$
1
$begingroup$
I will just mention that there is a discussion on meta about the newly created (umvue) tag. (It seemed reasonable to let the tag-creator know about this.)
$endgroup$
– Martin Sleziak
Jan 10 at 8:35
add a comment |
$begingroup$
I've started to learn methods of finding UMVUE distribution. I found some nice examples on this site.
I got stuck while evaluating UMVUE for Poisson distribution for a parametric function $g(theta)$.
How did the author get
$$sum_{t=o}^infty frac{h(t) n^t}{t!} theta^t = e^{n theta}g(theta)?$$
I thought
$$mathbb{E}(h(x)) = g(theta)$$
should be considered.
statistics parameter-estimation
edited Jan 12 at 11:47
Saad
19.7k92352
asked Jan 9 at 17:27
HendrraHendrra
1,163516
$endgroup$
I've started to learn methods of finding UMVUE distribution. I found some nice examples on this site.
I got stuck while evaluating UMVUE for Poisson distribution for a parametric function $g(theta)$.
How did the author get
$$sum_{t=o}^infty frac{h(t) n^t}{t!} theta^t = e^{n theta}g(theta)?$$
I thought
$$mathbb{E}(h(x)) = g(theta)$$
should be considered.
statistics parameter-estimation
statistics parameter-estimation
edited Jan 12 at 11:47
Saad
19.7k92352
asked Jan 9 at 17:27
HendrraHendrra
1,163516
edited Jan 12 at 11:47
Saad
19.7k92352
asked Jan 9 at 17:27
HendrraHendrra
1,163516
edited Jan 12 at 11:47
Saad
19.7k92352
edited Jan 12 at 11:47
Saad
19.7k92352
edited Jan 12 at 11:47
Saad
19.7k92352
19.7k92352
asked Jan 9 at 17:27
HendrraHendrra
1,163516
asked Jan 9 at 17:27
HendrraHendrra
1,163516
asked Jan 9 at 17:27
HendrraHendrra
1,163516
1,163516
1
$begingroup$
I will just mention that there is a discussion on meta about the newly created (umvue) tag. (It seemed reasonable to let the tag-creator know about this.)
$endgroup$
– Martin Sleziak
Jan 10 at 8:35
add a comment |
1
$begingroup$
I will just mention that there is a discussion on meta about the newly created (umvue) tag. (It seemed reasonable to let the tag-creator know about this.)
$endgroup$
– Martin Sleziak
Jan 10 at 8:35
1
1
$begingroup$
I will just mention that there is a discussion on meta about the newly created (umvue) tag. (It seemed reasonable to let the tag-creator know about this.)
$endgroup$
– Martin Sleziak
Jan 10 at 8:35
$begingroup$
I will just mention that there is a discussion on meta about the newly created (umvue) tag. (It seemed reasonable to let the tag-creator know about this.)
$endgroup$
– Martin Sleziak
Jan 10 at 8:35
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Observe
$$sum_{t=0}^{infty}dfrac{h(t)n^t}{t!}theta^t = sum_{t=0}^{infty}dfrac{h(t)(ntheta)^t}{t!}text{.}$$
Set $lambda = ntheta$. Then
$$sum_{t=0}^{infty}dfrac{h(t)(ntheta)^t}{t!} = sum_{t=0}^{infty}dfrac{h(t)lambda^t}{t!} = sum_{t=0}^{infty}dfrac{h(t)e^{-lambda}e^{lambda}lambda^t}{t!} = e^{lambda}sum_{t=0}^{infty}left[h(t)cdot dfrac{e^{-lambda}lambda^t}{t!}right]text{.}$$
Since $T sim P(lambda)$ (using the notation in your image), then
$$sum_{t=0}^{infty}left[h(t)cdot dfrac{e^{-lambda}lambda^t}{t!}right] = mathbb{E}[h(T)] = g(theta)$$
due to unbiasedness, hence
$$sum_{t=0}^{infty}dfrac{h(t)n^t}{t!}theta^t = e^{lambda}g(theta)=e^{ntheta}g(theta)text{.}$$
answered Jan 9 at 17:34
ClarinetistClarinetist
10.9k42778
$endgroup$
$begingroup$
Thank you! That's an awesome solution
$endgroup$
– Hendrra
Jan 9 at 17:42
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Observe
$$sum_{t=0}^{infty}dfrac{h(t)n^t}{t!}theta^t = sum_{t=0}^{infty}dfrac{h(t)(ntheta)^t}{t!}text{.}$$
Set $lambda = ntheta$. Then
$$sum_{t=0}^{infty}dfrac{h(t)(ntheta)^t}{t!} = sum_{t=0}^{infty}dfrac{h(t)lambda^t}{t!} = sum_{t=0}^{infty}dfrac{h(t)e^{-lambda}e^{lambda}lambda^t}{t!} = e^{lambda}sum_{t=0}^{infty}left[h(t)cdot dfrac{e^{-lambda}lambda^t}{t!}right]text{.}$$
Since $T sim P(lambda)$ (using the notation in your image), then
$$sum_{t=0}^{infty}left[h(t)cdot dfrac{e^{-lambda}lambda^t}{t!}right] = mathbb{E}[h(T)] = g(theta)$$
due to unbiasedness, hence
$$sum_{t=0}^{infty}dfrac{h(t)n^t}{t!}theta^t = e^{lambda}g(theta)=e^{ntheta}g(theta)text{.}$$
answered Jan 9 at 17:34
ClarinetistClarinetist
10.9k42778
$endgroup$
$begingroup$
Thank you! That's an awesome solution
$endgroup$
– Hendrra
Jan 9 at 17:42
add a comment |
$begingroup$
Observe
$$sum_{t=0}^{infty}dfrac{h(t)n^t}{t!}theta^t = sum_{t=0}^{infty}dfrac{h(t)(ntheta)^t}{t!}text{.}$$
Set $lambda = ntheta$. Then
$$sum_{t=0}^{infty}dfrac{h(t)(ntheta)^t}{t!} = sum_{t=0}^{infty}dfrac{h(t)lambda^t}{t!} = sum_{t=0}^{infty}dfrac{h(t)e^{-lambda}e^{lambda}lambda^t}{t!} = e^{lambda}sum_{t=0}^{infty}left[h(t)cdot dfrac{e^{-lambda}lambda^t}{t!}right]text{.}$$
Since $T sim P(lambda)$ (using the notation in your image), then
$$sum_{t=0}^{infty}left[h(t)cdot dfrac{e^{-lambda}lambda^t}{t!}right] = mathbb{E}[h(T)] = g(theta)$$
due to unbiasedness, hence
$$sum_{t=0}^{infty}dfrac{h(t)n^t}{t!}theta^t = e^{lambda}g(theta)=e^{ntheta}g(theta)text{.}$$
answered Jan 9 at 17:34
ClarinetistClarinetist
10.9k42778
$endgroup$
$begingroup$
Thank you! That's an awesome solution
$endgroup$
– Hendrra
Jan 9 at 17:42
add a comment |
$begingroup$
Observe
$$sum_{t=0}^{infty}dfrac{h(t)n^t}{t!}theta^t = sum_{t=0}^{infty}dfrac{h(t)(ntheta)^t}{t!}text{.}$$
Set $lambda = ntheta$. Then
$$sum_{t=0}^{infty}dfrac{h(t)(ntheta)^t}{t!} = sum_{t=0}^{infty}dfrac{h(t)lambda^t}{t!} = sum_{t=0}^{infty}dfrac{h(t)e^{-lambda}e^{lambda}lambda^t}{t!} = e^{lambda}sum_{t=0}^{infty}left[h(t)cdot dfrac{e^{-lambda}lambda^t}{t!}right]text{.}$$
Since $T sim P(lambda)$ (using the notation in your image), then
$$sum_{t=0}^{infty}left[h(t)cdot dfrac{e^{-lambda}lambda^t}{t!}right] = mathbb{E}[h(T)] = g(theta)$$
due to unbiasedness, hence
$$sum_{t=0}^{infty}dfrac{h(t)n^t}{t!}theta^t = e^{lambda}g(theta)=e^{ntheta}g(theta)text{.}$$
answered Jan 9 at 17:34
ClarinetistClarinetist
10.9k42778
$endgroup$
Observe
$$sum_{t=0}^{infty}dfrac{h(t)n^t}{t!}theta^t = sum_{t=0}^{infty}dfrac{h(t)(ntheta)^t}{t!}text{.}$$
Set $lambda = ntheta$. Then
$$sum_{t=0}^{infty}dfrac{h(t)(ntheta)^t}{t!} = sum_{t=0}^{infty}dfrac{h(t)lambda^t}{t!} = sum_{t=0}^{infty}dfrac{h(t)e^{-lambda}e^{lambda}lambda^t}{t!} = e^{lambda}sum_{t=0}^{infty}left[h(t)cdot dfrac{e^{-lambda}lambda^t}{t!}right]text{.}$$
Since $T sim P(lambda)$ (using the notation in your image), then
$$sum_{t=0}^{infty}left[h(t)cdot dfrac{e^{-lambda}lambda^t}{t!}right] = mathbb{E}[h(T)] = g(theta)$$
due to unbiasedness, hence
$$sum_{t=0}^{infty}dfrac{h(t)n^t}{t!}theta^t = e^{lambda}g(theta)=e^{ntheta}g(theta)text{.}$$
answered Jan 9 at 17:34
ClarinetistClarinetist
10.9k42778
edited Jan 9 at 17:46
answered Jan 9 at 17:34
ClarinetistClarinetist
10.9k42778
answered Jan 9 at 17:34
ClarinetistClarinetist
10.9k42778
answered Jan 9 at 17:34
ClarinetistClarinetist
10.9k42778
10.9k42778
$begingroup$
Thank you! That's an awesome solution
$endgroup$
– Hendrra
Jan 9 at 17:42
add a comment |
$begingroup$
Thank you! That's an awesome solution
$endgroup$
– Hendrra
Jan 9 at 17:42
$begingroup$
Thank you! That's an awesome solution
$endgroup$
– Hendrra
Jan 9 at 17:42
$begingroup$
Thank you! That's an awesome solution
$endgroup$
– Hendrra
Jan 9 at 17:42
add a comment |
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$begingroup$
I will just mention that there is a discussion on meta about the newly created (umvue) tag. (It seemed reasonable to let the tag-creator know about this.)
$endgroup$
– Martin Sleziak
Jan 10 at 8:35