Method of Moments estimation of a Poisson($theta$)
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Let $X_1,ldots,X_n$ ~ Poi($theta$). Calculate the estimatorios of moments for $theta$ in the next cases:
Remember that $E(X_1)=Var(X_1)=theta$
(a) $hat{theta}_1$ by equating the first non-central theoretical moment with the first non-central empirical moment.
(b) $hat{theta}_2$ by equating the theoretical variance with the empirical variance.
(c) $hat{theta}_3$ by equating the second non-central theoretical moment with the second non-central empirical moment.
I just want to know if i did it correctly. It seems too easy.
(a) $hat{theta}_1=overline{x}=frac{1}{n}sum^{n}_{i=1}x_i$
(b) $hat{theta}_2=overline{x}=frac{1}{n}sum^{n}_{i=1}(x_i-overline{x}_1)^2$
(c) $hat{theta}_3^2+hat{theta}_3=frac{1}{n}sum^{n}_{i=1}x_i^2$
statistics statistical-inference
asked Nov 28 at 0:30
pin_r
124
add a comment |
up vote
0
down vote
favorite
Let $X_1,ldots,X_n$ ~ Poi($theta$). Calculate the estimatorios of moments for $theta$ in the next cases:
Remember that $E(X_1)=Var(X_1)=theta$
(a) $hat{theta}_1$ by equating the first non-central theoretical moment with the first non-central empirical moment.
(b) $hat{theta}_2$ by equating the theoretical variance with the empirical variance.
(c) $hat{theta}_3$ by equating the second non-central theoretical moment with the second non-central empirical moment.
I just want to know if i did it correctly. It seems too easy.
(a) $hat{theta}_1=overline{x}=frac{1}{n}sum^{n}_{i=1}x_i$
(b) $hat{theta}_2=overline{x}=frac{1}{n}sum^{n}_{i=1}(x_i-overline{x}_1)^2$
(c) $hat{theta}_3^2+hat{theta}_3=frac{1}{n}sum^{n}_{i=1}x_i^2$
statistics statistical-inference
asked Nov 28 at 0:30
pin_r
124
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X_1,ldots,X_n$ ~ Poi($theta$). Calculate the estimatorios of moments for $theta$ in the next cases:
Remember that $E(X_1)=Var(X_1)=theta$
(a) $hat{theta}_1$ by equating the first non-central theoretical moment with the first non-central empirical moment.
(b) $hat{theta}_2$ by equating the theoretical variance with the empirical variance.
(c) $hat{theta}_3$ by equating the second non-central theoretical moment with the second non-central empirical moment.
I just want to know if i did it correctly. It seems too easy.
(a) $hat{theta}_1=overline{x}=frac{1}{n}sum^{n}_{i=1}x_i$
(b) $hat{theta}_2=overline{x}=frac{1}{n}sum^{n}_{i=1}(x_i-overline{x}_1)^2$
(c) $hat{theta}_3^2+hat{theta}_3=frac{1}{n}sum^{n}_{i=1}x_i^2$
statistics statistical-inference
asked Nov 28 at 0:30
pin_r
124
Let $X_1,ldots,X_n$ ~ Poi($theta$). Calculate the estimatorios of moments for $theta$ in the next cases:
Remember that $E(X_1)=Var(X_1)=theta$
(a) $hat{theta}_1$ by equating the first non-central theoretical moment with the first non-central empirical moment.
(b) $hat{theta}_2$ by equating the theoretical variance with the empirical variance.
(c) $hat{theta}_3$ by equating the second non-central theoretical moment with the second non-central empirical moment.
I just want to know if i did it correctly. It seems too easy.
(a) $hat{theta}_1=overline{x}=frac{1}{n}sum^{n}_{i=1}x_i$
(b) $hat{theta}_2=overline{x}=frac{1}{n}sum^{n}_{i=1}(x_i-overline{x}_1)^2$
(c) $hat{theta}_3^2+hat{theta}_3=frac{1}{n}sum^{n}_{i=1}x_i^2$
statistics statistical-inference
statistics statistical-inference
asked Nov 28 at 0:30
pin_r
124
asked Nov 28 at 0:30
pin_r
124
edited Nov 28 at 3:45
asked Nov 28 at 0:30
pin_r
124
asked Nov 28 at 0:30
pin_r
124
asked Nov 28 at 0:30
pin_r
124
124
add a comment |
add a comment |
1 Answer
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As for the empirical variance, $mu_1$ cannot be part of the formula, since actually $mu_1=theta$, and then it is not an estimator for $theta$. You should use
$$hat theta_2=frac1n sum(x_i-bar x)^2,$$
instead.
As for $hat theta_3$, that's just how you start; now solve for $hat theta_3$ (it is a quadratic equation), and then you'll have the estimator.
answered Nov 28 at 0:38
Alejandro Nasif Salum
3,989117
My translate mistake, so.. $hat{theta}_{1,2}=frac{-1+-sqrt{1-2sum x_i^2}}{2}$
– pin_r
Nov 28 at 3:51
Is not $-2$ but $+4$, and you have to choose from both solutions the one that gives a positive value (since $theta >0$). That is $$hattheta_3=frac{-1+sqrt{1+frac4n sum x_i^2}}2.$$
– Alejandro Nasif Salum
Nov 28 at 4:44
add a comment |
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1 Answer
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1 Answer
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up vote
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down vote
As for the empirical variance, $mu_1$ cannot be part of the formula, since actually $mu_1=theta$, and then it is not an estimator for $theta$. You should use
$$hat theta_2=frac1n sum(x_i-bar x)^2,$$
instead.
As for $hat theta_3$, that's just how you start; now solve for $hat theta_3$ (it is a quadratic equation), and then you'll have the estimator.
answered Nov 28 at 0:38
Alejandro Nasif Salum
3,989117
My translate mistake, so.. $hat{theta}_{1,2}=frac{-1+-sqrt{1-2sum x_i^2}}{2}$
– pin_r
Nov 28 at 3:51
Is not $-2$ but $+4$, and you have to choose from both solutions the one that gives a positive value (since $theta >0$). That is $$hattheta_3=frac{-1+sqrt{1+frac4n sum x_i^2}}2.$$
– Alejandro Nasif Salum
Nov 28 at 4:44
add a comment |
up vote
0
down vote
As for the empirical variance, $mu_1$ cannot be part of the formula, since actually $mu_1=theta$, and then it is not an estimator for $theta$. You should use
$$hat theta_2=frac1n sum(x_i-bar x)^2,$$
instead.
As for $hat theta_3$, that's just how you start; now solve for $hat theta_3$ (it is a quadratic equation), and then you'll have the estimator.
answered Nov 28 at 0:38
Alejandro Nasif Salum
3,989117
My translate mistake, so.. $hat{theta}_{1,2}=frac{-1+-sqrt{1-2sum x_i^2}}{2}$
– pin_r
Nov 28 at 3:51
Is not $-2$ but $+4$, and you have to choose from both solutions the one that gives a positive value (since $theta >0$). That is $$hattheta_3=frac{-1+sqrt{1+frac4n sum x_i^2}}2.$$
– Alejandro Nasif Salum
Nov 28 at 4:44
add a comment |
up vote
0
down vote
up vote
0
down vote
As for the empirical variance, $mu_1$ cannot be part of the formula, since actually $mu_1=theta$, and then it is not an estimator for $theta$. You should use
$$hat theta_2=frac1n sum(x_i-bar x)^2,$$
instead.
As for $hat theta_3$, that's just how you start; now solve for $hat theta_3$ (it is a quadratic equation), and then you'll have the estimator.
answered Nov 28 at 0:38
Alejandro Nasif Salum
3,989117
As for the empirical variance, $mu_1$ cannot be part of the formula, since actually $mu_1=theta$, and then it is not an estimator for $theta$. You should use
$$hat theta_2=frac1n sum(x_i-bar x)^2,$$
instead.
As for $hat theta_3$, that's just how you start; now solve for $hat theta_3$ (it is a quadratic equation), and then you'll have the estimator.
answered Nov 28 at 0:38
Alejandro Nasif Salum
3,989117
edited Nov 28 at 4:45
answered Nov 28 at 0:38
Alejandro Nasif Salum
3,989117
answered Nov 28 at 0:38
Alejandro Nasif Salum
3,989117
answered Nov 28 at 0:38
Alejandro Nasif Salum
3,989117
3,989117
My translate mistake, so.. $hat{theta}_{1,2}=frac{-1+-sqrt{1-2sum x_i^2}}{2}$
– pin_r
Nov 28 at 3:51
Is not $-2$ but $+4$, and you have to choose from both solutions the one that gives a positive value (since $theta >0$). That is $$hattheta_3=frac{-1+sqrt{1+frac4n sum x_i^2}}2.$$
– Alejandro Nasif Salum
Nov 28 at 4:44
add a comment |
My translate mistake, so.. $hat{theta}_{1,2}=frac{-1+-sqrt{1-2sum x_i^2}}{2}$
– pin_r
Nov 28 at 3:51
Is not $-2$ but $+4$, and you have to choose from both solutions the one that gives a positive value (since $theta >0$). That is $$hattheta_3=frac{-1+sqrt{1+frac4n sum x_i^2}}2.$$
– Alejandro Nasif Salum
Nov 28 at 4:44
My translate mistake, so.. $hat{theta}_{1,2}=frac{-1+-sqrt{1-2sum x_i^2}}{2}$
– pin_r
Nov 28 at 3:51
My translate mistake, so.. $hat{theta}_{1,2}=frac{-1+-sqrt{1-2sum x_i^2}}{2}$
– pin_r
Nov 28 at 3:51
Is not $-2$ but $+4$, and you have to choose from both solutions the one that gives a positive value (since $theta >0$). That is $$hattheta_3=frac{-1+sqrt{1+frac4n sum x_i^2}}2.$$
– Alejandro Nasif Salum
Nov 28 at 4:44
Is not $-2$ but $+4$, and you have to choose from both solutions the one that gives a positive value (since $theta >0$). That is $$hattheta_3=frac{-1+sqrt{1+frac4n sum x_i^2}}2.$$
– Alejandro Nasif Salum
Nov 28 at 4:44
add a comment |
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