Does pointwise convergence of estimator imply consistency












1












$begingroup$


Let $n in mathbb N$ and $Omega=mathbb N^{n}, mathcal{F}=2^{Omega},mathcal{P}:={P_{vartheta}:=operatorname{Geom}(vartheta)^{otimes n}:0<vartheta<1}$



Find the Estimator $hat{vartheta}:Omegato (0,infty)$ where $forall omega in Omega:P_{hat{vartheta}(x)}({omega})=max_{vartheta}P_{vartheta}({omega})$
using the function $f: vartheta mapsto log{(P_{vartheta}({omega}))}$



And then show that the estimator is consistent.



My idea:



$log({P_{vartheta}({omega})})=log(prod_{i=1}^{n}(1-vartheta)^{omega_{i}-1}vartheta)$



Then define $S:=sum_{i=1}^{n}omega_{i}$ and see that



$log(prod_{i=1}^{n}(1-vartheta)^{omega_{i}-1}vartheta)=log((1-vartheta)^{S-n}vartheta^n)=(S-n)log(1-vartheta)+nlog(vartheta)$



It follows that $f'(vartheta)=frac{n}{vartheta}-frac{S-n}{1-vartheta}$ and $f'(vartheta)=0 iff vartheta = frac{n}{S}$ and since $f^{''}(vartheta)<0$ the function is maximized at $vartheta = frac{n}{S}$



So our estimator $hat{vartheta}=frac{n}{S}$.



Now onto my actual problem, on showing that a estimator is consistent. My understanding of a consistent estimator $hat{vartheta}$ of $vartheta$ is that



For any $P_{vartheta} in mathcal{P}$, $hat{vartheta}xrightarrow{n to infty}vartheta(P_{vartheta})$



But how can I test whether $hat{vartheta}$ converges to a parameter if I do not know what parameter $vartheta$ is supposed to be?



Additional questions:



$1.$ Does my definition of consistent estimator: $hat{vartheta}xrightarrow{n to infty}vartheta(P)$ mean that $hat{vartheta}$ converges to $vartheta(P)$ pointwise and thereby almost everywhere?



$2.$ Since I am supposed to choose any $P_{vartheta}in mathcal{P}$, my probability measure already depends on my choice of parameter $vartheta$, so therefore I cannot choose any $P in mathcal{P}$, can I?



$3.$ Does pointwise convergence of an estimator imply cosistent estimator?






probability probability-theory statistics random-variables estimation







share|cite|improve this question









$endgroup$












  • $begingroup$
    Typically, consistency means convergence in probability (so you may use the WLLN for i.i.d sequences).
    $endgroup$
    – d.k.o.
    Jan 25 at 21:14










  • $begingroup$
    So if I can show that $forall omega in Omega,hat{vartheta}_{n}(omega)xrightarrow{n to infty} vartheta(P)(omega)$ for any $P in mathcal{P}$ then this would be convergence almost certainly and thereby convergence in probability?
    $endgroup$
    – MinaThuma
    Jan 25 at 23:47












  • $begingroup$
    How are you going to show pointwise convergence?
    $endgroup$
    – d.k.o.
    Jan 25 at 23:55
















1












$begingroup$


Let $n in mathbb N$ and $Omega=mathbb N^{n}, mathcal{F}=2^{Omega},mathcal{P}:={P_{vartheta}:=operatorname{Geom}(vartheta)^{otimes n}:0<vartheta<1}$



Find the Estimator $hat{vartheta}:Omegato (0,infty)$ where $forall omega in Omega:P_{hat{vartheta}(x)}({omega})=max_{vartheta}P_{vartheta}({omega})$
using the function $f: vartheta mapsto log{(P_{vartheta}({omega}))}$



And then show that the estimator is consistent.



My idea:



$log({P_{vartheta}({omega})})=log(prod_{i=1}^{n}(1-vartheta)^{omega_{i}-1}vartheta)$



Then define $S:=sum_{i=1}^{n}omega_{i}$ and see that



$log(prod_{i=1}^{n}(1-vartheta)^{omega_{i}-1}vartheta)=log((1-vartheta)^{S-n}vartheta^n)=(S-n)log(1-vartheta)+nlog(vartheta)$



It follows that $f'(vartheta)=frac{n}{vartheta}-frac{S-n}{1-vartheta}$ and $f'(vartheta)=0 iff vartheta = frac{n}{S}$ and since $f^{''}(vartheta)<0$ the function is maximized at $vartheta = frac{n}{S}$



So our estimator $hat{vartheta}=frac{n}{S}$.



Now onto my actual problem, on showing that a estimator is consistent. My understanding of a consistent estimator $hat{vartheta}$ of $vartheta$ is that



For any $P_{vartheta} in mathcal{P}$, $hat{vartheta}xrightarrow{n to infty}vartheta(P_{vartheta})$



But how can I test whether $hat{vartheta}$ converges to a parameter if I do not know what parameter $vartheta$ is supposed to be?



Additional questions:



$1.$ Does my definition of consistent estimator: $hat{vartheta}xrightarrow{n to infty}vartheta(P)$ mean that $hat{vartheta}$ converges to $vartheta(P)$ pointwise and thereby almost everywhere?



$2.$ Since I am supposed to choose any $P_{vartheta}in mathcal{P}$, my probability measure already depends on my choice of parameter $vartheta$, so therefore I cannot choose any $P in mathcal{P}$, can I?



$3.$ Does pointwise convergence of an estimator imply cosistent estimator?






probability probability-theory statistics random-variables estimation







share|cite|improve this question









$endgroup$












  • $begingroup$
    Typically, consistency means convergence in probability (so you may use the WLLN for i.i.d sequences).
    $endgroup$
    – d.k.o.
    Jan 25 at 21:14










  • $begingroup$
    So if I can show that $forall omega in Omega,hat{vartheta}_{n}(omega)xrightarrow{n to infty} vartheta(P)(omega)$ for any $P in mathcal{P}$ then this would be convergence almost certainly and thereby convergence in probability?
    $endgroup$
    – MinaThuma
    Jan 25 at 23:47












  • $begingroup$
    How are you going to show pointwise convergence?
    $endgroup$
    – d.k.o.
    Jan 25 at 23:55














1












1








1





$begingroup$


Let $n in mathbb N$ and $Omega=mathbb N^{n}, mathcal{F}=2^{Omega},mathcal{P}:={P_{vartheta}:=operatorname{Geom}(vartheta)^{otimes n}:0<vartheta<1}$



Find the Estimator $hat{vartheta}:Omegato (0,infty)$ where $forall omega in Omega:P_{hat{vartheta}(x)}({omega})=max_{vartheta}P_{vartheta}({omega})$
using the function $f: vartheta mapsto log{(P_{vartheta}({omega}))}$



And then show that the estimator is consistent.



My idea:



$log({P_{vartheta}({omega})})=log(prod_{i=1}^{n}(1-vartheta)^{omega_{i}-1}vartheta)$



Then define $S:=sum_{i=1}^{n}omega_{i}$ and see that



$log(prod_{i=1}^{n}(1-vartheta)^{omega_{i}-1}vartheta)=log((1-vartheta)^{S-n}vartheta^n)=(S-n)log(1-vartheta)+nlog(vartheta)$



It follows that $f'(vartheta)=frac{n}{vartheta}-frac{S-n}{1-vartheta}$ and $f'(vartheta)=0 iff vartheta = frac{n}{S}$ and since $f^{''}(vartheta)<0$ the function is maximized at $vartheta = frac{n}{S}$



So our estimator $hat{vartheta}=frac{n}{S}$.



Now onto my actual problem, on showing that a estimator is consistent. My understanding of a consistent estimator $hat{vartheta}$ of $vartheta$ is that



For any $P_{vartheta} in mathcal{P}$, $hat{vartheta}xrightarrow{n to infty}vartheta(P_{vartheta})$



But how can I test whether $hat{vartheta}$ converges to a parameter if I do not know what parameter $vartheta$ is supposed to be?



Additional questions:



$1.$ Does my definition of consistent estimator: $hat{vartheta}xrightarrow{n to infty}vartheta(P)$ mean that $hat{vartheta}$ converges to $vartheta(P)$ pointwise and thereby almost everywhere?



$2.$ Since I am supposed to choose any $P_{vartheta}in mathcal{P}$, my probability measure already depends on my choice of parameter $vartheta$, so therefore I cannot choose any $P in mathcal{P}$, can I?



$3.$ Does pointwise convergence of an estimator imply cosistent estimator?






probability probability-theory statistics random-variables estimation







share|cite|improve this question









$endgroup$




Let $n in mathbb N$ and $Omega=mathbb N^{n}, mathcal{F}=2^{Omega},mathcal{P}:={P_{vartheta}:=operatorname{Geom}(vartheta)^{otimes n}:0<vartheta<1}$



Find the Estimator $hat{vartheta}:Omegato (0,infty)$ where $forall omega in Omega:P_{hat{vartheta}(x)}({omega})=max_{vartheta}P_{vartheta}({omega})$
using the function $f: vartheta mapsto log{(P_{vartheta}({omega}))}$



And then show that the estimator is consistent.



My idea:



$log({P_{vartheta}({omega})})=log(prod_{i=1}^{n}(1-vartheta)^{omega_{i}-1}vartheta)$



Then define $S:=sum_{i=1}^{n}omega_{i}$ and see that



$log(prod_{i=1}^{n}(1-vartheta)^{omega_{i}-1}vartheta)=log((1-vartheta)^{S-n}vartheta^n)=(S-n)log(1-vartheta)+nlog(vartheta)$



It follows that $f'(vartheta)=frac{n}{vartheta}-frac{S-n}{1-vartheta}$ and $f'(vartheta)=0 iff vartheta = frac{n}{S}$ and since $f^{''}(vartheta)<0$ the function is maximized at $vartheta = frac{n}{S}$



So our estimator $hat{vartheta}=frac{n}{S}$.



Now onto my actual problem, on showing that a estimator is consistent. My understanding of a consistent estimator $hat{vartheta}$ of $vartheta$ is that



For any $P_{vartheta} in mathcal{P}$, $hat{vartheta}xrightarrow{n to infty}vartheta(P_{vartheta})$



But how can I test whether $hat{vartheta}$ converges to a parameter if I do not know what parameter $vartheta$ is supposed to be?



Additional questions:



$1.$ Does my definition of consistent estimator: $hat{vartheta}xrightarrow{n to infty}vartheta(P)$ mean that $hat{vartheta}$ converges to $vartheta(P)$ pointwise and thereby almost everywhere?



$2.$ Since I am supposed to choose any $P_{vartheta}in mathcal{P}$, my probability measure already depends on my choice of parameter $vartheta$, so therefore I cannot choose any $P in mathcal{P}$, can I?



$3.$ Does pointwise convergence of an estimator imply cosistent estimator?






probability probability-theory statistics random-variables estimation




probability probability-theory statistics random-variables estimation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 25 at 15:54









MinaThumaMinaThuma

1968




1968












  • $begingroup$
    Typically, consistency means convergence in probability (so you may use the WLLN for i.i.d sequences).
    $endgroup$
    – d.k.o.
    Jan 25 at 21:14










  • $begingroup$
    So if I can show that $forall omega in Omega,hat{vartheta}_{n}(omega)xrightarrow{n to infty} vartheta(P)(omega)$ for any $P in mathcal{P}$ then this would be convergence almost certainly and thereby convergence in probability?
    $endgroup$
    – MinaThuma
    Jan 25 at 23:47












  • $begingroup$
    How are you going to show pointwise convergence?
    $endgroup$
    – d.k.o.
    Jan 25 at 23:55


















  • $begingroup$
    Typically, consistency means convergence in probability (so you may use the WLLN for i.i.d sequences).
    $endgroup$
    – d.k.o.
    Jan 25 at 21:14










  • $begingroup$
    So if I can show that $forall omega in Omega,hat{vartheta}_{n}(omega)xrightarrow{n to infty} vartheta(P)(omega)$ for any $P in mathcal{P}$ then this would be convergence almost certainly and thereby convergence in probability?
    $endgroup$
    – MinaThuma
    Jan 25 at 23:47












  • $begingroup$
    How are you going to show pointwise convergence?
    $endgroup$
    – d.k.o.
    Jan 25 at 23:55
















$begingroup$
Typically, consistency means convergence in probability (so you may use the WLLN for i.i.d sequences).
$endgroup$
– d.k.o.
Jan 25 at 21:14




$begingroup$
Typically, consistency means convergence in probability (so you may use the WLLN for i.i.d sequences).
$endgroup$
– d.k.o.
Jan 25 at 21:14












$begingroup$
So if I can show that $forall omega in Omega,hat{vartheta}_{n}(omega)xrightarrow{n to infty} vartheta(P)(omega)$ for any $P in mathcal{P}$ then this would be convergence almost certainly and thereby convergence in probability?
$endgroup$
– MinaThuma
Jan 25 at 23:47






$begingroup$
So if I can show that $forall omega in Omega,hat{vartheta}_{n}(omega)xrightarrow{n to infty} vartheta(P)(omega)$ for any $P in mathcal{P}$ then this would be convergence almost certainly and thereby convergence in probability?
$endgroup$
– MinaThuma
Jan 25 at 23:47














$begingroup$
How are you going to show pointwise convergence?
$endgroup$
– d.k.o.
Jan 25 at 23:55




$begingroup$
How are you going to show pointwise convergence?
$endgroup$
– d.k.o.
Jan 25 at 23:55










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