# Does pointwise convergence of estimator imply consistency

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 $$\hat{\vartheta}:\Omega\to (0,\infty)$$ 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

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

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?

• Typically, consistency means convergence in probability (so you may use the WLLN for i.i.d sequences). – d.k.o. Jan 25 at 21:14
• 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? – MinaThuma Jan 25 at 23:47
• How are you going to show pointwise convergence? – d.k.o. Jan 25 at 23:55