This is a follow up question to this: Conceptual question about estimators
I am still stuck at showing consistency for the second part.
A random number generator produces uniformly distributed random numbers on the intervall $[0,a]$ where $a>0$ is unknown. We can draw $n$ independent $\mathcal U_{[0,a]}$ random numbers $X_1,...,X_n$ for the estimation. Is:
$$\begin{equation*} \hat \theta =\max{\{X_1,X_2,\ldots,X_n\}} \end{equation*} $$
a consistent estimator? I know that I need to show:
$$\lim_{n \to \infty} \Pr\left(\vert \hat{\theta}-a\vert > \varepsilon\right)=0$$
But I don't really know how. I can show that:
$$E[\hat \theta]=\frac{n}{n+1}a$$
and therefore $E[\hat \theta]-a=\frac{n}{n+1}a-a \not= 0 \implies \text{biased}$
But I am not sure if that helps me show consistency.
Clearly $\hat{\theta}$ depends on $n$, so I denote it as $\hat{\theta}_{n}$. Observe that $\hat{\theta}_{n}\leq a,$ so $\left|\hat{\theta}_{n}-a\right|>\varepsilon$ iff $\hat{\theta}_{n}<a-\varepsilon$ iff $X_{k}<a-\varepsilon$ for $k=1,\ldots,n$. It follows that \begin{eqnarray*} & & \{\omega\in\Omega\mid\left|\hat{\theta}_{n}(\omega)-a\right|>\varepsilon\}\\ & = & \cap_{k=1}^{n}\{\omega\in\Omega\mid X_{k}(\omega)<a-\varepsilon\}. \end{eqnarray*} Since $X_{1},X_{2},\ldots,X_{n}$ are i.i.d., we have \begin{eqnarray*} & & P\left(\{\omega\in\Omega\mid\left|\hat{\theta}_{n}(\omega)-a\right|>\varepsilon\}\right)\\ & = & \prod_{k=1}^{n}P\left(\{\omega\in\Omega\mid X_{k}(\omega)<a-\varepsilon\}\right)\\ & = & \prod_{k=1}^{n}\frac{a-\varepsilon}{a}\\ & = & \left(\frac{a-\varepsilon}{a}\right)^{n}\\ & \rightarrow & 0 \end{eqnarray*} as $n\rightarrow\infty$.
Therefore $\hat{\theta}$ is a consistent estimator for $a$.
Remark: Actually, $\hat{\theta}_{n}\rightarrow a$ pointwisely a.e.. For, we have already proved that $\hat{\theta}_{n}\rightarrow a$ in probability. By a result due to Vitali, there exists a subsequence $\left(\hat{\theta}_{n_{k}}\right)_{k}$ such that $\hat{\theta}_{n_{k}}\rightarrow a$ pointwisely a.e. as $k\rightarrow\infty$. Let $\Omega_{0}=\{\omega\in\Omega\mid\hat{\theta}_{n_{k}}(\omega)\rightarrow a\}$, then $P(\Omega_{0})=1$. Let $\omega\in\Omega_{0}$. Observe that $(\hat{\theta}_{n}(\omega))_{n}$ is monotonic increasing and bounded above by $a$, so $\lim_{n\rightarrow\infty}\hat{\theta}_{n}(\omega)$ exists. In particular, $\lim_{n\rightarrow\infty}\hat{\theta}_{n}(\omega)=\lim_{k\rightarrow\infty}\hat{\theta}_{n_{k}}(\omega)=a$. This shows that $\hat{\theta}_{n}\rightarrow a$ a.e..