0
$\begingroup$

I have a small confusion about showing strong consistency:

Definition : $T_n$ is strongly consistent for $\theta$ iff $T_n \rightarrow ^{a.s.} \theta$

So how do we show that $\frac{1}{n+a} \sum_{i=1}^{n} x_i$ is a strongly consistent estimator for $\mu$ ? Where a is a fixed known constant

Say we know $\frac{1}{n} \sum_{i=1}^{n} x_i$ is strongly consistent for $\mu$

$|\frac{1}{n} \sum_{i=1}^{n} x_i - \frac{1}{n+a} \sum_{i=1}^{n} x_i| < \epsilon \implies |\frac{a}{n(n+a)} \sum_{i=1}^{n} x_i| < \epsilon$
Given this : $lim_{n \rightarrow \infty}P(|\frac{a}{n(n+a)} \sum_{i=1}^{n} x_i| < \epsilon) = 1 \space \space \forall \epsilon$.

IS this enough? IS there a sleek way to do this?

Thank you for advice!

$\endgroup$
  • $\begingroup$ what is $a$? Though it may be immaterial to the proof itself, it is better to define every symbol in the problem clearly. $\endgroup$ – Zhanxiong Dec 17 '17 at 21:03
2
$\begingroup$

The claim can be proved by using the definition of convergence almost surely. To spell out details, it's helpful to introduce the probability space $(\Omega, \mathscr{F}, P)$ on which the random variables $X_1, X_2, \ldots$ are defined. Technically, instead of "subtract then add", it may be handier to consider "multiply then divide".

Since $\dfrac{1}{n}\sum_{i = 1}^n X_i$ converges to $\mu$ almost surely, there exists a set $S \in \mathscr{F}$ with $P(S) = 1$ such that for each $\omega \in S$, $$\frac{1}{n}\sum_{i = 1}^n X_i(\omega) \to \mu$$ as $n \to \infty$.

Consequently, for each $\omega \in S$, $$\frac{1}{n + a}\sum_{i = 1}^n X_i(\omega) = \frac{n}{n + a}\times \frac{1}{n}\sum_{i = 1}^nX_i(\omega) \to 1 \times \mu = \mu$$ as $n \to \infty$ (provided $a$ is a fixed number independent of $n$). Note this precisely means $\dfrac{1}{n + a}\sum_{i = 1}^n X_i$ converges to $\mu$ almost surely and the proof is complete.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.