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!

  • $\begingroup$ 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

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.

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