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Let be a random variable $X$ with normal distribution $X\sim N (\mu, \sigma^2)$ and observations $x_1, x_2, · · ·, x_N$ come from a simple random sample.

Prove that $\hat{\mu} = \sum_{n=1}^N \frac{x_n}{N-1}$ is a consistent estimator of the mean.

First, we know that $$ \sum_{n=1}^N \frac{x_n}{N} \xrightarrow{P} \mu. $$

In the other hand, $\lim_{n \to\infty}\frac{N}{N-1} = 1. $

Then by Slutsky theorem, we have:

$$ \sum_{n=1}^N \frac{x_n}{N-1} \xrightarrow{P} \mu $$

This is correct?

Gabriel Granda
  • 129

1 Answers1

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For any $c>0$

$$\sum _{i=1}^n\frac{X_i}n \rightarrow ^p \mu\\ \frac {n}{n-c}\rightarrow ^p 1$$

So

$$\frac{\sum _{i=1}^n X_i}{n-c}\rightarrow^p \mu$$

including for $c=1$. Looks right to me.

Vons
  • 11,004