<p>The definition of &quot;consistent&quot; is that the estimator approaches the truth as the sample size grows large. The parameter of interest is presumably <span class="math-container">$\mu_y$</span>, the population expected value of <span class="math-container">$y$</span>.</p>
<p>In this case, you just need to make assumptions on <span class="math-container">$y_i$</span> to determine if there is consistency.</p>
<p>If <span class="math-container">$\mu_y$</span> exists (not all distributions have a finite mean) and your data are independent and identically distributed, then by Kolmogorov's second strong law of large numbers,</p>
<p><span class="math-container">$$\frac{1}{n}\sum_{i=1}^ny_i \rightarrow^{a.s.} \mu_y $$</span></p>
<p>In English, the estimator is consistent. The superscript &quot;a.s.&quot; denotes almost sure convergence.</p>

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2022-10-30T19:20:08.100Z
Thank you for answering all of my questions!
Sera
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2022-10-30T17:40:52.137Z
2022-10-30T17:40:52.137Z
Michael Gmeiner
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<p>Is it consistent? How can we derive whether an estimator of the parameter is consistent or not?</p>

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2022-10-30T16:15:55.410Z
It is inconsistent for the variance
Bertrand
<econometrics><statistics>
Consistency of an estimator
Sera
# Consistency of an estimator

(Asked by: username_0 on 2022-10-30)

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Is it consistent? How can we derive whether an estimator of the parameter is consistent or not?

**Tags:** econometrics, statistics
> **Comments:**
> **username_1**: It is inconsistent for the variance  

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## 1 answers

### Answer by username_2 on 2022-10-30. Score: 3 | Accepted answer

The definition of "consistent" is that the estimator approaches the truth as the sample size grows large. The parameter of interest is presumably $\mu\_y$, the population expected value of $y$.

In this case, you just need to make assumptions on $y\_i$ to determine if there is consistency.

If $\mu\_y$ exists (not all distributions have a finite mean) and your data are independent and identically distributed, then by Kolmogorov's second strong law of large numbers,

$$\frac{1}{n}\sum\_{i=1}^ny\_i \rightarrow^{a.s.} \mu\_y $$

In English, the estimator is consistent. The superscript "a.s." denotes almost sure convergence.

> **Comments:**
> **username_0**: Thank you for answering all of my questions!  

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