Let $X_1,...,X_n$ be an iid (independent and identically distributed) sample with mean $ \mu $ and variance $\sigma^2$.

We can use this conclusion :$ (n-1)S^2 = \sum_{i=1}^n (Xi-\overline X)^2 = \sum_{i=1}^n (Xi-\mu ) ^2 - n(\mu-\overline X) ^2 $

Suppose that $\mathbb E(X_i-\mu)^4<\infty$, and use the Weak Law of Large Numbers to show that

$ S^2 \rightarrow \sigma^2$ (in probability) as $n \rightarrow \infty $

My questions are:

  1. How do I go about showing that?

  2. Why suppose $\mathbb E(X_i-\mu)^4<\infty$? (The second may be included in the first)


closed as off-topic by Alexander Gruber Apr 28 at 8:30

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  • 1
    $\begingroup$ What have you tried so far? $\endgroup$ – Olly Reynolds Apr 22 at 12:14
  • $\begingroup$ @Jamesodare I really had no idea . $\endgroup$ – gong.y Apr 22 at 12:25

Hint: Since $\ \left(X_i-\mu\right)^2, i=1,2,\dots $ are iid random variables, the weak law of large numbers applies to them, provided they have finite variance. That is, under the stated condition, $\ \frac{1}{n}\sum_{i=1}^n (Xi-\mu )^2\ =\left(1-\frac{1}{n}\right)S^2+$ $\left(\mu-\overline X\right)^2\ $ converges in probability to $\ \mathbb{E}\left(Xi-\mu \right)^2=\sigma^2\ $, and you also know that $\ \left(\mu-\overline X\right)^2 $ converges to $\ 0\ $ in probability. So, what is the relation between the variance of $\ \left(X_i-\mu\right)^2\ $ and $\ \mathbb{E}\left(Xi-\mu \right)^4\ $?


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