# How to use the Weak Law of Large Numbers to show this? [closed]

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