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I tried to solve a problem two different ways and I got different results.

Let $( X_i )_{i \in \mathbb{N}}$ be a series of independent, identically distributed random variables, with $\mathbb{E}[X_i] = 1$ and $\mathbb{V}[X_i] = 1$

Determine

$$ \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right) $$

Here are the two approaches that I tried.

Central limit theorem \begin{align*} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right) \\ = {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i - 1) \leq \sqrt{n} - \frac{n}{\sqrt{n}}\right) \\ = {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i - 1) \leq 0\right) \\ = {} & \Phi_{0,1}(0) = \frac{1}{2} \end{align*}

Law of large numbers

\begin{align*} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right) \\ = {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{n} \sum_{i=1}^n X_i \leq 1 \right) \\ \geq {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{n} \sum_{i=1}^n X_i = 1\right) \\ = {} & 1 \end{align*} according to the strong law of large numbers. This then means that $$ \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right) = 1 $$

What am I doing wrong here? My understanding is that the CLT solution is correct, but I don't see what I did wrong with applying the law of large numbers either.

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If e.g. $X_1$ has continuous distribution then so has $\sum_{i=1}^nX_i$ for every $n$.

Consequence: $$\mathsf P(\frac1n\sum_{i=1}^nX_i=1)=0\text{ for every }n$$ so that also: $$\lim_{n\to\infty}\mathsf P(\frac1n\sum_{i=1}^nX_i=1)=0\neq1$$

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  • $\begingroup$ How does this relate to the strong law of large numbers? $\endgroup$ – Alexguitar Jun 24 '18 at 13:03
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    $\begingroup$ That law says here that $P(\lim_{n\to\infty}\frac1n\sum_{i=1}^nX_i=1)=1$. That is essentially different from $\lim_{n\to\infty}P(\frac1n\sum_{i=1}^nX_i=1)=1$. $\endgroup$ – drhab Jun 24 '18 at 13:07
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What the (strong) law of large numbers says: $$ \require{cancel} \begin{align} \text{right: } & \Pr\left( \lim_{n\to\infty} \frac 1 n \sum_{i=1}^n X_i = 1\right) = 1. \\ \\ \text{wrong: } & \xcancel{\lim_{n\to\infty} \Pr\left( \frac 1 n \sum_{i=1}^n X_i = 1\right) = 1.} \end{align} $$

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    $\begingroup$ I think it is worth noting that the wrong interpretation of the law of large numbers is quite clearly wrong if you think about it: just think of a series of die rolls. The chance of getting exactly the average value is rapidly dropping as you keep rolling. And that is for discrete variables. For continuous, it is zero right from the start. $\endgroup$ – tomasz Jun 24 '18 at 20:36

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