Proof verification: Using Chebyshev inequality and weak law of large numbers.

Question

Exercise. Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space, $A \in \mathcal{A}$ and $X_N : \Omega^N \to \mathbb{R}$ a random variable such that $$X_N(\omega_1, \dots, \omega_N) := \frac{1}{N} \sum_{n=1}^{N} \chi_{A}(\omega_n)$$ Show that $\lim_{N \to \infty} \mathbb{P}[|X_N - \mathbb{P}(A)| > \sigma] = 0$ for any $\sigma > 0$.

Attempt. Fix $N$. Then we can write $X_N =\frac{1}{N} (X^1 + \dots + X^N)$ with $X^i : \Omega^N \to \mathbb{R}$ such that $X^{i}$ is equal to $1$ if $\omega_i \in A$ and is zero otherwise. So we have $$ \mathbb{P}(X^i = 1) = \sum_{\omega \in \Omega^N \\ \omega_i \in A } p_{\Omega^N}(\omega) = \sum_{\omega \in A} p_{\Omega}(\omega) = \mathbb{P}(A)$$ Then we get that $\mathbb{E}(X_N) = \frac{1}{N} \big(\mathbb{E}(X^1) + \dots + \mathbb{E}(X^N)\big) = \frac{1}{N} \cdot N \cdot \sum_{\omega \in A} p_{\Omega}(\omega) = \mathbb{P}(A)$ and $\mathbb{V}(X_N) = \mathbb{P}(A) - \mathbb{P}(A)^2$.

Applying Chebyshev inequality, we get the bound $$\mathbb{P}[|X_N - \mathbb{E}(A)| > \sigma] = \mathbb{P}[|X_N - \mathbb{P}(A)| > \sigma] \leq \frac{1}{\sigma^{2}}\mathbb{V}(X_N) = \frac{1}{\sigma^2}\mathbb{P}(A) - \mathbb{P}(A)^{2}$$

Now fix $\sigma > 0$. Since $X^1 + \dots + X^N = \#\text{ Count of $\omega$'s in $A$ } \text{gotten in $N$ tosses}$, $X_N$ is the average and we can apply the weak law of large numbers to get $$\lim_{N \to \infty} \mathbb{P}[|X_N - \mathbb{P}\{X^{i} = 1 \}| > \sigma] = \lim_{N \to \infty} \mathbb{P}[|X_N - \mathbb{P}(A)| > \sigma] = 0.$$

Please let me know if this solution is total garbage! thanks.

Answer 1

The problem is ill-posted.

For, $[|X_{N}-P(A)|>\sigma]$ is a abbreviation of $\{(\omega_{1},\ldots,\omega_{N})\in\Omega^{N}\mid|X_{N}(\omega_{1},\ldots,\omega_{N})-P(A)|>\sigma\},$ which is a subset of $\Omega^{N}$. If $B\subseteq\Omega^{N}$, it is meaningless to talk about $P(B)$. Are you talking about $(P\times\cdots\times P)(B)$ instead?