v1.5.0(00734247): "Testing"

Definition (Continuous function)

A function $f: X \rightarrow Y$ between topological spaces is continuous if for every open set $V \subseteq Y$ , the preimage $f^{-1}(V) \subseteq X$ is open in $X$ .

Explanation (Equivalent formulations)

For metric spaces, continuity can be characterized by: $\forall \varepsilon > 0, \exists \delta > 0$ such that $d_X(x,y) < \delta \implies d_Y(f(x),f(y)) < \varepsilon$ . This captures the intuition that points close to each other in $X$ map to points close to each other in $Y$ .

Theorem (Law of Large Numbers)

Let $X_1, X_2, \ldots$ be a sequence of independent and identically distributed random variables with expected value $\mathbb{E}[X_i] = \mu < \infty$ . Then for any $\varepsilon > 0$ :

\lim_{n \to \infty} P\left(\left|\frac{1}{n}\sum_{i=1}^{n}X_i - \mu\right| > \varepsilon\right) = 0

Proof (Proof)

We’ll prove this using Chebyshev’s inequality. Let $S_n = \sum_{i=1}^{n}X_i$ and $\sigma^2 = \text{Var}(X_i)$ . By Chebyshev’s inequality:

P\left(\left|\frac{S_n}{n} - \mu\right| \geq \varepsilon\right) \leq \frac{\text{Var}(S_n/n)}{\varepsilon^2}

Since the variables are independent, we have:

\text{Var}(S_n/n) = \frac{\text{Var}(S_n)}{n^2} = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}

Substituting this into our inequality:

P\left(\left|\frac{S_n}{n} - \mu\right| \geq \varepsilon\right) \leq \frac{\sigma^2}{n\varepsilon^2}

As $n \to \infty$ , the right side approaches 0, which proves the theorem.

Lemma (Monotone Convergence Theorem)

Let $(f_n)$ be a sequence of non-negative measurable functions such that $f_n(x) \leq f_{n+1}(x)$ for all $n \in \mathbb{N}$ and almost all $x$ . Define $f(x) = \lim_{n \to \infty} f_n(x)$ . Then:

\lim_{n \to \infty} \int f_n \, d\mu = \int f \, d\mu

Proof (Proof)

Let $g_n = f_n \cdot \chi_E$ where $E = \{x : f(x) < \infty\}$ . By Fatou’s lemma:

\int f \, d\mu = \int \lim_{n \to \infty} g_n \, d\mu \leq \liminf_{n \to \infty} \int g_n \, d\mu = \liminf_{n \to \infty} \int f_n \, d\mu

For the reverse inequality, note that $f_n \leq f$ for all $n$ , so $\int f_n \, d\mu \leq \int f \, d\mu$ . Taking the limit:

\limsup_{n \to \infty} \int f_n \, d\mu \leq \int f \, d\mu

Combining these inequalities:

\int f \, d\mu \leq \liminf_{n \to \infty} \int f_n \, d\mu \leq \limsup_{n \to \infty} \int f_n \, d\mu \leq \int f \, d\mu

Therefore, $\lim_{n \to \infty} \int f_n \, d\mu = \int f \, d\mu$ .

Exercise (Finding the derivative of a product function)

Calculate the derivative of $f(x) = x^3 \sin(x)$ using the product rule.

Answer

\frac{d}{dx}[x^3 \sin(x)] = 3x^2 \sin(x) + x^3 \cos(x)

Problem (Convergence of arithmetic means)

Prove that if a sequence $(a_n)$ converges to $L$ , then the sequence of arithmetic means $(\frac{a_1 + a_2 + \ldots + a_n}{n})$ also converges to $L$ .

Solution (Detailed proof)

Let $\varepsilon > 0$ be given. Since $(a_n)$ converges to $L$ , there exists $N \in \mathbb{N}$ such that $|a_n - L| < \frac{\varepsilon}{2}$ for all $n \geq N$ . Let $S_n = \frac{a_1 + a_2 + \ldots + a_n}{n}$ be the sequence of arithmetic means.

We can split $S_n$ as follows:

S_n = \frac{a_1 + a_2 + \ldots + a_N + a_{N+1} + \ldots + a_n}{n}

For $n > N$ , we have:

\begin{align*} |S_n - L| &= \left|\frac{a_1 + a_2 + \ldots + a_n}{n} - L\right| \\ &= \left|\frac{a_1 + a_2 + \ldots + a_n - nL}{n}\right| \\ &= \left|\frac{(a_1 - L) + (a_2 - L) + \ldots + (a_n - L)}{n}\right| \\ &\leq \frac{|a_1 - L| + |a_2 - L| + \ldots + |a_N - L| + |a_{N+1} - L| + \ldots + |a_n - L|}{n} \end{align*}

Let $M = \max\{|a_1 - L|, |a_2 - L|, \ldots, |a_N - L|\}$ . Then:

|S_n - L| \leq \frac{NM + (n-N)\frac{\varepsilon}{2}}{n} = \frac{NM}{n} + \frac{n-N}{n} \cdot \frac{\varepsilon}{2}

As $n \to \infty$ , $\frac{NM}{n} \to 0$ and $\frac{n-N}{n} \to 1$ . So for sufficiently large $n$ , we have:

|S_n - L| < \varepsilon

Therefore, the sequence of arithmetic means converges to $L$ .

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