SlutskysTheorem - crowlogic/arb4j GitHub Wiki
Slutsky's Theorem Proof (Modified)
Preliminaries
We want to prove that if $X_n \xrightarrow{d} X$ and $Y_n \xrightarrow{p} Y$, then $X_n + Y_n \xrightarrow{d} X + Y$.
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Convergence in Distribution: $X_n \xrightarrow{d} X$ means that for all continuous bounded functions $g$, $\mathbb{E}[g(X_n)] \to \mathbb{E}[g(X)]$ as $n \to \infty$.
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Convergence in Probability: $Y_n \xrightarrow{p} Y$ implies that for any $\epsilon > 0$, $P(|Y_n - Y| > \epsilon) \to 0$ as $n \to \infty$.
Main Proof
To prove $X_n + Y_n \xrightarrow{d} X + Y$, we need to show:
$$\lim_{n \to \infty} \mathbb{E}[g(X_n + Y_n)] = \mathbb{E}[g(X + Y)]$$
Step 1: Taylor Expansion
Expand $g(X_n + Y_n)$ around $X_n + Y$ using a first-order Taylor series:
$$g(X_n + Y_n) = g(X_n + Y) + (Y_n - Y)g'(X_n + \theta(Y_n - Y))$$
where $\theta$ lies between $Y$ and $Y_n$.
Step 2: Expectation
Take expectation of both sides and apply linearity:
$$\mathbb{E}[g(X_n + Y_n)] = \mathbb{E}[g(X_n + Y)] + \mathbb{E}[(Y_n - Y)g'(X_n + \theta(Y_n - Y))]$$
Step 3: Taking Limits
Since $X_n \xrightarrow{d} X$ and $Y_n \xrightarrow{p} Y$:
$$\lim_{n \to \infty} \mathbb{E}[g(X_n + Y)] = \mathbb{E}[g(X + Y)]$$
And, $Y_n \xrightarrow{p} Y$ implies that $\mathbb{E}[(Y_n - Y)g'(X_n + \theta(Y_n - Y))] \to 0$ (by Dominated Convergence Theorem).
Step 4: Conclusion
Combine these limits to conclude:
$$\lim_{n \to \infty} \mathbb{E}[g(X_n + Y_n)] = \mathbb{E}[g(X + Y)]$$
This completes the proof for the sum. The proofs for the other parts (difference, product, division) are similar.