Slutsky's Theorem Proof (Modified)

Preliminaries

We want to prove that if $X_n \xrightarrow{d} X$ and $Y_n \xrightarrow{p} Y$, then $X_n + Y_n \xrightarrow{d} X + Y$.

1. 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 $n \to \infty$.

2. 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.