Slutsky's Theorem

Definition

Definition

Let \(X_{n} \xrightarrow d X, \ A_{n} \to a, \ B_{n} \to b\) for constants \(a\) and \(b\). Then

\[ A_{n}X_{N} +B_{n} \xrightarrow d aX+b. \]

Here, \(A_{n}, B_{n}\) can be R.V's but must converge to constants.

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Proof

Proof

This is a proof in the special case where we assume that \(a_{n}\) is a sequence of real numbers, \(a>0\), and that every point is a continuity point of the CDF of \(aX\).

The goal is to show that \(a_{n}X_{n} \xrightarrow d aX\).

We get for large enough \(n\), \(a_{n} > 0\) and:

\[ \text{Pr}(a_{n}X_{n} \leq 0) = \text{Pr}(X_{n} \leq 0) \to \text{Pr}(X \leq 0) = \text{Pr}(aX \leq 0). \]

Next, for any \(t\) choose an \(n\) large enough so that \(\left| \frac{a}{a_{n}} -1\right| \leq \epsilon / |t|\), assume \(t>0\):

\[ \begin{align} \text{Pr}(a_{n}X_{n} \leq t) &= \text{Pr}\left( X_{n} \leq \frac{a}{a_{n}} \frac{t}{a} \right) & (1)\\ &\leq \text{Pr}\left( X_{n} \leq \left( 1+\frac{\epsilon}{t} \right) \frac{t}{a}\right) & (2) \\ &= \text{Pr}\left( X_{n} \leq \frac{t+ \epsilon}{a} \right) \to \text{Pr}\left( X \leq \frac{t+ \epsilon}{a} \right) & (3) \\ &=\text{Pr}(aX \leq t + \epsilon) & (4) \end{align} \]

This gives an upper bound. To find the lower bound:

\[ \begin{align} \text{Pr}(a_{n}X_{n} \leq t) &= \text{Pr}\left( X_{n} \leq \frac{a}{a_{n}} \frac{t}{a}\right) & (5) \\ &\geq \text{Pr}\left( X_{n} \leq \left( 1-\frac{\epsilon}{t} \right) \frac{t}{a}\right) & (6) \\ &= \text{Pr}\left( X_{n} \leq \frac{t- \epsilon}{a} \right) \to \text{Pr}\left( X \leq \frac{t-\epsilon}{a} \right) & (7) \\ &= \text{Pr}(aX \leq t-\epsilon) & (8) \end{align} \]

Thus,

\[ \text{Pr}(aX \leq t -\epsilon) \leq \lim\inf_{n} \text{Pr}(a_{n}X_{n} \leq t) \leq \lim \sup_{n}\text{Pr}(a_{n}X_{n} \leq t) \leq \text{Pr}(aX \leq t+ \epsilon) \]

Taking \(\epsilon \to 0\) proves the result.