Taking a Limit of a Probability

[Yagoda]

Homework Statement

I want to show that if I have a consistent sequence of estimators W_n for \theta, i.e. \lim_{n \rightarrow \infty} P(|W_n - \theta| < \epsilon) = 1, then U_n = a_nW_n + b_n is also a consistent sequence of estimators for \theta where \lim_{n \rightarrow \infty}a_n = 1 and \lim_{n \rightarrow \infty}b_n = 0.

Homework Equations

The Attempt at a Solution

We are looking at \lim_{n \rightarrow \infty} P(|a_nW_n + b_n - \theta| < \epsilon), which is equivalent to \lim_{n \rightarrow \infty} P(-\epsilon < a_nW_n + b_n - \theta < \epsilon) or \lim_{n \rightarrow \infty} P(-\epsilon - b_n< a_nW_n - \theta < \epsilon - b_n).
My question is how I can apply what I know about the limits of W_n, a_n, b_n in this expression. While I would like to be able to say that since the a_n's go to 1 and the b_n's to 0 I can apply the consistency of Wn, but I don't know if that is acceptable.

 

[haruspex]

Use the definition of limit for the a and b sequences. Given ϵ > 0 there exists N such that...

 

[Ray Vickson]

Homework Statement

I want to show that if I have a consistent sequence of estimators W_n for \theta, i.e. \lim_{n \rightarrow \infty} P(|W_n - \theta| < \epsilon) = 1, then U_n = a_nW_n + b_n is also a consistent sequence of estimators for \theta where \lim_{n \rightarrow \infty}a_n = 1 and \lim_{n \rightarrow \infty}b_n = 0.

Homework Equations

The Attempt at a Solution

We are looking at \lim_{n \rightarrow \infty} P(|a_nW_n + b_n - \theta| < \epsilon), which is equivalent to \lim_{n \rightarrow \infty} P(-\epsilon < a_nW_n + b_n - \theta < \epsilon) or \lim_{n \rightarrow \infty} P(-\epsilon - b_n< a_nW_n - \theta < \epsilon - b_n).
My question is how I can apply what I know about the limits of W_n, a_n, b_n in this expression. While I would like to be able to say that since the a_n's go to 1 and the b_n's to 0 I can apply the consistency of Wn, but I don't know if that is acceptable.

\{-\epsilon - b_n &lt; a_n W_n - \theta &lt; \epsilon -b_n \}<br /> = \left\{ \frac{-\epsilon - b_n + \theta (1 -a_n)}{a_n} <br /> &lt; W_n - \theta &lt; \frac{ \epsilon - b_n + \theta (1-a_n)}{a_n} \right\}