# Taking a Limit of a Probability

1. Feb 10, 2014

### Yagoda

1. The problem statement, all variables and given/known data 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 $\lim_{n \rightarrow \infty}b_n = 0.$

2. Relevant equations

3. 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 an's go to 1 and the bn's to 0 I can apply the consistency of Wn, but I don't know if that is acceptable.

2. Feb 10, 2014

### haruspex

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

3. Feb 10, 2014

### Ray Vickson

$$\{-\epsilon - b_n < a_n W_n - \theta < \epsilon -b_n \} = \left\{ \frac{-\epsilon - b_n + \theta (1 -a_n)}{a_n} < W_n - \theta < \frac{ \epsilon - b_n + \theta (1-a_n)}{a_n} \right\}$$