Probability of absolute difference

[safina]

Homework Statement

Let there be m independent samples, each of size n, and are drawn from a population of size N using a design D(S,P). Based on the ith sample i = 1, 2, ..., m, let $$\hat{\theta_{i}}$$(s) denote an unbiased estimator of $$\theta$$. Let $$\hat{\theta}$$ = $$\frac{1}{m}$$$$\sum^{m}_{i=1}\hat{\theta_{i}}(s)$$.
Show that for any $$\epsilon$$ > 0, P$$\left\{$$|$$\hat{\theta_{i}}$$(s) -$$\theta$$| > $$\epsilon\right\}$$ $$\leq$$ $$\frac{V[\hat{\theta_{i}}(s)]}{\epsilon^{2}}$$.

Homework Equations

The Attempt at a Solution

 

[lippyka]

Since \hat{\theta_{i}}(s) is unbiased, we can write E[\hat{\theta_{i}}(s)] = \theta. So, P\left\{|\hat{\theta_{i}}(s) -\theta| > \epsilon\right\} = P\left\{\hat{\theta_{i}}(s) -\theta > \epsilon\right\} + P\left\{\theta - \hat{\theta_{i}}(s) > \epsilon\right\} = P\left\{\hat{\theta_{i}}(s) -E[\hat{\theta_{i}}(s)] > \epsilon\right\} + P\left\{E[\hat{\theta_{i}}(s)] - \hat{\theta_{i}}(s) > \epsilon\right\}= 2P\left\{\hat{\theta_{i}}(s) -E[\hat{\theta_{i}}(s)] > \epsilon\right\}. Now, by Chebychev's Inequality, 2P\left\{\hat{\theta_{i}}(s) -E[\hat{\theta_{i}}(s)] > \epsilon\right\} \leq \frac{2V[\hat{\theta_{i}}(s)]}{\epsilon^{2}}. Therefore,P\left\{|\hat{\theta_{i}}(s) -\theta| > \epsilon\right\} \leq \frac{V[\hat{\theta_{i}}(s)]}{\epsilon^{2}}.