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I have a random sample of size $X_1, X_2, .., X_n$ following $U(0,2)$. I need to prove that $X_{(n)}$ which is the maximum ordered statistics will converge to $2$ in probability and almost surely.

I know that I can just prove this to converge to 2 almost surely and from there I can make a statement that since it converges almost surely, it will converge in probability and in distribution too.

But the definitions of convergence in probability and almost sure convergence looks identical to me. I could prove that this this maximum ordered statistic will converge to $2$ in probability using definition as follows:

$$\lim_{n \to \infty} P(|X_n-2|<\epsilon) = \lim_{n \to\infty} P(-\epsilon+2<X_n<\epsilon+2) = \lim_{n\to\infty} P(-\epsilon+2<X_n<2)$$

The CDF of the largest observation is $(\frac{x}{2})^n$. Hence, the above probability can be written as:

$$\lim_{n\to\infty} P(-\epsilon+2<X_n<\epsilon+2) = \lim_{n\to\infty}1 - F(2 - \epsilon) = 1$$

Hence, I could see that the largest observation will converge in probability.

How do I go about proving the almost sure convergence. I know the definition. I don't want to write it down again but I cannot separate between this definition and the one I have read for almost sure. It just puts the limit inside the probability statement unlike this definition which puts it before probability.

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Theorem 7.5 from here.
$$\sum_{n=1}^{\infty}P[|X_{(n)}-2|>\epsilon]=\sum_{n=1}^{\infty}P[X_{(n)}<2-\epsilon] =\sum_{n=1}^{\infty}\left(\frac{2-\epsilon}2\right)^n=\frac{2-\epsilon}{\epsilon}$$

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  • $\begingroup$ Hello John. Thanks for the answer. I understand the defition there and also understand your steps. Now, we need to show that the result is less than $\infty$. How can we do that? Since, $\epsilon$ is a small number and hence $\frac{2}{\epsilon}$ will be large. Something large minus 1 is still large and how do we know that it is less than $\infty$ $\endgroup$ – userNoOne Feb 20 at 14:36
  • $\begingroup$ It is large but still finite. That's all that is required in the theorem. It does not have to have finite limit. $\endgroup$ – John L Feb 20 at 18:10
  • $\begingroup$ Okay, Thank you for the reply. It is finite because the terms involved in the final expression involves constants only. Hence, finite. Let me know if I am thinking in wrong direction. $\endgroup$ – userNoOne Feb 21 at 5:08

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