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Let $$X_n=\begin{cases} 1/n, &\text{ with prob } 1-1/n\\ n^2, &\text{ with prob } 1/n.\end{cases}.$$ $$Y_{n}=\begin{cases} 0, &\text{ with prob } 1-1/n\\ n, &\text{ with prob } 1/n.\end{cases}.$$

We want to show that $X_n \rightarrow 0$ in probability, which means showing that $\lim_{n\rightarrow\infty}P(|X_n-0|>\epsilon)=0$ for all $\epsilon>0$. And similarly for showing $Y_n \rightarrow 0$ in probability.

The solutions I've seen to these questions go as follows:

Whenever $\epsilon>1/n$, we have $0\leq P(|X_n-0|>\epsilon)=P(X_n>1/n)=P(X_n=n^2)=1/n$. The result follows by letting $n\rightarrow \infty$ and using the Sandwich theorem.

And for the second one (https://youtu.be/x4q6H6lxFFE?t=594):

Whenever $\epsilon<n$, we have $0\leq P(|Y_n-0|\geq \epsilon)=1/n$. The result follows by letting $n\rightarrow \infty$ and using the Sandwich theorem.

Simple enough, but why are we allowed to put conditions on $\epsilon$? I thought $\epsilon$ was just any positive number. Also even though the examples are really similar, in one case it's $\epsilon>1/n$ and in the other it's $\epsilon<n$. Is that just because one definition allows for $> \epsilon$ and the other is $\geq\epsilon$?

Edit: On second glance I'm confused as to why the second proof has $\epsilon<n$? Shouldn't it just be $\epsilon>0$ as then $P(|Y_n|>\epsilon)=P(Y_n>0)=P(Y_n=n)=1/n$?

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This is not a condition to $\epsilon$. This is a condition to $n$. Read it as follows:

Whenever $n>1/\epsilon$, we have $$0\leq P(|X_n−0|>\epsilon)=P(X_n>1/n)=P(X_n=n^2)=1/n$$

For any $\epsilon>0$ and for $n\to\infty$ the moment when $n$ became greater than $1/\epsilon$ will definitely come. Moreover, since we find limits, we are not interested in the behaviour of first elements in the sequence of $P(|X_n−0|>\epsilon)$. We can only check how this probability looks like for sufficiently large $n$.

For the second question: Sure, $\epsilon>0$ by definition of convergence in probability, and for the case when the values of $Y_n$ are either zero or $n$, to have $Y_n>\epsilon$ iff $Y_n=n$ you should have $n>\epsilon$.

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  • $\begingroup$ Thanks a lot! And ah, I had just edited my post before your reply when I realised $\epsilon>n$ would make no sense. Can we not just have the $\epsilon>0$ condition though as in my last line? $\endgroup$ – Alex.F Mar 26 '17 at 1:16
  • $\begingroup$ I edited the answer correspondingly. $\endgroup$ – NCh Mar 26 '17 at 1:18
  • $\begingroup$ Am I right in thinking $n>\epsilon$ and $\epsilon>0$ are equivalent conditions in the $Y_n$ case then? $\endgroup$ – Alex.F Mar 26 '17 at 1:28
  • $\begingroup$ None. Look at answer: $P(Y_n>\epsilon)=P(Y_n=n)$ iff $n > \epsilon$. $\endgroup$ – NCh Mar 26 '17 at 1:33
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    $\begingroup$ You are right. The author of the solution subconsciously assumed that $\epsilon$ is small, and for this case $n>1/\epsilon$ is autometically greater than $\sqrt{\epsilon}$. The reason to do so exists: if you prove that $P(|X_n-0|>\epsilon)\to 0$ for small $\epsilon$, it then holds for large $\epsilon$ due to monotonicity of probability. $\endgroup$ – NCh Mar 26 '17 at 2:38

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