Example of covergence in probability not constant

[newbie]

Hi, this is not homework. I am studying for an exam and try to build up some intuition. Can someone give me an example of a situation where a sequence of random variables converges in probability to a random variable but the variable to which they converge is not a constant.

[Giallo]

I don't know if you mean what I understood, but take any infinite sequence of variables X_i with the same mean m and variance o², then the central limit theorem says that the sum of those variables converges to a normal distribution N(m,o²).

[Dason]

It might sound trivial but it does the trick:

Let X ~ Bin(n, p). Let X_i = X for all i. Then {X_i} converges in probability to X.

Edit:

For a slightly less trivial example consider X_i = X + 1/i. Then {X_i} converges in probability to X as well.

@Giallo - That sounds like convergence in distribution and not necessarily in probability.

[newbie]

Thanks, I think that would be consistent with our definition of convergence in distribution. For my class convergence in probability is \mathop {\lim }\limits_{n \to \infty } P\left( {\left| {{\xi _n} - \xi } \right| > \varepsilon } \right) = 0\forall \varepsilon > 0 where \xi_n,\xi are random variables.

[newbie]

Our posts crossed in mid air. my response above was to the first response.

[newbie]

Ok, thanks. I must be missing some important fact. Using your second example. Let's say X is a Bernoulli trial with p of success = 1/2. Then it would seem that for n large, you still have p=1/2 of |\xi_n-\xi|>\epsilon for epsilon less than 1. Is there another way of writing the definition where we refer to the Borel sets in the range?

[Dason]

No - the probability of the difference being larger than epsilon goes to 0 since for an epsilon we can find an n such that for i>n, 1/i < epsilon. So in that case

|X_i - X| = |X + 1/i - X| = |1/i| and we can get that to be smaller than any epsilon if we let i get large enough. So it does in fact converge in probability.

Edit: I think the hard part to grasp here is that I'm not taking X_i to be some unspecified bernoulli + 1/i. I'm taking them to be the random variable we are converging to plus 1/i.

[newbie]

Thanks for responding but I'm still a little confused. Is this equivalent to \left| {{X_i}\left( \omega \right) - X\left( \omega \right)} \right| = \left| {X\left( \omega \right) + 1/i - X\left( \omega \right)} \right|\forall \omega \in \Omega.
Sorry if I am being dense.

[Dason]

That's exactly it. I was getting a little lazy with the notation and wasn't sure how deep into the rabbits hole you've gone. So yes that is it and you should see that the X(\omega) cancels and you're just left with 1/i in there which you can make arbitrarily small so we have convergence in probability.

Really convergence in probability is mostly used for convergence to constants though and is then used along with Slutsky's theorem to do something more interesting with convergence in distribution.

[newbie]

Thanks, got it now.