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convergence in probability
Let $\{X_{i}\}$ be a sequence of random variables defined on a probability space $(\Omega,\mathcal{F},P)$ taking values in a separable metric space $(Y,d)$, where $d$ is the metric. Then we say the sequence $X_{i}$ converges in probability or converges in measure to a random variable $X$ if for every $\varepsilon>0$,
$\lim_{{i\rightarrow\infty}}P(d(X_{i},X)\geq\varepsilon)=0.$ 
We denote convergence in probability of $X_{i}$ to $X$ by
$X_{i}\stackrel{pr}{\longrightarrow}X.$ 
Equivalently, $X_{i}\stackrel{pr}{\longrightarrow}X$ iff every subsequence of $\{X_{i}\}$ contains a subsequence which converges to $X$ almost surely.
Remarks.

Unlike ordinary convergence, $X_{i}\stackrel{pr}{\longrightarrow}X$ and $X_{i}\stackrel{pr}{\longrightarrow}Y$ only implies that $X=Y$ almost surely.

The need for separability on $Y$ is to ensure that the metric, $d(X_{i},X)$, is a random variable, for all random variables $X_{i}$ and $X$.

Convergence almost surely implies convergence in probability but not conversely.
References
 1 R. M. Dudley, Real Analysis and Probability, Cambridge University Press (2002).
 2 W. Feller, An Introduction to Probability Theory and Its Applications. Vol. 1, Wiley, 3rd ed. (1968).
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Comments
question for 'convergence in probability'
Does one really needs "separable metric space" for this definition? I wonder why?

knowledge can become a science
only with a help of mathematics
Re: question for 'convergence in probability'
In a nutshell, the reason is to make the definition of "convergence in probability" meaningful. In order for the definition to make sense, we want to turn the "distance" between two random variables into a random variable.
Recall that a second countable space is separable (and Lindelof). Conversely, if the separable space is a metric space, it is second countable, which can be proved via an argument using the axiom of choice.
Next, it can be shown that the Borel sigma algebra generated by the product of two second countable spaces via the product topology can be in fact generated by the products of the Borel subsets from the individual spaces.
Finally, since the metric from X*X to R is continuous by definition, it is measurable according to the previous paragraph (the inverse image of a Borel set in R is a Borel set in X*X), and hence a random variable.
Chi
Re: question for 'convergence in probability'
Thanks for the answer! I learned that definition for random variables with values in R, and then it is clear that there is no problem of the <"distance" between two random variables [being] a random variable>.
The only thing which is not so clear for me is why from
> it can be shown that the Borel sigma algebra
> generated by the product of two second countable spaces
> via the product topology can be in fact
> generated by the products of the Borel subsets
> from the individual spaces
it follows that
> the metric from X*X to R is measurable
> (the inverse image of a Borel set in R is a Borel set in X*X)
I feel that this is extremely stupid question, but could you probably just give some more explanation for it? Is it just because the metric is continuous? Thanks ;)
Serg.

knowledge can become a science
only with a help of mathematics
Re: question for 'convergence in probability'
No problem at all. Yes continuity will help.
You can show that a function f from a measurabe space (X,A) to a measurable space (Y,B) is measurable iff f^{1}(U) is in A, for any U in the smallest generating set C of B (B is the sigma algebra generated by C).
In our example, B is the Borel sigma algebra, so it's generated by all open intervals in R. Since f is continuous, f^{1}(U) is open for any open interval U in R. So f^{1}(U) is of course a Borel set in the Borel sigma algebra generated by the open sets in the product space of X*X. So f is measurable (for the Borel sigma algebra).
But this is not enough. A random variable defined on the product of probability spaces requires that it is measurable on the product sigma algebra (a sigma algebra generated by products of sets, each of which a member in its respective sigma algebra), not just Borel sigma algebra. However, if the spaces are separable, the two notions of sigma algebras coicide (the third paragraph in my first email).
I hope this is not causing more confusion.
Chi
Re: question for 'convergence in probability'
> I hope this is not causing more confusion
Thanks, now it is clear. Probably it could be a good idea to write in the entry, that for definition to be meaningfull one need to have that the set
${d(X_i,X)>\varepsilon}$
is an event, and this is really the case if values set Y is separable.
Thanks once more!
Serg.

knowledge can become a science
only with a help of mathematics
Re: question for 'convergence in probability'
Thanks. I will add a little bit of information in the entry. However, I am not going to prove the whole thing. I personally don't like to add too many proofs to the encyclopedia. I think this is where good reference books come in (and I will add the reference also to the entry).
Chi