1. ## Convergence in probability

Let be $X_1, X_2, ..., X_n, Y_1, Y_2, ..., Y_n$ random variables on the same $(\Omega, \mathbf{F}, \mathbb{P})$ probabilty field. Assume that $X_n \buildrel P \over \rightarrow X$ and $Y_n \buildrel P \over \rightarrow Y$ (denotes convergence in probabilty).
Prove that:
a) $X_nY_n$ $\buildrel P \over \rightarrow XY$.
b) If $Y_n \neq 0, Y \neq 0$ (with probabilty 1), then $X_n/Y_n$ $\buildrel P \over \rightarrow X/Y$.

I would really appreciate if you could help me.

2. ## Re: Convergence in probability

You might try one of the standard mathematical tricks: Do nothing, but do it in a smart way!

In this case one such trick is to write

$0 = X_{n}Y-X_{n}Y$

and insert this in the salient difference

$|X_{n}Y_{n}-XY| \ .$

Invoking the Triangle inequality you can write

$|X_{n}Y_{n}-XY| \leq |X_{n}-X||Y| + |Y_{n}-Y||X_{n}| \ .$

If the left-hand side of this inequality is larger than some small positive number $\varepsilon\ ,$ then at least one of the two terms on the right-hand side of the inequality is greater than $\varepsilon/2\ .$

3. ## Re: Convergence in probability

Originally Posted by Alibiki
You might try one of the standard mathematical tricks: Do nothing, but do it in a smart way!

In this case one such trick is to write

$0 = X_{n}Y-X_{n}Y$

and insert this in the salient difference

$|X_{n}Y_{n}-XY| \ .$

Invoking the Triangle inequality you can write

$|X_{n}Y_{n}-XY| \leq |X_{n}-X||Y| + |Y_{n}-Y||X_{n}| \ .$

If the left-hand side of this inequality is larger than some small positive number $\varepsilon\ ,$ then at least one of the two terms on the right-hand side of the inequality is greater than $\varepsilon/2\ .$