I'm struggling with the following problem
a) Show that if $X_n \overset{p}\to X$ (convergence in probability) and $Y$ any random variable $X_nY \overset{p}\to XY$
b) Find a case where the previous result isn't true if we change convergence in probability by convergence in distribution.
Thanks in advance for your help.
a), direct method:
For any $c > 0$, you can show $$P(|X_n Y - X Y| > \epsilon, |Y| \ge c) \to 0.$$
Fix $\delta > 0$. For an appropriately chosen $c > 0$, we have $$P(|X_n Y - XY| > \epsilon) \le P(|X_n Y - XY| > \epsilon, |Y| \ge c) + P(0 < |Y| < c) \le \frac{\delta}{2} + \frac{\delta}{2} = \delta$$ for all large $n$.
a), alternate method:
The random vector $(X_n, Y)$ converges to $(X,Y)$ in probability. Thus by the continuous mapping theorem applied to this random vector, $X_n Y \overset{p}{\to} XY$.
b): Let $X$ be a nonzero symmetric random variable, and let $X_n := -X$ and $Y := X$. We have $X_n \overset{d}{\to} -X$ because $X$ is symmetric. However, $X_n Y = -X^2$ while $X Y = X^2$.