# Product in convergence

1)If $x_n \to 0$ in probability and $y_n \to 0$ in probability . prove that $x_ny_n \to 0$ in probability

2) if $x_n \to x$ in r-mean and $c_n \to c$ as $n \to \infty$ then $c_n x_n \to cx$ in r-mean

3) if $x_n \to x_0$ in distribution as $n \to \infty$ iff $c_n \to c$ as $n\to \infty$

what idea of this type of product of convergence

I attempt of this but there is no background of this 3 question please help me,, thanx befor what must i prove

in 1)need to show $p(|x_ny_n|)\geq \epsilon ) \to 0 \, as \,n\to$ $\infty$ as convergence in probability

• Ad 1) do you mean xy rather than 0 – zoli Dec 21 '17 at 18:48
• In addition to the typo zoli points out for statement (1), I notice statement (3) makes no sense as written, there must be typos there. Please fix the typos then begin work on statement 1 by writing definitions and stating what you want to show in terms of those definitions. – Michael Dec 21 '17 at 19:07
• no 1 ) i need to show $p(|x_n y_n|\geq \epsilon)$ $\to$ 0 as $n \to \infty$ @Michael – nikola Dec 21 '17 at 19:20
• For (1), I observe that if we fix $\epsilon>0$ then $|X_nY_n|\geq \epsilon$ implies either $|X_n|\geq \sqrt{\epsilon}$ or $|Y_n|\geq \sqrt{\epsilon}$. Can you work from there? [Note: The statement of part (3) still needs fixing. Assuming the convergence we want to show in part (2) is "in $r$-mean" the statement is not true unless we add an assumption like $E[|X_n|^r]\leq b$ for all $n$, for some constant $b$.] – Michael Dec 21 '17 at 19:26
• I agree with @Michael , (2) is not true. Take $\Omega=[0,1]$ with the Lebesgue measure. Take $$X(\omega)=X_n(\omega) =\omega^{-1}\mathbf{1}_{(0,1]}$$. Furthermore take $c_n=n^{-1}$. Then $X_n$ converges to $X$ in $r=1$ mean but $$E[|c_nX_n -cX|]=E[|c_nX_n|]=\infty$$ for all $n$ hence no convergence in $r=1$ mean – Shashi Dec 21 '17 at 19:40

1) $$P(|x_ny_n|>\epsilon)\leq P(\{|x_n|>\sqrt{\epsilon}\} \cup\{|y_n|>\sqrt{\epsilon}\})\leq P(|x_n|>\sqrt{\epsilon})+P(|y_n|>\sqrt{\epsilon}).$$