2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Ad 1) do you mean xy rather than 0 $\endgroup$ – zoli Dec 21 '17 at 18:48
  • $\begingroup$ 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. $\endgroup$ – Michael Dec 21 '17 at 19:07
  • 1
    $\begingroup$ no 1 ) i need to show $p(|x_n y_n|\geq \epsilon)$ $\to$ 0 as $n \to \infty$ @Michael $\endgroup$ – nikola Dec 21 '17 at 19:20
  • 1
    $\begingroup$ 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 $b$.] $\endgroup$ – Michael Dec 21 '17 at 19:26
  • 1
    $\begingroup$ 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 $\endgroup$ – Shashi Dec 21 '17 at 19:40
2
$\begingroup$

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

Can you finish it from here?

$\endgroup$
  • $\begingroup$ To spread holiday cheer, it may help to acknowledge others who have made earlier comments that are similar. [ I have given you a +1 just now. ] $\endgroup$ – Michael Dec 21 '17 at 19:48
  • $\begingroup$ hhhh ... my study is Accumulated ؛ iam from palestine there is no holiday for me after two weeks my exam will start@Michael but iam Surprised why my doctor take no 1 and 2 to prove it in homework $\endgroup$ – nikola Dec 21 '17 at 19:59
  • $\begingroup$ it seem that no 2 ) the correct formula it is math.stackexchange.com/questions/317706/… $\endgroup$ – nikola Dec 21 '17 at 20:11
  • $\begingroup$ @nikola : Best wishes on your exam! Hopefully you will get some vacation time after. $\endgroup$ – Michael Dec 22 '17 at 8:53
  • $\begingroup$ thnx alot ..... you are kind! $\endgroup$ – nikola Dec 22 '17 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.