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I am trying to come up with an example of a sequence of random variables $|Y_n|\leq 1$ converging in probability to $0$, but where the variance $Var(Y_n)$ does not converge to $0$ in limit. Is there a way I can use a transformation of a variable $Y_n ~ Bern\left(\frac{1}{n}\right)$?

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    $\begingroup$ Doesn't such example contradict dominated convergence theorem for convergence in probability? $\endgroup$ – Shashi Oct 17 at 20:57
  • $\begingroup$ Which part does it contradict? thanks! $\endgroup$ – user321627 Oct 17 at 21:25
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    $\begingroup$ I mean the variance should go to zero by DCT... So you cannot find an example $\endgroup$ – Shashi Oct 17 at 21:31
  • $\begingroup$ How could I prove the variance goes to zero by the DCT? $\endgroup$ – user321627 Oct 17 at 22:11
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Such example cannot exist.

Why? Well, let $$a_n:=\operatorname{Var}(Y_n) $$ Take a subsequence $a_{n_k} $, note that $Y_{n_k} $ still converges in probability to zero. It has a further subsequence $Y_{n_{k_j}}$ converging almost surely to zero. Since $|Y_{n_{k_j}}|\leq 1$ a.s., we have $|Y_{n_{k_j}}^2|\leq 1$ a.s.. Hence by DCT $$a_{n_{k_j}} = \mathbb E(Y_{n_{k_j}}^2)-\mathbb E(Y_{n_{k_j}}) ^2\to 0$$ as $j\to\infty$.

So every subsequence of $a_n$ has a further subsequence converging to zero. So the sequence $a_n$ converges to zero by this fact.

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Reversed binomial distribution described bellow worked.

EDIT: The $Y_m$ I constructed may not satisfy the condition "converging in probability to $0$". It may be modified but I couldn't find precisely what "converging in probability to $0$" means.

EDIT: I found the definition of "converging in probability to $0$" in the Wikipedia and that my example does not work.

(My trial was: Let $P( X_m=j/m ) = (1/4)^m * (2m)!/(m-j)!(m+j)!$ for $-m\leq j\leq m$,
for $-m< j<0$, $P( Y_m=-1-j/m ) = P( X_m=j/m )$
for $0<j< m, P( Y_m=1-j/m ) = P( X_m=j/m )$
for $j=-m,0,m, P( Y_m=j/m ) = P( X_m=j/m ) $)

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