New answers tagged weak-convergence
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$\Pr(X_n\leq x) \to \Pr(X\leq x)$ for $x\geq 0$ and $\Pr(X_n< x) \to \Pr(X\leq x)$ for $x<0$ imply convergence in distribution?
Fix an $x<0$.
$P(X_{n}\leq x)\leq P(X_{n}<x+\frac{1}{m})\xrightarrow{n\to\infty} P(X\leq x+\frac{1}{m})$ for each $m$
And hence you have $\lim\sup_{n\to\infty}P(X_{n}\leq x)\leq P(X\leq x+\frac{...
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3
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$\Pr(X_n\leq x) \to \Pr(X\leq x)$ for $x\geq 0$ and $\Pr(X_n< x) \to \Pr(X\leq x)$ for $x<0$ imply convergence in distribution?
Take an $x<0$, then
Step I: $\Pr(X_n = x) \to 0$.
To show this note that for every $\epsilon>0$,
$$
0\leq \Pr(X_n = x) \leq \Pr(X_n <x+\epsilon) - \Pr(X_n <x-\epsilon).
$$
Taking $\limsup$,...
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Clarifications about a weakly convergent seaqence
There is no reason behind this, as the claim as written in your question is wrong. (To see that this is not true, take $a_n=b_n$ weakly but not strongly converging to zero.)
In addition, nothing in ...
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1
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Accepted
Weak limit of a bilinear operator
The claim is false generally as indicated in the comments. However, we can show it holds under additional hypotheses on $X$ and/or $Y$.
A Banach space $B$ is said to have the Dunford-Pettis property (...
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