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Feb 20, 2020 at 17:57 answer added Thiago Bittencourt timeline score: 0
S Jan 29, 2020 at 17:23 history bounty ended an1lam
S Jan 29, 2020 at 17:23 history notice removed an1lam
Jan 28, 2020 at 14:03 vote accept an1lam
Jan 28, 2020 at 5:43 answer added Anon timeline score: 4
Jan 22, 2020 at 18:00 history tweeted twitter.com/StackStats/status/1220043606603550720
S Jan 22, 2020 at 16:34 history bounty started an1lam
S Jan 22, 2020 at 16:34 history notice added an1lam Draw attention
Dec 29, 2019 at 23:31 comment added Sean Roberson Precisely. I'll do a full solution when I can figure the second part.
Dec 29, 2019 at 22:28 comment added an1lam Oh I think I see now: you can let the $ Y $ variable in Cauchy-Schwarz just be $ 1 $ and then you get $$ \mathbb{E}(\lvert X_n - b \rvert) \leq \sqrt{\mathbb{E}(X_n - b)^2} $$ which goes to $ 0 $ by assumption.
Dec 26, 2019 at 20:49 comment added an1lam Maybe I'm just being dense but I don't see how the Cauchy-Schwarz inequality helps with the inequality chain you started. The probability version of Cauchy-Schwarz that I'm familiar with is $ \mathbb{E}(X^2Y^2) \leq \mathbb{E}(X^2) \mathbb{E}(Y^2) $. How does that relate to $ \mathbb{E} \lvert X_n - b \rvert \leq \dots $?
Dec 25, 2019 at 17:06 comment added an1lam Ah I see. I have to check but I think we can assume the limits exist (as in it's in the problem statement).
Dec 25, 2019 at 5:27 comment added Sean Roberson Also: your last line only holds if all limits exist and are finite. We don't know if $E(X_n)$ exists!
Dec 25, 2019 at 5:12 comment added Sean Roberson Just a hint: If $E(X_n) \to b$ in $L^2$, then we can have $\int |X_n - b| \ dP \leq \ldots$ Use Cauchy-Schwarz!
Dec 25, 2019 at 3:40 review Close votes
Dec 25, 2019 at 12:45
Dec 25, 2019 at 3:05 review First posts
Dec 25, 2019 at 3:20
Dec 25, 2019 at 3:00 history asked an1lam CC BY-SA 4.0