Convergence of the expected value of bounded random variables
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Let $X_n$ be a sequence of bounded random variable such that
$$
mathbb{E}left[X_nright]tomathbb{E}left[Xright]
$$
with $X$ a bounded random variable. Can I conclude that
$$
mathbb{E}left[X_n,Wright]tomathbb{E}left[X,Wright]
$$
for any bounded random variable $W$ ?
convergence random-variables expected-value
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add a comment |
$begingroup$
Let $X_n$ be a sequence of bounded random variable such that
$$
mathbb{E}left[X_nright]tomathbb{E}left[Xright]
$$
with $X$ a bounded random variable. Can I conclude that
$$
mathbb{E}left[X_n,Wright]tomathbb{E}left[X,Wright]
$$
for any bounded random variable $W$ ?
convergence random-variables expected-value
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@nicomezi in my counterexample we even have $X_nstackrel{d}{=}X$ for every $n$.
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– drhab
Dec 21 '18 at 11:26
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I meant $X_n overset{P} to X$ indeed, made a confusion. @drhab.
$endgroup$
– nicomezi
Dec 21 '18 at 11:31
add a comment |
$begingroup$
Let $X_n$ be a sequence of bounded random variable such that
$$
mathbb{E}left[X_nright]tomathbb{E}left[Xright]
$$
with $X$ a bounded random variable. Can I conclude that
$$
mathbb{E}left[X_n,Wright]tomathbb{E}left[X,Wright]
$$
for any bounded random variable $W$ ?
convergence random-variables expected-value
$endgroup$
Let $X_n$ be a sequence of bounded random variable such that
$$
mathbb{E}left[X_nright]tomathbb{E}left[Xright]
$$
with $X$ a bounded random variable. Can I conclude that
$$
mathbb{E}left[X_n,Wright]tomathbb{E}left[X,Wright]
$$
for any bounded random variable $W$ ?
convergence random-variables expected-value
convergence random-variables expected-value
asked Dec 21 '18 at 11:13
AlmostSureUserAlmostSureUser
331418
331418
$begingroup$
@nicomezi in my counterexample we even have $X_nstackrel{d}{=}X$ for every $n$.
$endgroup$
– drhab
Dec 21 '18 at 11:26
$begingroup$
I meant $X_n overset{P} to X$ indeed, made a confusion. @drhab.
$endgroup$
– nicomezi
Dec 21 '18 at 11:31
add a comment |
$begingroup$
@nicomezi in my counterexample we even have $X_nstackrel{d}{=}X$ for every $n$.
$endgroup$
– drhab
Dec 21 '18 at 11:26
$begingroup$
I meant $X_n overset{P} to X$ indeed, made a confusion. @drhab.
$endgroup$
– nicomezi
Dec 21 '18 at 11:31
$begingroup$
@nicomezi in my counterexample we even have $X_nstackrel{d}{=}X$ for every $n$.
$endgroup$
– drhab
Dec 21 '18 at 11:26
$begingroup$
@nicomezi in my counterexample we even have $X_nstackrel{d}{=}X$ for every $n$.
$endgroup$
– drhab
Dec 21 '18 at 11:26
$begingroup$
I meant $X_n overset{P} to X$ indeed, made a confusion. @drhab.
$endgroup$
– nicomezi
Dec 21 '18 at 11:31
$begingroup$
I meant $X_n overset{P} to X$ indeed, made a confusion. @drhab.
$endgroup$
– nicomezi
Dec 21 '18 at 11:31
add a comment |
2 Answers
2
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oldest
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No.
Throw a fair coin and for every $n$ let $X_n=1$ if it lands on heads and $X_n=0$ otherwise.
Let $X=1$ if it lands on tails and $X=0$ otherwise.
Then let $W=X$ so that $X_nW=0$ and $XW=X$.
$endgroup$
add a comment |
$begingroup$
Let $X$ have uniform distribution on $(-1,1)$, $X_n=-X$ for all $n$ and $W=X$. Then $EX_n W=-1/3$ for all $n$ and $EXW=1/3$ even though $EX_n =0 to 0=EX$.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No.
Throw a fair coin and for every $n$ let $X_n=1$ if it lands on heads and $X_n=0$ otherwise.
Let $X=1$ if it lands on tails and $X=0$ otherwise.
Then let $W=X$ so that $X_nW=0$ and $XW=X$.
$endgroup$
add a comment |
$begingroup$
No.
Throw a fair coin and for every $n$ let $X_n=1$ if it lands on heads and $X_n=0$ otherwise.
Let $X=1$ if it lands on tails and $X=0$ otherwise.
Then let $W=X$ so that $X_nW=0$ and $XW=X$.
$endgroup$
add a comment |
$begingroup$
No.
Throw a fair coin and for every $n$ let $X_n=1$ if it lands on heads and $X_n=0$ otherwise.
Let $X=1$ if it lands on tails and $X=0$ otherwise.
Then let $W=X$ so that $X_nW=0$ and $XW=X$.
$endgroup$
No.
Throw a fair coin and for every $n$ let $X_n=1$ if it lands on heads and $X_n=0$ otherwise.
Let $X=1$ if it lands on tails and $X=0$ otherwise.
Then let $W=X$ so that $X_nW=0$ and $XW=X$.
answered Dec 21 '18 at 11:22
drhabdrhab
104k545136
104k545136
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$begingroup$
Let $X$ have uniform distribution on $(-1,1)$, $X_n=-X$ for all $n$ and $W=X$. Then $EX_n W=-1/3$ for all $n$ and $EXW=1/3$ even though $EX_n =0 to 0=EX$.
$endgroup$
add a comment |
$begingroup$
Let $X$ have uniform distribution on $(-1,1)$, $X_n=-X$ for all $n$ and $W=X$. Then $EX_n W=-1/3$ for all $n$ and $EXW=1/3$ even though $EX_n =0 to 0=EX$.
$endgroup$
add a comment |
$begingroup$
Let $X$ have uniform distribution on $(-1,1)$, $X_n=-X$ for all $n$ and $W=X$. Then $EX_n W=-1/3$ for all $n$ and $EXW=1/3$ even though $EX_n =0 to 0=EX$.
$endgroup$
Let $X$ have uniform distribution on $(-1,1)$, $X_n=-X$ for all $n$ and $W=X$. Then $EX_n W=-1/3$ for all $n$ and $EXW=1/3$ even though $EX_n =0 to 0=EX$.
answered Dec 21 '18 at 11:51
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
74.4k53270
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$begingroup$
@nicomezi in my counterexample we even have $X_nstackrel{d}{=}X$ for every $n$.
$endgroup$
– drhab
Dec 21 '18 at 11:26
$begingroup$
I meant $X_n overset{P} to X$ indeed, made a confusion. @drhab.
$endgroup$
– nicomezi
Dec 21 '18 at 11:31