If $X_n$ converges in distribution to $X$ and $E[X^2]$ is finite, could we have $E[X_n^2]$ is finite?
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I only know 1) $X_n$ converges in distribution to $X$; 2) $X_n$ is bounded by a constant $M$, and 3) $mathbb{E}[X^2]$ is finite, could I get $mathbb{E}[X_n^2]$ is finite?
Moreover, if $mathbb{E}[X^k]$ is finite, could we have $mathbb{E}[X_n^k]$ is finite?
I know that we CANNOT get $mathbb{E}[X_n]rightarrow mathbb{E}[X]$. But could we get $mathbb{E}[X_n]$ is finite if $mathbb{E}[X]$ is finite?
Thank you.
probability-distributions convergence
asked Jan 11 at 5:03
Alex LiuAlex Liu
31
$endgroup$
add a comment |
$begingroup$
I only know 1) $X_n$ converges in distribution to $X$; 2) $X_n$ is bounded by a constant $M$, and 3) $mathbb{E}[X^2]$ is finite, could I get $mathbb{E}[X_n^2]$ is finite?
Moreover, if $mathbb{E}[X^k]$ is finite, could we have $mathbb{E}[X_n^k]$ is finite?
I know that we CANNOT get $mathbb{E}[X_n]rightarrow mathbb{E}[X]$. But could we get $mathbb{E}[X_n]$ is finite if $mathbb{E}[X]$ is finite?
Thank you.
probability-distributions convergence
asked Jan 11 at 5:03
Alex LiuAlex Liu
31
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$begingroup$
The question doesn't make sense unless you drop condition 2).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:58
$begingroup$
ah...that's right. Is that possible to estimate $lim_{nrightarrow infty}mathbb{E}[X_n]$ under three conditions? I only know how to calculate $mathbb{E}[X]$
$endgroup$
– Alex Liu
Jan 11 at 6:09
add a comment |
$begingroup$
I only know 1) $X_n$ converges in distribution to $X$; 2) $X_n$ is bounded by a constant $M$, and 3) $mathbb{E}[X^2]$ is finite, could I get $mathbb{E}[X_n^2]$ is finite?
Moreover, if $mathbb{E}[X^k]$ is finite, could we have $mathbb{E}[X_n^k]$ is finite?
I know that we CANNOT get $mathbb{E}[X_n]rightarrow mathbb{E}[X]$. But could we get $mathbb{E}[X_n]$ is finite if $mathbb{E}[X]$ is finite?
Thank you.
probability-distributions convergence
asked Jan 11 at 5:03
Alex LiuAlex Liu
31
$endgroup$
I only know 1) $X_n$ converges in distribution to $X$; 2) $X_n$ is bounded by a constant $M$, and 3) $mathbb{E}[X^2]$ is finite, could I get $mathbb{E}[X_n^2]$ is finite?
Moreover, if $mathbb{E}[X^k]$ is finite, could we have $mathbb{E}[X_n^k]$ is finite?
I know that we CANNOT get $mathbb{E}[X_n]rightarrow mathbb{E}[X]$. But could we get $mathbb{E}[X_n]$ is finite if $mathbb{E}[X]$ is finite?
Thank you.
probability-distributions convergence
probability-distributions convergence
asked Jan 11 at 5:03
Alex LiuAlex Liu
31
asked Jan 11 at 5:03
Alex LiuAlex Liu
31
asked Jan 11 at 5:03
Alex LiuAlex Liu
31
asked Jan 11 at 5:03
Alex LiuAlex Liu
31
asked Jan 11 at 5:03
Alex LiuAlex Liu
31
31
$begingroup$
The question doesn't make sense unless you drop condition 2).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:58
$begingroup$
ah...that's right. Is that possible to estimate $lim_{nrightarrow infty}mathbb{E}[X_n]$ under three conditions? I only know how to calculate $mathbb{E}[X]$
$endgroup$
– Alex Liu
Jan 11 at 6:09
add a comment |
$begingroup$
The question doesn't make sense unless you drop condition 2).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:58
$begingroup$
ah...that's right. Is that possible to estimate $lim_{nrightarrow infty}mathbb{E}[X_n]$ under three conditions? I only know how to calculate $mathbb{E}[X]$
$endgroup$
– Alex Liu
Jan 11 at 6:09
$begingroup$
The question doesn't make sense unless you drop condition 2).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:58
$begingroup$
The question doesn't make sense unless you drop condition 2).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:58
$begingroup$
ah...that's right. Is that possible to estimate $lim_{nrightarrow infty}mathbb{E}[X_n]$ under three conditions? I only know how to calculate $mathbb{E}[X]$
$endgroup$
– Alex Liu
Jan 11 at 6:09
$begingroup$
ah...that's right. Is that possible to estimate $lim_{nrightarrow infty}mathbb{E}[X_n]$ under three conditions? I only know how to calculate $mathbb{E}[X]$
$endgroup$
– Alex Liu
Jan 11 at 6:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$E[X_n^k]leq E[M^k]=M^k<infty$$
answered Jan 11 at 5:10
Zachary SelkZachary Selk
593311
$endgroup$
add a comment |
$begingroup$
The revised statement is false. Let $Y$ be any random variable with $EY^{2}=infty$, $X=0$ and $X_n=frac Y n$. Then $X_n to X$ in distribution (in fact almost surely) and $EX^{2}=0<infty$ but $EX_n^{2}=infty$ for all $n$.
answered Jan 11 at 6:13
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
$$E[X_n^k]leq E[M^k]=M^k<infty$$
answered Jan 11 at 5:10
Zachary SelkZachary Selk
593311
$endgroup$
add a comment |
$begingroup$
$$E[X_n^k]leq E[M^k]=M^k<infty$$
answered Jan 11 at 5:10
Zachary SelkZachary Selk
593311
$endgroup$
add a comment |
$begingroup$
$$E[X_n^k]leq E[M^k]=M^k<infty$$
answered Jan 11 at 5:10
Zachary SelkZachary Selk
593311
$endgroup$
$$E[X_n^k]leq E[M^k]=M^k<infty$$
answered Jan 11 at 5:10
Zachary SelkZachary Selk
593311
answered Jan 11 at 5:10
Zachary SelkZachary Selk
593311
answered Jan 11 at 5:10
Zachary SelkZachary Selk
593311
answered Jan 11 at 5:10
Zachary SelkZachary Selk
593311
593311
add a comment |
add a comment |
$begingroup$
The revised statement is false. Let $Y$ be any random variable with $EY^{2}=infty$, $X=0$ and $X_n=frac Y n$. Then $X_n to X$ in distribution (in fact almost surely) and $EX^{2}=0<infty$ but $EX_n^{2}=infty$ for all $n$.
answered Jan 11 at 6:13
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
add a comment |
$begingroup$
The revised statement is false. Let $Y$ be any random variable with $EY^{2}=infty$, $X=0$ and $X_n=frac Y n$. Then $X_n to X$ in distribution (in fact almost surely) and $EX^{2}=0<infty$ but $EX_n^{2}=infty$ for all $n$.
answered Jan 11 at 6:13
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
add a comment |
$begingroup$
The revised statement is false. Let $Y$ be any random variable with $EY^{2}=infty$, $X=0$ and $X_n=frac Y n$. Then $X_n to X$ in distribution (in fact almost surely) and $EX^{2}=0<infty$ but $EX_n^{2}=infty$ for all $n$.
answered Jan 11 at 6:13
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
The revised statement is false. Let $Y$ be any random variable with $EY^{2}=infty$, $X=0$ and $X_n=frac Y n$. Then $X_n to X$ in distribution (in fact almost surely) and $EX^{2}=0<infty$ but $EX_n^{2}=infty$ for all $n$.
answered Jan 11 at 6:13
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Jan 11 at 6:13
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Jan 11 at 6:13
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Jan 11 at 6:13
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
55.4k42057
add a comment |
add a comment |
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$begingroup$
The question doesn't make sense unless you drop condition 2).
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:58
$begingroup$
ah...that's right. Is that possible to estimate $lim_{nrightarrow infty}mathbb{E}[X_n]$ under three conditions? I only know how to calculate $mathbb{E}[X]$
$endgroup$
– Alex Liu
Jan 11 at 6:09