A sequence of random variables, such that CDF converges
$begingroup$
Given a sequence of random variables $X_n$, is it true that for some
of its subsequences $X_{n(k)}$ there is a random variable $X$ such
that $F_{X_{n(k)}}(t) → F_{X}(t)$ for all $t ∈ mathbb{R}$. (F is a CDF.)
I guess it's true, but have I trouble proving that, and don't even understand how to approach that. Any input would be greatly appreciated!
probability-theory
asked Dec 23 '18 at 21:20
PaprikaPaprika
113
$endgroup$
add a comment |
$begingroup$
Given a sequence of random variables $X_n$, is it true that for some
of its subsequences $X_{n(k)}$ there is a random variable $X$ such
that $F_{X_{n(k)}}(t) → F_{X}(t)$ for all $t ∈ mathbb{R}$. (F is a CDF.)
I guess it's true, but have I trouble proving that, and don't even understand how to approach that. Any input would be greatly appreciated!
probability-theory
asked Dec 23 '18 at 21:20
PaprikaPaprika
113
$endgroup$
add a comment |
$begingroup$
Given a sequence of random variables $X_n$, is it true that for some
of its subsequences $X_{n(k)}$ there is a random variable $X$ such
that $F_{X_{n(k)}}(t) → F_{X}(t)$ for all $t ∈ mathbb{R}$. (F is a CDF.)
I guess it's true, but have I trouble proving that, and don't even understand how to approach that. Any input would be greatly appreciated!
probability-theory
asked Dec 23 '18 at 21:20
PaprikaPaprika
113
$endgroup$
Given a sequence of random variables $X_n$, is it true that for some
of its subsequences $X_{n(k)}$ there is a random variable $X$ such
that $F_{X_{n(k)}}(t) → F_{X}(t)$ for all $t ∈ mathbb{R}$. (F is a CDF.)
I guess it's true, but have I trouble proving that, and don't even understand how to approach that. Any input would be greatly appreciated!
probability-theory
probability-theory
asked Dec 23 '18 at 21:20
PaprikaPaprika
113
asked Dec 23 '18 at 21:20
PaprikaPaprika
113
edited Dec 23 '18 at 21:25
Paprika
asked Dec 23 '18 at 21:20
PaprikaPaprika
113
asked Dec 23 '18 at 21:20
PaprikaPaprika
113
asked Dec 23 '18 at 21:20
PaprikaPaprika
113
113
add a comment |
add a comment |
1 Answer
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$begingroup$
No, it is not true as stated. For instance, if $P(X_n=n)=1$ for all $n$, your supposed $X$ would have cdf. $F_X(t)=0$ for all $t$.
But it is close to something true & important: Prokhorov's Theorem, which covers the case where the $X_n$ are ``tight'', that is, for every $epsilon$ there is a $K$ such that $P|X_n|>K)<epsilon$ for all $n$. (Which of course is not satisfied by the example in the above paragraph.)
answered Dec 23 '18 at 21:26
kimchi loverkimchi lover
10.8k31128
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1 Answer
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$begingroup$
No, it is not true as stated. For instance, if $P(X_n=n)=1$ for all $n$, your supposed $X$ would have cdf. $F_X(t)=0$ for all $t$.
But it is close to something true & important: Prokhorov's Theorem, which covers the case where the $X_n$ are ``tight'', that is, for every $epsilon$ there is a $K$ such that $P|X_n|>K)<epsilon$ for all $n$. (Which of course is not satisfied by the example in the above paragraph.)
answered Dec 23 '18 at 21:26
kimchi loverkimchi lover
10.8k31128
$endgroup$
add a comment |
$begingroup$
No, it is not true as stated. For instance, if $P(X_n=n)=1$ for all $n$, your supposed $X$ would have cdf. $F_X(t)=0$ for all $t$.
But it is close to something true & important: Prokhorov's Theorem, which covers the case where the $X_n$ are ``tight'', that is, for every $epsilon$ there is a $K$ such that $P|X_n|>K)<epsilon$ for all $n$. (Which of course is not satisfied by the example in the above paragraph.)
answered Dec 23 '18 at 21:26
kimchi loverkimchi lover
10.8k31128
$endgroup$
add a comment |
$begingroup$
No, it is not true as stated. For instance, if $P(X_n=n)=1$ for all $n$, your supposed $X$ would have cdf. $F_X(t)=0$ for all $t$.
But it is close to something true & important: Prokhorov's Theorem, which covers the case where the $X_n$ are ``tight'', that is, for every $epsilon$ there is a $K$ such that $P|X_n|>K)<epsilon$ for all $n$. (Which of course is not satisfied by the example in the above paragraph.)
answered Dec 23 '18 at 21:26
kimchi loverkimchi lover
10.8k31128
$endgroup$
No, it is not true as stated. For instance, if $P(X_n=n)=1$ for all $n$, your supposed $X$ would have cdf. $F_X(t)=0$ for all $t$.
But it is close to something true & important: Prokhorov's Theorem, which covers the case where the $X_n$ are ``tight'', that is, for every $epsilon$ there is a $K$ such that $P|X_n|>K)<epsilon$ for all $n$. (Which of course is not satisfied by the example in the above paragraph.)
answered Dec 23 '18 at 21:26
kimchi loverkimchi lover
10.8k31128
answered Dec 23 '18 at 21:26
kimchi loverkimchi lover
10.8k31128
answered Dec 23 '18 at 21:26
kimchi loverkimchi lover
10.8k31128
answered Dec 23 '18 at 21:26
kimchi loverkimchi lover
10.8k31128
10.8k31128
add a comment |
add a comment |
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