If $E[h(M_n)]to h(x)$ for all bounded and continuous functions $h$ then $M_nto x$ in probability
Let $M_n$ be a sequence of random variables and let $xinmathbb{R}$ such that
$$mathbb{E}[h(M_n)]to h(x)$$
as $ntoinfty$ for all bounded and continuous functions $h$.
How to show that for all $delta>0$ we have $mathbb{P}(|M_n-x|>delta)to0$?
What I thought:
Convergence of $mathbb{E}[h(M_n)]$ means that for all $epsilon>0$ there is an $Ngeq0$ such that $mathbb{E}[h(M_n)-h(x)]<epsilon$.
Further we hav $mathbb{P}(|M_n-x|>delta)=mathbb{E}[1_{|M_n-x|>delta}]$, however the indicator function is not continuous. How do I fix this?
probability-theory random-variables weak-convergence
edited Dec 4 '18 at 18:34
Did
246k23221456
asked Dec 4 '18 at 9:51
Yaroslav MlynárYaroslav Mlynár
253
add a comment |
Let $M_n$ be a sequence of random variables and let $xinmathbb{R}$ such that
$$mathbb{E}[h(M_n)]to h(x)$$
as $ntoinfty$ for all bounded and continuous functions $h$.
How to show that for all $delta>0$ we have $mathbb{P}(|M_n-x|>delta)to0$?
What I thought:
Convergence of $mathbb{E}[h(M_n)]$ means that for all $epsilon>0$ there is an $Ngeq0$ such that $mathbb{E}[h(M_n)-h(x)]<epsilon$.
Further we hav $mathbb{P}(|M_n-x|>delta)=mathbb{E}[1_{|M_n-x|>delta}]$, however the indicator function is not continuous. How do I fix this?
probability-theory random-variables weak-convergence
edited Dec 4 '18 at 18:34
Did
246k23221456
asked Dec 4 '18 at 9:51
Yaroslav MlynárYaroslav Mlynár
253
add a comment |
Let $M_n$ be a sequence of random variables and let $xinmathbb{R}$ such that
$$mathbb{E}[h(M_n)]to h(x)$$
as $ntoinfty$ for all bounded and continuous functions $h$.
How to show that for all $delta>0$ we have $mathbb{P}(|M_n-x|>delta)to0$?
What I thought:
Convergence of $mathbb{E}[h(M_n)]$ means that for all $epsilon>0$ there is an $Ngeq0$ such that $mathbb{E}[h(M_n)-h(x)]<epsilon$.
Further we hav $mathbb{P}(|M_n-x|>delta)=mathbb{E}[1_{|M_n-x|>delta}]$, however the indicator function is not continuous. How do I fix this?
probability-theory random-variables weak-convergence
edited Dec 4 '18 at 18:34
Did
246k23221456
asked Dec 4 '18 at 9:51
Yaroslav MlynárYaroslav Mlynár
253
Let $M_n$ be a sequence of random variables and let $xinmathbb{R}$ such that
$$mathbb{E}[h(M_n)]to h(x)$$
as $ntoinfty$ for all bounded and continuous functions $h$.
How to show that for all $delta>0$ we have $mathbb{P}(|M_n-x|>delta)to0$?
What I thought:
Convergence of $mathbb{E}[h(M_n)]$ means that for all $epsilon>0$ there is an $Ngeq0$ such that $mathbb{E}[h(M_n)-h(x)]<epsilon$.
Further we hav $mathbb{P}(|M_n-x|>delta)=mathbb{E}[1_{|M_n-x|>delta}]$, however the indicator function is not continuous. How do I fix this?
probability-theory random-variables weak-convergence
probability-theory random-variables weak-convergence
edited Dec 4 '18 at 18:34
Did
246k23221456
asked Dec 4 '18 at 9:51
Yaroslav MlynárYaroslav Mlynár
253
edited Dec 4 '18 at 18:34
Did
246k23221456
asked Dec 4 '18 at 9:51
Yaroslav MlynárYaroslav Mlynár
253
edited Dec 4 '18 at 18:34
Did
246k23221456
edited Dec 4 '18 at 18:34
Did
246k23221456
edited Dec 4 '18 at 18:34
Did
246k23221456
246k23221456
asked Dec 4 '18 at 9:51
Yaroslav MlynárYaroslav Mlynár
253
asked Dec 4 '18 at 9:51
Yaroslav MlynárYaroslav Mlynár
253
asked Dec 4 '18 at 9:51
Yaroslav MlynárYaroslav Mlynár
253
253
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The stated assumption implies that $M_nxrightarrow{d}x$ and, thus, $M_nxrightarrow{p} x$.
Direct proof. For any $delta>0$,
begin{align}
mathsf{P}(|M_n-x|>delta)&=mathsf{P}(M_n>x+delta)+mathsf{P}(M_n<x-delta) \
&le 1+mathsf{E}h_{delta}(M_n-x)-mathsf{E}h_{delta}(M_n+delta-x),tag{1}
end{align}
where $h_delta(v):=1wedge(0vee v/delta)$. Since $h_{delta}$ is a continuous bounded function, $(1)$ converges to
$$
1+h_{delta}(0)-h_{delta}(delta)=0.
$$
answered Dec 4 '18 at 10:06
d.k.o.d.k.o.
8,622528
But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
– Yaroslav Mlynár
Dec 4 '18 at 10:21
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The stated assumption implies that $M_nxrightarrow{d}x$ and, thus, $M_nxrightarrow{p} x$.
Direct proof. For any $delta>0$,
begin{align}
mathsf{P}(|M_n-x|>delta)&=mathsf{P}(M_n>x+delta)+mathsf{P}(M_n<x-delta) \
&le 1+mathsf{E}h_{delta}(M_n-x)-mathsf{E}h_{delta}(M_n+delta-x),tag{1}
end{align}
where $h_delta(v):=1wedge(0vee v/delta)$. Since $h_{delta}$ is a continuous bounded function, $(1)$ converges to
$$
1+h_{delta}(0)-h_{delta}(delta)=0.
$$
answered Dec 4 '18 at 10:06
d.k.o.d.k.o.
8,622528
But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
– Yaroslav Mlynár
Dec 4 '18 at 10:21
add a comment |
The stated assumption implies that $M_nxrightarrow{d}x$ and, thus, $M_nxrightarrow{p} x$.
Direct proof. For any $delta>0$,
begin{align}
mathsf{P}(|M_n-x|>delta)&=mathsf{P}(M_n>x+delta)+mathsf{P}(M_n<x-delta) \
&le 1+mathsf{E}h_{delta}(M_n-x)-mathsf{E}h_{delta}(M_n+delta-x),tag{1}
end{align}
where $h_delta(v):=1wedge(0vee v/delta)$. Since $h_{delta}$ is a continuous bounded function, $(1)$ converges to
$$
1+h_{delta}(0)-h_{delta}(delta)=0.
$$
answered Dec 4 '18 at 10:06
d.k.o.d.k.o.
8,622528
But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
– Yaroslav Mlynár
Dec 4 '18 at 10:21
add a comment |
The stated assumption implies that $M_nxrightarrow{d}x$ and, thus, $M_nxrightarrow{p} x$.
Direct proof. For any $delta>0$,
begin{align}
mathsf{P}(|M_n-x|>delta)&=mathsf{P}(M_n>x+delta)+mathsf{P}(M_n<x-delta) \
&le 1+mathsf{E}h_{delta}(M_n-x)-mathsf{E}h_{delta}(M_n+delta-x),tag{1}
end{align}
where $h_delta(v):=1wedge(0vee v/delta)$. Since $h_{delta}$ is a continuous bounded function, $(1)$ converges to
$$
1+h_{delta}(0)-h_{delta}(delta)=0.
$$
answered Dec 4 '18 at 10:06
d.k.o.d.k.o.
8,622528
The stated assumption implies that $M_nxrightarrow{d}x$ and, thus, $M_nxrightarrow{p} x$.
Direct proof. For any $delta>0$,
begin{align}
mathsf{P}(|M_n-x|>delta)&=mathsf{P}(M_n>x+delta)+mathsf{P}(M_n<x-delta) \
&le 1+mathsf{E}h_{delta}(M_n-x)-mathsf{E}h_{delta}(M_n+delta-x),tag{1}
end{align}
where $h_delta(v):=1wedge(0vee v/delta)$. Since $h_{delta}$ is a continuous bounded function, $(1)$ converges to
$$
1+h_{delta}(0)-h_{delta}(delta)=0.
$$
answered Dec 4 '18 at 10:06
d.k.o.d.k.o.
8,622528
edited Dec 4 '18 at 18:14
answered Dec 4 '18 at 10:06
d.k.o.d.k.o.
8,622528
answered Dec 4 '18 at 10:06
d.k.o.d.k.o.
8,622528
answered Dec 4 '18 at 10:06
d.k.o.d.k.o.
8,622528
8,622528
But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
– Yaroslav Mlynár
Dec 4 '18 at 10:21
add a comment |
But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
– Yaroslav Mlynár
Dec 4 '18 at 10:21
But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
– Yaroslav Mlynár
Dec 4 '18 at 10:21
But I do not want to use convergence in distribution, I would prefer to prove this without using the implications you give, i.e. without the portmanteau lemma
– Yaroslav Mlynár
Dec 4 '18 at 10:21
add a comment |
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