Characterization of convergence in probability
$begingroup$
I have the following proposition:
Let $ (Omega, mathcal{F}, mathbb{P} ) $ be a probability space and $(X,d)$ a metric space. Let $ { f_h }_{h >0} $ be s.t. $f_h : Omega to X$ and $f: Omega to X$. Then
$$f_h overset{mathbb{P}}{to} f Leftrightarrow int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} overset{h to 0}{longrightarrow} 0$$
In the proof I have this equality
$$int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} = int_0^1 mathbb{P}(d(f_h,f)>s)ds $$
Why is it true?
probability convergence
asked Dec 11 '18 at 8:28
Bremen000Bremen000
429210
$endgroup$
add a comment |
$begingroup$
I have the following proposition:
Let $ (Omega, mathcal{F}, mathbb{P} ) $ be a probability space and $(X,d)$ a metric space. Let $ { f_h }_{h >0} $ be s.t. $f_h : Omega to X$ and $f: Omega to X$. Then
$$f_h overset{mathbb{P}}{to} f Leftrightarrow int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} overset{h to 0}{longrightarrow} 0$$
In the proof I have this equality
$$int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} = int_0^1 mathbb{P}(d(f_h,f)>s)ds $$
Why is it true?
probability convergence
asked Dec 11 '18 at 8:28
Bremen000Bremen000
429210
$endgroup$
add a comment |
$begingroup$
I have the following proposition:
Let $ (Omega, mathcal{F}, mathbb{P} ) $ be a probability space and $(X,d)$ a metric space. Let $ { f_h }_{h >0} $ be s.t. $f_h : Omega to X$ and $f: Omega to X$. Then
$$f_h overset{mathbb{P}}{to} f Leftrightarrow int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} overset{h to 0}{longrightarrow} 0$$
In the proof I have this equality
$$int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} = int_0^1 mathbb{P}(d(f_h,f)>s)ds $$
Why is it true?
probability convergence
asked Dec 11 '18 at 8:28
Bremen000Bremen000
429210
$endgroup$
I have the following proposition:
Let $ (Omega, mathcal{F}, mathbb{P} ) $ be a probability space and $(X,d)$ a metric space. Let $ { f_h }_{h >0} $ be s.t. $f_h : Omega to X$ and $f: Omega to X$. Then
$$f_h overset{mathbb{P}}{to} f Leftrightarrow int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} overset{h to 0}{longrightarrow} 0$$
In the proof I have this equality
$$int_{Omega} [1 wedge d(f_h, f)] d mathbb{P} = int_0^1 mathbb{P}(d(f_h,f)>s)ds $$
Why is it true?
probability convergence
probability convergence
asked Dec 11 '18 at 8:28
Bremen000Bremen000
429210
asked Dec 11 '18 at 8:28
Bremen000Bremen000
429210
asked Dec 11 '18 at 8:28
Bremen000Bremen000
429210
asked Dec 11 '18 at 8:28
Bremen000Bremen000
429210
asked Dec 11 '18 at 8:28
Bremen000Bremen000
429210
429210
add a comment |
add a comment |
1 Answer
1
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$begingroup$
For any non-negative measurable function $X$ we have $int_{Omega} XdP=int_0^{infty} P{X>t} dt$ $,,$ (1). In this case the integrand on RHS is $0$ for $t>1$. In fact $P{1wedge d(f_h,f) >s}=P{ d(f_h,f) >s}$ or $0$ according as $s <1$ or $s >1$.
Proof of (1): by Fubini's Theorem $int_0^{infty} P{X>t} dt=int_0^{infty} int_{Omega} I_{{t<X}} dP= int_{Omega} int_0^{infty}I_{{t<X}} dP=int_{Omega} XdP$.
answered Dec 11 '18 at 8:31
Kavi Rama MurthyKavi Rama Murthy
59k42161
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
For any non-negative measurable function $X$ we have $int_{Omega} XdP=int_0^{infty} P{X>t} dt$ $,,$ (1). In this case the integrand on RHS is $0$ for $t>1$. In fact $P{1wedge d(f_h,f) >s}=P{ d(f_h,f) >s}$ or $0$ according as $s <1$ or $s >1$.
Proof of (1): by Fubini's Theorem $int_0^{infty} P{X>t} dt=int_0^{infty} int_{Omega} I_{{t<X}} dP= int_{Omega} int_0^{infty}I_{{t<X}} dP=int_{Omega} XdP$.
answered Dec 11 '18 at 8:31
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
add a comment |
$begingroup$
For any non-negative measurable function $X$ we have $int_{Omega} XdP=int_0^{infty} P{X>t} dt$ $,,$ (1). In this case the integrand on RHS is $0$ for $t>1$. In fact $P{1wedge d(f_h,f) >s}=P{ d(f_h,f) >s}$ or $0$ according as $s <1$ or $s >1$.
Proof of (1): by Fubini's Theorem $int_0^{infty} P{X>t} dt=int_0^{infty} int_{Omega} I_{{t<X}} dP= int_{Omega} int_0^{infty}I_{{t<X}} dP=int_{Omega} XdP$.
answered Dec 11 '18 at 8:31
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
add a comment |
$begingroup$
For any non-negative measurable function $X$ we have $int_{Omega} XdP=int_0^{infty} P{X>t} dt$ $,,$ (1). In this case the integrand on RHS is $0$ for $t>1$. In fact $P{1wedge d(f_h,f) >s}=P{ d(f_h,f) >s}$ or $0$ according as $s <1$ or $s >1$.
Proof of (1): by Fubini's Theorem $int_0^{infty} P{X>t} dt=int_0^{infty} int_{Omega} I_{{t<X}} dP= int_{Omega} int_0^{infty}I_{{t<X}} dP=int_{Omega} XdP$.
answered Dec 11 '18 at 8:31
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
For any non-negative measurable function $X$ we have $int_{Omega} XdP=int_0^{infty} P{X>t} dt$ $,,$ (1). In this case the integrand on RHS is $0$ for $t>1$. In fact $P{1wedge d(f_h,f) >s}=P{ d(f_h,f) >s}$ or $0$ according as $s <1$ or $s >1$.
Proof of (1): by Fubini's Theorem $int_0^{infty} P{X>t} dt=int_0^{infty} int_{Omega} I_{{t<X}} dP= int_{Omega} int_0^{infty}I_{{t<X}} dP=int_{Omega} XdP$.
answered Dec 11 '18 at 8:31
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Dec 11 '18 at 8:31
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Dec 11 '18 at 8:31
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Dec 11 '18 at 8:31
Kavi Rama MurthyKavi Rama Murthy
59k42161
59k42161
add a comment |
add a comment |
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