Mixing
show that $X_n overset{mathrm{P}}{to} X iff d(X_n,X) to 0 $
$begingroup$
let $L^0$ be the space on all random variables then :
$$begin{align}
d : &L^0 times L^0 &to &[0,1] \
&(X,Y) &mapsto &mathbb{E}[frac{|X-Y|}{1+|X - Y|}]
end{align}$$
defines a well defined metric on $L^0$
proof of first implication : considering the continuous bounded function $phi(t) = frac{|t|}{1+|t|}$ and using the fact that $E[phi(X_n-X)] to 0$ in case $X_n -X overset{mathrm{d}}{to} 0$
then $$X_n overset{mathrm{P}}{to} X implies X_n - X overset{mathrm{P}}{to} 0 implies X_n - X overset{mathrm{d}}{to} 0 implies d(X_n,X) to 0 $$
for second I wanted to use the fact that the convergence in $L^1$ is preserved under composition with continuous function coupled with the fact that $$frac{frac{t}{1 + t}}{1 -frac{t}{1 + t}} = t$$
but $y mapsto frac{y}{1-y}$ isn't continuous at $1$.
hints ?
probability-theory convergence metric-spaces
asked Jan 25 at 14:45
rapidracimrapidracim
1,7241419
$endgroup$
add a comment |
$begingroup$
let $L^0$ be the space on all random variables then :
$$begin{align}
d : &L^0 times L^0 &to &[0,1] \
&(X,Y) &mapsto &mathbb{E}[frac{|X-Y|}{1+|X - Y|}]
end{align}$$
defines a well defined metric on $L^0$
proof of first implication : considering the continuous bounded function $phi(t) = frac{|t|}{1+|t|}$ and using the fact that $E[phi(X_n-X)] to 0$ in case $X_n -X overset{mathrm{d}}{to} 0$
then $$X_n overset{mathrm{P}}{to} X implies X_n - X overset{mathrm{P}}{to} 0 implies X_n - X overset{mathrm{d}}{to} 0 implies d(X_n,X) to 0 $$
for second I wanted to use the fact that the convergence in $L^1$ is preserved under composition with continuous function coupled with the fact that $$frac{frac{t}{1 + t}}{1 -frac{t}{1 + t}} = t$$
but $y mapsto frac{y}{1-y}$ isn't continuous at $1$.
hints ?
probability-theory convergence metric-spaces
asked Jan 25 at 14:45
rapidracimrapidracim
1,7241419
$endgroup$
add a comment |
$begingroup$
let $L^0$ be the space on all random variables then :
$$begin{align}
d : &L^0 times L^0 &to &[0,1] \
&(X,Y) &mapsto &mathbb{E}[frac{|X-Y|}{1+|X - Y|}]
end{align}$$
defines a well defined metric on $L^0$
proof of first implication : considering the continuous bounded function $phi(t) = frac{|t|}{1+|t|}$ and using the fact that $E[phi(X_n-X)] to 0$ in case $X_n -X overset{mathrm{d}}{to} 0$
then $$X_n overset{mathrm{P}}{to} X implies X_n - X overset{mathrm{P}}{to} 0 implies X_n - X overset{mathrm{d}}{to} 0 implies d(X_n,X) to 0 $$
for second I wanted to use the fact that the convergence in $L^1$ is preserved under composition with continuous function coupled with the fact that $$frac{frac{t}{1 + t}}{1 -frac{t}{1 + t}} = t$$
but $y mapsto frac{y}{1-y}$ isn't continuous at $1$.
hints ?
probability-theory convergence metric-spaces
asked Jan 25 at 14:45
rapidracimrapidracim
1,7241419
$endgroup$
let $L^0$ be the space on all random variables then :
$$begin{align}
d : &L^0 times L^0 &to &[0,1] \
&(X,Y) &mapsto &mathbb{E}[frac{|X-Y|}{1+|X - Y|}]
end{align}$$
defines a well defined metric on $L^0$
proof of first implication : considering the continuous bounded function $phi(t) = frac{|t|}{1+|t|}$ and using the fact that $E[phi(X_n-X)] to 0$ in case $X_n -X overset{mathrm{d}}{to} 0$
then $$X_n overset{mathrm{P}}{to} X implies X_n - X overset{mathrm{P}}{to} 0 implies X_n - X overset{mathrm{d}}{to} 0 implies d(X_n,X) to 0 $$
for second I wanted to use the fact that the convergence in $L^1$ is preserved under composition with continuous function coupled with the fact that $$frac{frac{t}{1 + t}}{1 -frac{t}{1 + t}} = t$$
but $y mapsto frac{y}{1-y}$ isn't continuous at $1$.
hints ?
probability-theory convergence metric-spaces
probability-theory convergence metric-spaces
asked Jan 25 at 14:45
rapidracimrapidracim
1,7241419
asked Jan 25 at 14:45
rapidracimrapidracim
1,7241419
asked Jan 25 at 14:45
rapidracimrapidracim
1,7241419
asked Jan 25 at 14:45
rapidracimrapidracim
1,7241419
asked Jan 25 at 14:45
rapidracimrapidracim
1,7241419
1,7241419
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Since $tmapsto frac{t}{1+t}$ is increasing, we find by Chebyshev's inequality
$$
Bbb P(|X_n-X|geepsilon) =Bbb Pleft(frac{|X_n-X|}{1+|X_n-X|}ge frac{epsilon}{1+epsilon}right)le (1+epsilon)/epsiloncdot d(X_n,X) to 0.
$$
answered Jan 25 at 15:11
SongSong
18.3k21549
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Since $tmapsto frac{t}{1+t}$ is increasing, we find by Chebyshev's inequality
$$
Bbb P(|X_n-X|geepsilon) =Bbb Pleft(frac{|X_n-X|}{1+|X_n-X|}ge frac{epsilon}{1+epsilon}right)le (1+epsilon)/epsiloncdot d(X_n,X) to 0.
$$
answered Jan 25 at 15:11
SongSong
18.3k21549
$endgroup$
add a comment |
$begingroup$
Since $tmapsto frac{t}{1+t}$ is increasing, we find by Chebyshev's inequality
$$
Bbb P(|X_n-X|geepsilon) =Bbb Pleft(frac{|X_n-X|}{1+|X_n-X|}ge frac{epsilon}{1+epsilon}right)le (1+epsilon)/epsiloncdot d(X_n,X) to 0.
$$
answered Jan 25 at 15:11
SongSong
18.3k21549
$endgroup$
add a comment |
$begingroup$
Since $tmapsto frac{t}{1+t}$ is increasing, we find by Chebyshev's inequality
$$
Bbb P(|X_n-X|geepsilon) =Bbb Pleft(frac{|X_n-X|}{1+|X_n-X|}ge frac{epsilon}{1+epsilon}right)le (1+epsilon)/epsiloncdot d(X_n,X) to 0.
$$
answered Jan 25 at 15:11
SongSong
18.3k21549
$endgroup$
Since $tmapsto frac{t}{1+t}$ is increasing, we find by Chebyshev's inequality
$$
Bbb P(|X_n-X|geepsilon) =Bbb Pleft(frac{|X_n-X|}{1+|X_n-X|}ge frac{epsilon}{1+epsilon}right)le (1+epsilon)/epsiloncdot d(X_n,X) to 0.
$$
answered Jan 25 at 15:11
SongSong
18.3k21549
answered Jan 25 at 15:11
SongSong
18.3k21549
answered Jan 25 at 15:11
SongSong
18.3k21549
answered Jan 25 at 15:11
SongSong
18.3k21549
18.3k21549
add a comment |
add a comment |
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