Problem with truthfulness of equivalence.Upper bound of a set (equivalence problem)Equivalence in...
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Problem with truthfulness of equivalence.
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$begingroup$
Let $X_n$ be a sequance of random variables. I wonder about this equivalence :
$${displaystyle {overset {}{X_{n} {xrightarrow {p}} 0 }}} Leftrightarrow E(X_n)rightarrow0 $$
"$Rightarrow$"
If ${displaystyle {overset {}{X_{n} {xrightarrow {p}} 0 }}}$ then for very big $n$ $(X_n)$ only take values very close to zero, so Expected value of $(X_n)$ is also arbitrarily close to zero. So it has to be : $E(X_n)rightarrow0.$
"$Leftarrow$"
Let's take $X_n$ ~ $U[-1,1]$. Then for any $n$, we have $E(X_n)=0, so ;E(X_n)rightarrow0.$
But $P(|X_n|<varepsilon) neq1 $, so $X_n$ dosen't converge to $0$ in probability.
Am i thinning correctly ?
probability-theory proof-verification convergence
asked Mar 14 at 20:40
LouisLouis
195
$endgroup$
add a comment |
$begingroup$
Let $X_n$ be a sequance of random variables. I wonder about this equivalence :
$${displaystyle {overset {}{X_{n} {xrightarrow {p}} 0 }}} Leftrightarrow E(X_n)rightarrow0 $$
"$Rightarrow$"
If ${displaystyle {overset {}{X_{n} {xrightarrow {p}} 0 }}}$ then for very big $n$ $(X_n)$ only take values very close to zero, so Expected value of $(X_n)$ is also arbitrarily close to zero. So it has to be : $E(X_n)rightarrow0.$
"$Leftarrow$"
Let's take $X_n$ ~ $U[-1,1]$. Then for any $n$, we have $E(X_n)=0, so ;E(X_n)rightarrow0.$
But $P(|X_n|<varepsilon) neq1 $, so $X_n$ dosen't converge to $0$ in probability.
Am i thinning correctly ?
probability-theory proof-verification convergence
asked Mar 14 at 20:40
LouisLouis
195
$endgroup$
$begingroup$
But i didn't prove by example
$endgroup$
– Louis
Mar 14 at 20:45
$begingroup$
I wanted to show that the "$Leftarrow$" implication is not true.
$endgroup$
– Louis
Mar 14 at 20:49
$begingroup$
By counterexample.
$endgroup$
– Louis
Mar 14 at 20:49
add a comment |
$begingroup$
Let $X_n$ be a sequance of random variables. I wonder about this equivalence :
$${displaystyle {overset {}{X_{n} {xrightarrow {p}} 0 }}} Leftrightarrow E(X_n)rightarrow0 $$
"$Rightarrow$"
If ${displaystyle {overset {}{X_{n} {xrightarrow {p}} 0 }}}$ then for very big $n$ $(X_n)$ only take values very close to zero, so Expected value of $(X_n)$ is also arbitrarily close to zero. So it has to be : $E(X_n)rightarrow0.$
"$Leftarrow$"
Let's take $X_n$ ~ $U[-1,1]$. Then for any $n$, we have $E(X_n)=0, so ;E(X_n)rightarrow0.$
But $P(|X_n|<varepsilon) neq1 $, so $X_n$ dosen't converge to $0$ in probability.
Am i thinning correctly ?
probability-theory proof-verification convergence
asked Mar 14 at 20:40
LouisLouis
195
$endgroup$
Let $X_n$ be a sequance of random variables. I wonder about this equivalence :
$${displaystyle {overset {}{X_{n} {xrightarrow {p}} 0 }}} Leftrightarrow E(X_n)rightarrow0 $$
"$Rightarrow$"
If ${displaystyle {overset {}{X_{n} {xrightarrow {p}} 0 }}}$ then for very big $n$ $(X_n)$ only take values very close to zero, so Expected value of $(X_n)$ is also arbitrarily close to zero. So it has to be : $E(X_n)rightarrow0.$
"$Leftarrow$"
Let's take $X_n$ ~ $U[-1,1]$. Then for any $n$, we have $E(X_n)=0, so ;E(X_n)rightarrow0.$
But $P(|X_n|<varepsilon) neq1 $, so $X_n$ dosen't converge to $0$ in probability.
Am i thinning correctly ?
probability-theory proof-verification convergence
probability-theory proof-verification convergence
asked Mar 14 at 20:40
LouisLouis
195
asked Mar 14 at 20:40
LouisLouis
195
asked Mar 14 at 20:40
LouisLouis
195
asked Mar 14 at 20:40
LouisLouis
195
asked Mar 14 at 20:40
LouisLouis
195
195
$begingroup$
But i didn't prove by example
$endgroup$
– Louis
Mar 14 at 20:45
$begingroup$
I wanted to show that the "$Leftarrow$" implication is not true.
$endgroup$
– Louis
Mar 14 at 20:49
$begingroup$
By counterexample.
$endgroup$
– Louis
Mar 14 at 20:49
add a comment |
$begingroup$
But i didn't prove by example
$endgroup$
– Louis
Mar 14 at 20:45
$begingroup$
I wanted to show that the "$Leftarrow$" implication is not true.
$endgroup$
– Louis
Mar 14 at 20:49
$begingroup$
By counterexample.
$endgroup$
– Louis
Mar 14 at 20:49
$begingroup$
But i didn't prove by example
$endgroup$
– Louis
Mar 14 at 20:45
$begingroup$
But i didn't prove by example
$endgroup$
– Louis
Mar 14 at 20:45
$begingroup$
I wanted to show that the "$Leftarrow$" implication is not true.
$endgroup$
– Louis
Mar 14 at 20:49
$begingroup$
I wanted to show that the "$Leftarrow$" implication is not true.
$endgroup$
– Louis
Mar 14 at 20:49
$begingroup$
By counterexample.
$endgroup$
– Louis
Mar 14 at 20:49
$begingroup$
By counterexample.
$endgroup$
– Louis
Mar 14 at 20:49
add a comment |
1 Answer
1
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oldest
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$begingroup$
Take $X_n=nI(Zin [0, 1/n])$ where $Zsim text{Unif}[0, 1]$. Then $X_nto 0$ almost surely, and hence in probability yet $EX_n=1$ for all $n$ so $EX_nnotto 0$.
answered Mar 14 at 21:01
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Take $X_n=nI(Zin [0, 1/n])$ where $Zsim text{Unif}[0, 1]$. Then $X_nto 0$ almost surely, and hence in probability yet $EX_n=1$ for all $n$ so $EX_nnotto 0$.
answered Mar 14 at 21:01
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
add a comment |
$begingroup$
Take $X_n=nI(Zin [0, 1/n])$ where $Zsim text{Unif}[0, 1]$. Then $X_nto 0$ almost surely, and hence in probability yet $EX_n=1$ for all $n$ so $EX_nnotto 0$.
answered Mar 14 at 21:01
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
add a comment |
$begingroup$
Take $X_n=nI(Zin [0, 1/n])$ where $Zsim text{Unif}[0, 1]$. Then $X_nto 0$ almost surely, and hence in probability yet $EX_n=1$ for all $n$ so $EX_nnotto 0$.
answered Mar 14 at 21:01
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
Take $X_n=nI(Zin [0, 1/n])$ where $Zsim text{Unif}[0, 1]$. Then $X_nto 0$ almost surely, and hence in probability yet $EX_n=1$ for all $n$ so $EX_nnotto 0$.
answered Mar 14 at 21:01
Foobaz JohnFoobaz John
22.8k41452
answered Mar 14 at 21:01
Foobaz JohnFoobaz John
22.8k41452
answered Mar 14 at 21:01
Foobaz JohnFoobaz John
22.8k41452
answered Mar 14 at 21:01
Foobaz JohnFoobaz John
22.8k41452
22.8k41452
add a comment |
add a comment |
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$begingroup$
But i didn't prove by example
$endgroup$
– Louis
Mar 14 at 20:45
$begingroup$
I wanted to show that the "$Leftarrow$" implication is not true.
$endgroup$
– Louis
Mar 14 at 20:49
$begingroup$
By counterexample.
$endgroup$
– Louis
Mar 14 at 20:49