Mixing
Convergence in probability- sign inequality
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We know, that $lim_{ ntoinfty } Pleft{ left| X_n-Xright| ge epsilon right}=0$ for each $epsilon >0$.
When I can write $lim_{ ntoinfty } Pleft{ left| X_n-Xright| > epsilon right}=0$ and why? (another inequality sign)
convergence random-variables
asked Jan 30 at 16:16
pawelKpawelK
618
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add a comment |
$begingroup$
We know, that $lim_{ ntoinfty } Pleft{ left| X_n-Xright| ge epsilon right}=0$ for each $epsilon >0$.
When I can write $lim_{ ntoinfty } Pleft{ left| X_n-Xright| > epsilon right}=0$ and why? (another inequality sign)
convergence random-variables
asked Jan 30 at 16:16
pawelKpawelK
618
$endgroup$
$begingroup$
These are two equivalent definitions of convergence in probability.
$endgroup$
– Mark
Jan 30 at 16:19
add a comment |
$begingroup$
We know, that $lim_{ ntoinfty } Pleft{ left| X_n-Xright| ge epsilon right}=0$ for each $epsilon >0$.
When I can write $lim_{ ntoinfty } Pleft{ left| X_n-Xright| > epsilon right}=0$ and why? (another inequality sign)
convergence random-variables
asked Jan 30 at 16:16
pawelKpawelK
618
$endgroup$
We know, that $lim_{ ntoinfty } Pleft{ left| X_n-Xright| ge epsilon right}=0$ for each $epsilon >0$.
When I can write $lim_{ ntoinfty } Pleft{ left| X_n-Xright| > epsilon right}=0$ and why? (another inequality sign)
convergence random-variables
convergence random-variables
asked Jan 30 at 16:16
pawelKpawelK
618
asked Jan 30 at 16:16
pawelKpawelK
618
asked Jan 30 at 16:16
pawelKpawelK
618
asked Jan 30 at 16:16
pawelKpawelK
618
asked Jan 30 at 16:16
pawelKpawelK
618
618
$begingroup$
These are two equivalent definitions of convergence in probability.
$endgroup$
– Mark
Jan 30 at 16:19
add a comment |
$begingroup$
These are two equivalent definitions of convergence in probability.
$endgroup$
– Mark
Jan 30 at 16:19
$begingroup$
These are two equivalent definitions of convergence in probability.
$endgroup$
– Mark
Jan 30 at 16:19
$begingroup$
These are two equivalent definitions of convergence in probability.
$endgroup$
– Mark
Jan 30 at 16:19
add a comment |
1 Answer
1
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$begingroup$
Since $epsilon>0$ is arbitrary, the equivalence of these assertions follows from the line of inclusions
$${|X_n-X|> epsilon } subset {|X_n-X|ge epsilon }subset {|X_n-X|> epsilon/2 }$$
and the monotonicity of probability measures.
answered Jan 30 at 16:26
Mars PlasticMars Plastic
1,477122
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Since $epsilon>0$ is arbitrary, the equivalence of these assertions follows from the line of inclusions
$${|X_n-X|> epsilon } subset {|X_n-X|ge epsilon }subset {|X_n-X|> epsilon/2 }$$
and the monotonicity of probability measures.
answered Jan 30 at 16:26
Mars PlasticMars Plastic
1,477122
$endgroup$
add a comment |
$begingroup$
Since $epsilon>0$ is arbitrary, the equivalence of these assertions follows from the line of inclusions
$${|X_n-X|> epsilon } subset {|X_n-X|ge epsilon }subset {|X_n-X|> epsilon/2 }$$
and the monotonicity of probability measures.
answered Jan 30 at 16:26
Mars PlasticMars Plastic
1,477122
$endgroup$
add a comment |
$begingroup$
Since $epsilon>0$ is arbitrary, the equivalence of these assertions follows from the line of inclusions
$${|X_n-X|> epsilon } subset {|X_n-X|ge epsilon }subset {|X_n-X|> epsilon/2 }$$
and the monotonicity of probability measures.
answered Jan 30 at 16:26
Mars PlasticMars Plastic
1,477122
$endgroup$
Since $epsilon>0$ is arbitrary, the equivalence of these assertions follows from the line of inclusions
$${|X_n-X|> epsilon } subset {|X_n-X|ge epsilon }subset {|X_n-X|> epsilon/2 }$$
and the monotonicity of probability measures.
answered Jan 30 at 16:26
Mars PlasticMars Plastic
1,477122
edited Jan 30 at 16:37
answered Jan 30 at 16:26
Mars PlasticMars Plastic
1,477122
answered Jan 30 at 16:26
Mars PlasticMars Plastic
1,477122
answered Jan 30 at 16:26
Mars PlasticMars Plastic
1,477122
1,477122
add a comment |
add a comment |
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$begingroup$
These are two equivalent definitions of convergence in probability.
$endgroup$
– Mark
Jan 30 at 16:19