Mixing
Convergence in probability towards a constant
$begingroup$
Prove that if a sequence $X_1,X_2..$ of random variables satisfies the following conditions: $$1.lim_{ntoinfty}mathbb{E}(X_n)=a,space ainmathbb{R}$$
$$2.lim_{ntoinfty}mathbb{V}(X_n)=0$$
Then this sequence converges in probability towards $a$: $$lim_{ntoinfty}mathbb{P}(|X_n-a|geqepsilon) = 0$$
convergence probability-limit-theorems expected-value
asked Jan 11 at 23:29
AromaTheLoopAromaTheLoop
444
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add a comment |
$begingroup$
Prove that if a sequence $X_1,X_2..$ of random variables satisfies the following conditions: $$1.lim_{ntoinfty}mathbb{E}(X_n)=a,space ainmathbb{R}$$
$$2.lim_{ntoinfty}mathbb{V}(X_n)=0$$
Then this sequence converges in probability towards $a$: $$lim_{ntoinfty}mathbb{P}(|X_n-a|geqepsilon) = 0$$
convergence probability-limit-theorems expected-value
asked Jan 11 at 23:29
AromaTheLoopAromaTheLoop
444
$endgroup$
add a comment |
$begingroup$
Prove that if a sequence $X_1,X_2..$ of random variables satisfies the following conditions: $$1.lim_{ntoinfty}mathbb{E}(X_n)=a,space ainmathbb{R}$$
$$2.lim_{ntoinfty}mathbb{V}(X_n)=0$$
Then this sequence converges in probability towards $a$: $$lim_{ntoinfty}mathbb{P}(|X_n-a|geqepsilon) = 0$$
convergence probability-limit-theorems expected-value
asked Jan 11 at 23:29
AromaTheLoopAromaTheLoop
444
$endgroup$
Prove that if a sequence $X_1,X_2..$ of random variables satisfies the following conditions: $$1.lim_{ntoinfty}mathbb{E}(X_n)=a,space ainmathbb{R}$$
$$2.lim_{ntoinfty}mathbb{V}(X_n)=0$$
Then this sequence converges in probability towards $a$: $$lim_{ntoinfty}mathbb{P}(|X_n-a|geqepsilon) = 0$$
convergence probability-limit-theorems expected-value
convergence probability-limit-theorems expected-value
asked Jan 11 at 23:29
AromaTheLoopAromaTheLoop
444
asked Jan 11 at 23:29
AromaTheLoopAromaTheLoop
444
asked Jan 11 at 23:29
AromaTheLoopAromaTheLoop
444
asked Jan 11 at 23:29
AromaTheLoopAromaTheLoop
444
asked Jan 11 at 23:29
AromaTheLoopAromaTheLoop
444
444
add a comment |
add a comment |
2 Answers
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$begingroup$
Fix $varepsilon>0$. Note that
$$
|X_n-a|leq |X_n-EX_n|+|EX_n-a|
$$
whence
$$
P(|X_n-a|gevarepsilon)leq P(|X_n-EX_n|gevarepsilon/2)+P(|EX_n-a|gevarepsilon/2) tag{1}
$$
The first term on the RHS of (1) goes to zero by the assumption that $text{Var}(X_n)to 0$ combined with Chebeshev's inequality. The second term on the RHS of (1) goes to zero by the assumption that $EX_nto a$ .
answered Jan 12 at 0:08
Foobaz JohnFoobaz John
22.1k41352
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$begingroup$
$E(X_n-a)^{2}=EX_n^{2}+a^{2}-2aEX_n =var(X_n)+(EX_n)^{2}+a^{2}-2aEX_n to 0+a^{2}+a^{2}-2a^{2}=0$. Hence $P{|X_n-a| geq epsilon} leq frac {E(X_n-a)^{2}} {epsilon^{2}} to 0$.
answered Jan 11 at 23:33
Kavi Rama MurthyKavi Rama Murthy
59.1k42161
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2 Answers
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2 Answers
2
active
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$begingroup$
Fix $varepsilon>0$. Note that
$$
|X_n-a|leq |X_n-EX_n|+|EX_n-a|
$$
whence
$$
P(|X_n-a|gevarepsilon)leq P(|X_n-EX_n|gevarepsilon/2)+P(|EX_n-a|gevarepsilon/2) tag{1}
$$
The first term on the RHS of (1) goes to zero by the assumption that $text{Var}(X_n)to 0$ combined with Chebeshev's inequality. The second term on the RHS of (1) goes to zero by the assumption that $EX_nto a$ .
answered Jan 12 at 0:08
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
$begingroup$
Fix $varepsilon>0$. Note that
$$
|X_n-a|leq |X_n-EX_n|+|EX_n-a|
$$
whence
$$
P(|X_n-a|gevarepsilon)leq P(|X_n-EX_n|gevarepsilon/2)+P(|EX_n-a|gevarepsilon/2) tag{1}
$$
The first term on the RHS of (1) goes to zero by the assumption that $text{Var}(X_n)to 0$ combined with Chebeshev's inequality. The second term on the RHS of (1) goes to zero by the assumption that $EX_nto a$ .
answered Jan 12 at 0:08
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
$begingroup$
Fix $varepsilon>0$. Note that
$$
|X_n-a|leq |X_n-EX_n|+|EX_n-a|
$$
whence
$$
P(|X_n-a|gevarepsilon)leq P(|X_n-EX_n|gevarepsilon/2)+P(|EX_n-a|gevarepsilon/2) tag{1}
$$
The first term on the RHS of (1) goes to zero by the assumption that $text{Var}(X_n)to 0$ combined with Chebeshev's inequality. The second term on the RHS of (1) goes to zero by the assumption that $EX_nto a$ .
answered Jan 12 at 0:08
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
Fix $varepsilon>0$. Note that
$$
|X_n-a|leq |X_n-EX_n|+|EX_n-a|
$$
whence
$$
P(|X_n-a|gevarepsilon)leq P(|X_n-EX_n|gevarepsilon/2)+P(|EX_n-a|gevarepsilon/2) tag{1}
$$
The first term on the RHS of (1) goes to zero by the assumption that $text{Var}(X_n)to 0$ combined with Chebeshev's inequality. The second term on the RHS of (1) goes to zero by the assumption that $EX_nto a$ .
answered Jan 12 at 0:08
Foobaz JohnFoobaz John
22.1k41352
answered Jan 12 at 0:08
Foobaz JohnFoobaz John
22.1k41352
answered Jan 12 at 0:08
Foobaz JohnFoobaz John
22.1k41352
answered Jan 12 at 0:08
Foobaz JohnFoobaz John
22.1k41352
22.1k41352
add a comment |
add a comment |
$begingroup$
$E(X_n-a)^{2}=EX_n^{2}+a^{2}-2aEX_n =var(X_n)+(EX_n)^{2}+a^{2}-2aEX_n to 0+a^{2}+a^{2}-2a^{2}=0$. Hence $P{|X_n-a| geq epsilon} leq frac {E(X_n-a)^{2}} {epsilon^{2}} to 0$.
answered Jan 11 at 23:33
Kavi Rama MurthyKavi Rama Murthy
59.1k42161
$endgroup$
add a comment |
$begingroup$
$E(X_n-a)^{2}=EX_n^{2}+a^{2}-2aEX_n =var(X_n)+(EX_n)^{2}+a^{2}-2aEX_n to 0+a^{2}+a^{2}-2a^{2}=0$. Hence $P{|X_n-a| geq epsilon} leq frac {E(X_n-a)^{2}} {epsilon^{2}} to 0$.
answered Jan 11 at 23:33
Kavi Rama MurthyKavi Rama Murthy
59.1k42161
$endgroup$
add a comment |
$begingroup$
$E(X_n-a)^{2}=EX_n^{2}+a^{2}-2aEX_n =var(X_n)+(EX_n)^{2}+a^{2}-2aEX_n to 0+a^{2}+a^{2}-2a^{2}=0$. Hence $P{|X_n-a| geq epsilon} leq frac {E(X_n-a)^{2}} {epsilon^{2}} to 0$.
answered Jan 11 at 23:33
Kavi Rama MurthyKavi Rama Murthy
59.1k42161
$endgroup$
$E(X_n-a)^{2}=EX_n^{2}+a^{2}-2aEX_n =var(X_n)+(EX_n)^{2}+a^{2}-2aEX_n to 0+a^{2}+a^{2}-2a^{2}=0$. Hence $P{|X_n-a| geq epsilon} leq frac {E(X_n-a)^{2}} {epsilon^{2}} to 0$.
answered Jan 11 at 23:33
Kavi Rama MurthyKavi Rama Murthy
59.1k42161
answered Jan 11 at 23:33
Kavi Rama MurthyKavi Rama Murthy
59.1k42161
answered Jan 11 at 23:33
Kavi Rama MurthyKavi Rama Murthy
59.1k42161
answered Jan 11 at 23:33
Kavi Rama MurthyKavi Rama Murthy
59.1k42161
59.1k42161
add a comment |
add a comment |
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