Bound for probability with almost sure convergence
Let $(X_n)_n$ be random variables which converge almost surely to a constant $x in mathbb R$, i.e. $X_n xrightarrow{n to infty} x$ a.s. Let $Y$ be another random variable.
Question: Can I say something like $$mathrm{Pr}(Y geq X_n) leq mathrm{Pr} left(Y geq frac{x}{2} right)$$ for $n$ large enough?
probability probability-theory convergence stochastic-processes random-variables
asked Nov 27 at 8:39
Kariani
51
add a comment |
Let $(X_n)_n$ be random variables which converge almost surely to a constant $x in mathbb R$, i.e. $X_n xrightarrow{n to infty} x$ a.s. Let $Y$ be another random variable.
Question: Can I say something like $$mathrm{Pr}(Y geq X_n) leq mathrm{Pr} left(Y geq frac{x}{2} right)$$ for $n$ large enough?
probability probability-theory convergence stochastic-processes random-variables
asked Nov 27 at 8:39
Kariani
51
This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44
add a comment |
Let $(X_n)_n$ be random variables which converge almost surely to a constant $x in mathbb R$, i.e. $X_n xrightarrow{n to infty} x$ a.s. Let $Y$ be another random variable.
Question: Can I say something like $$mathrm{Pr}(Y geq X_n) leq mathrm{Pr} left(Y geq frac{x}{2} right)$$ for $n$ large enough?
probability probability-theory convergence stochastic-processes random-variables
asked Nov 27 at 8:39
Kariani
51
Let $(X_n)_n$ be random variables which converge almost surely to a constant $x in mathbb R$, i.e. $X_n xrightarrow{n to infty} x$ a.s. Let $Y$ be another random variable.
Question: Can I say something like $$mathrm{Pr}(Y geq X_n) leq mathrm{Pr} left(Y geq frac{x}{2} right)$$ for $n$ large enough?
probability probability-theory convergence stochastic-processes random-variables
probability probability-theory convergence stochastic-processes random-variables
asked Nov 27 at 8:39
Kariani
51
asked Nov 27 at 8:39
Kariani
51
asked Nov 27 at 8:39
Kariani
51
asked Nov 27 at 8:39
Kariani
51
asked Nov 27 at 8:39
Kariani
51
51
This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44
add a comment |
This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44
This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44
This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44
add a comment |
1 Answer
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No, you cannot have such an inequality. Let $X_n$ take the values $0$ and $1$ with probabilities $frac 1 {2^{n}}$ and $1-frac 1 {2^{n}}$ respectively. Since $sum P(X_n=0)<infty$ Borel cantalli Lemma shows that $X_n to 1$ almost surely. Now take $Y=frac 1 4$. Then $P(Ygeq X_n) =frac 1 {2^{n}}$ and $P(Y geq frac 1 2)=0$.
answered Nov 27 at 8:46
Kavi Rama Murthy
49.9k31854
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
No, you cannot have such an inequality. Let $X_n$ take the values $0$ and $1$ with probabilities $frac 1 {2^{n}}$ and $1-frac 1 {2^{n}}$ respectively. Since $sum P(X_n=0)<infty$ Borel cantalli Lemma shows that $X_n to 1$ almost surely. Now take $Y=frac 1 4$. Then $P(Ygeq X_n) =frac 1 {2^{n}}$ and $P(Y geq frac 1 2)=0$.
answered Nov 27 at 8:46
Kavi Rama Murthy
49.9k31854
add a comment |
No, you cannot have such an inequality. Let $X_n$ take the values $0$ and $1$ with probabilities $frac 1 {2^{n}}$ and $1-frac 1 {2^{n}}$ respectively. Since $sum P(X_n=0)<infty$ Borel cantalli Lemma shows that $X_n to 1$ almost surely. Now take $Y=frac 1 4$. Then $P(Ygeq X_n) =frac 1 {2^{n}}$ and $P(Y geq frac 1 2)=0$.
answered Nov 27 at 8:46
Kavi Rama Murthy
49.9k31854
add a comment |
No, you cannot have such an inequality. Let $X_n$ take the values $0$ and $1$ with probabilities $frac 1 {2^{n}}$ and $1-frac 1 {2^{n}}$ respectively. Since $sum P(X_n=0)<infty$ Borel cantalli Lemma shows that $X_n to 1$ almost surely. Now take $Y=frac 1 4$. Then $P(Ygeq X_n) =frac 1 {2^{n}}$ and $P(Y geq frac 1 2)=0$.
answered Nov 27 at 8:46
Kavi Rama Murthy
49.9k31854
No, you cannot have such an inequality. Let $X_n$ take the values $0$ and $1$ with probabilities $frac 1 {2^{n}}$ and $1-frac 1 {2^{n}}$ respectively. Since $sum P(X_n=0)<infty$ Borel cantalli Lemma shows that $X_n to 1$ almost surely. Now take $Y=frac 1 4$. Then $P(Ygeq X_n) =frac 1 {2^{n}}$ and $P(Y geq frac 1 2)=0$.
answered Nov 27 at 8:46
Kavi Rama Murthy
49.9k31854
answered Nov 27 at 8:46
Kavi Rama Murthy
49.9k31854
answered Nov 27 at 8:46
Kavi Rama Murthy
49.9k31854
answered Nov 27 at 8:46
Kavi Rama Murthy
49.9k31854
49.9k31854
add a comment |
add a comment |
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This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44