$lim_{xrightarrow+infty}...
$begingroup$
I have met with a probability problem which I have no idea to deal with. It says:
Let $alpha>0$, $betageq0$, prove:
$lim_{xrightarrow+infty} x^{alpha+beta}mathbb{P}(|X|>x)=0Leftrightarrowlim_{xrightarrow+infty}x^{alpha}mathbb{E}||X|^{beta}cdot1_{{|X|>x}}|=0$.
The sufficiency is easy to prove. However, I have no ideas how to prove the necessity. Can you give me some advice? Any hints would be appreciated.
Thanks for your time and patience.
probability-theory probability-limit-theorems
asked Dec 30 '18 at 2:48
CsnakeCsnake
111
$endgroup$
add a comment |
$begingroup$
I have met with a probability problem which I have no idea to deal with. It says:
Let $alpha>0$, $betageq0$, prove:
$lim_{xrightarrow+infty} x^{alpha+beta}mathbb{P}(|X|>x)=0Leftrightarrowlim_{xrightarrow+infty}x^{alpha}mathbb{E}||X|^{beta}cdot1_{{|X|>x}}|=0$.
The sufficiency is easy to prove. However, I have no ideas how to prove the necessity. Can you give me some advice? Any hints would be appreciated.
Thanks for your time and patience.
probability-theory probability-limit-theorems
asked Dec 30 '18 at 2:48
CsnakeCsnake
111
$endgroup$
add a comment |
$begingroup$
I have met with a probability problem which I have no idea to deal with. It says:
Let $alpha>0$, $betageq0$, prove:
$lim_{xrightarrow+infty} x^{alpha+beta}mathbb{P}(|X|>x)=0Leftrightarrowlim_{xrightarrow+infty}x^{alpha}mathbb{E}||X|^{beta}cdot1_{{|X|>x}}|=0$.
The sufficiency is easy to prove. However, I have no ideas how to prove the necessity. Can you give me some advice? Any hints would be appreciated.
Thanks for your time and patience.
probability-theory probability-limit-theorems
asked Dec 30 '18 at 2:48
CsnakeCsnake
111
$endgroup$
I have met with a probability problem which I have no idea to deal with. It says:
Let $alpha>0$, $betageq0$, prove:
$lim_{xrightarrow+infty} x^{alpha+beta}mathbb{P}(|X|>x)=0Leftrightarrowlim_{xrightarrow+infty}x^{alpha}mathbb{E}||X|^{beta}cdot1_{{|X|>x}}|=0$.
The sufficiency is easy to prove. However, I have no ideas how to prove the necessity. Can you give me some advice? Any hints would be appreciated.
Thanks for your time and patience.
probability-theory probability-limit-theorems
probability-theory probability-limit-theorems
asked Dec 30 '18 at 2:48
CsnakeCsnake
111
asked Dec 30 '18 at 2:48
CsnakeCsnake
111
asked Dec 30 '18 at 2:48
CsnakeCsnake
111
asked Dec 30 '18 at 2:48
CsnakeCsnake
111
asked Dec 30 '18 at 2:48
CsnakeCsnake
111
111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can simplify the statement by writing $Y$ for $|X|^{beta}$ and repalcing $x^{beta} $ by $t$. The statement then becomes the following: $t^{1+s}P(Y>t) to 0$ iff $t^{s}EYI_{Y>t} to 0$ as $t to infty$.[Here $s=frac {alpha} {beta}$. I will leave the case $beta=0$ to you]. As noted by you the íf'part follows easily by Chebychev's Inequality. For the ónly if'part write $t^{s}EYI_{Y>t} to 0$ as $t^{s}int_t^{infty} P(Y>u)du$ which is less than or equal to $t^{s}int_t^{infty} epsilon u^{-1-s}du=epsilon$ for $t$ sufficiently large.
answered Dec 30 '18 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
$endgroup$
$begingroup$
Oh, I get your idea. The key point is to rewrite $mathbb{E}[Ycdot 1_{{Y>t}}]$. The case where $beta=0$ is an identity. Thanks a lot.
$endgroup$
– Csnake
Dec 30 '18 at 6:02
add a comment |
$begingroup$
Suppose
$lim_{xto infty} x^{alpha+beta}P(|X|>x) = 0$.
Note that
begin{align}
E[|X|^beta cdot 1_{|X|>x}]
&= P(|X|>x)E[|X|^beta | |X| > x] \
&= x^beta P(|X|>x) E[|X/x|^beta | |X/x|>1] \
&le x^beta P(|X|>x) E[|X|^beta], quad text{for large $x$}.
end{align}
Note that for large $x$, $|X/x| < |X|$, and thus
$|X/x|^beta < |X|^beta$.
Assuming $E[|X|^beta] < infty$,
$$
lim_{x to infty} x^alpha E[|X|^beta cdot 1_{|X|>x}] = 0.
$$
answered Dec 30 '18 at 3:53
induction601induction601
1,307314
$endgroup$
$begingroup$
Thanks. But I do not know why the first equation holds. In fact, on the right hand $mathbb{E}[|X|^{beta}| |X|>x]$ is a random variable, but the left hand is a scalar, though a similar formula holds when $mathbb{E}$ changes to $mathbb{P}$.
$endgroup$
– Csnake
Dec 30 '18 at 5:40
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can simplify the statement by writing $Y$ for $|X|^{beta}$ and repalcing $x^{beta} $ by $t$. The statement then becomes the following: $t^{1+s}P(Y>t) to 0$ iff $t^{s}EYI_{Y>t} to 0$ as $t to infty$.[Here $s=frac {alpha} {beta}$. I will leave the case $beta=0$ to you]. As noted by you the íf'part follows easily by Chebychev's Inequality. For the ónly if'part write $t^{s}EYI_{Y>t} to 0$ as $t^{s}int_t^{infty} P(Y>u)du$ which is less than or equal to $t^{s}int_t^{infty} epsilon u^{-1-s}du=epsilon$ for $t$ sufficiently large.
answered Dec 30 '18 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
$endgroup$
$begingroup$
Oh, I get your idea. The key point is to rewrite $mathbb{E}[Ycdot 1_{{Y>t}}]$. The case where $beta=0$ is an identity. Thanks a lot.
$endgroup$
– Csnake
Dec 30 '18 at 6:02
add a comment |
$begingroup$
You can simplify the statement by writing $Y$ for $|X|^{beta}$ and repalcing $x^{beta} $ by $t$. The statement then becomes the following: $t^{1+s}P(Y>t) to 0$ iff $t^{s}EYI_{Y>t} to 0$ as $t to infty$.[Here $s=frac {alpha} {beta}$. I will leave the case $beta=0$ to you]. As noted by you the íf'part follows easily by Chebychev's Inequality. For the ónly if'part write $t^{s}EYI_{Y>t} to 0$ as $t^{s}int_t^{infty} P(Y>u)du$ which is less than or equal to $t^{s}int_t^{infty} epsilon u^{-1-s}du=epsilon$ for $t$ sufficiently large.
answered Dec 30 '18 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
$endgroup$
$begingroup$
Oh, I get your idea. The key point is to rewrite $mathbb{E}[Ycdot 1_{{Y>t}}]$. The case where $beta=0$ is an identity. Thanks a lot.
$endgroup$
– Csnake
Dec 30 '18 at 6:02
add a comment |
$begingroup$
You can simplify the statement by writing $Y$ for $|X|^{beta}$ and repalcing $x^{beta} $ by $t$. The statement then becomes the following: $t^{1+s}P(Y>t) to 0$ iff $t^{s}EYI_{Y>t} to 0$ as $t to infty$.[Here $s=frac {alpha} {beta}$. I will leave the case $beta=0$ to you]. As noted by you the íf'part follows easily by Chebychev's Inequality. For the ónly if'part write $t^{s}EYI_{Y>t} to 0$ as $t^{s}int_t^{infty} P(Y>u)du$ which is less than or equal to $t^{s}int_t^{infty} epsilon u^{-1-s}du=epsilon$ for $t$ sufficiently large.
answered Dec 30 '18 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
$endgroup$
You can simplify the statement by writing $Y$ for $|X|^{beta}$ and repalcing $x^{beta} $ by $t$. The statement then becomes the following: $t^{1+s}P(Y>t) to 0$ iff $t^{s}EYI_{Y>t} to 0$ as $t to infty$.[Here $s=frac {alpha} {beta}$. I will leave the case $beta=0$ to you]. As noted by you the íf'part follows easily by Chebychev's Inequality. For the ónly if'part write $t^{s}EYI_{Y>t} to 0$ as $t^{s}int_t^{infty} P(Y>u)du$ which is less than or equal to $t^{s}int_t^{infty} epsilon u^{-1-s}du=epsilon$ for $t$ sufficiently large.
answered Dec 30 '18 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
answered Dec 30 '18 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
answered Dec 30 '18 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
answered Dec 30 '18 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
69.3k53169
$begingroup$
Oh, I get your idea. The key point is to rewrite $mathbb{E}[Ycdot 1_{{Y>t}}]$. The case where $beta=0$ is an identity. Thanks a lot.
$endgroup$
– Csnake
Dec 30 '18 at 6:02
add a comment |
$begingroup$
Oh, I get your idea. The key point is to rewrite $mathbb{E}[Ycdot 1_{{Y>t}}]$. The case where $beta=0$ is an identity. Thanks a lot.
$endgroup$
– Csnake
Dec 30 '18 at 6:02
$begingroup$
Oh, I get your idea. The key point is to rewrite $mathbb{E}[Ycdot 1_{{Y>t}}]$. The case where $beta=0$ is an identity. Thanks a lot.
$endgroup$
– Csnake
Dec 30 '18 at 6:02
$begingroup$
Oh, I get your idea. The key point is to rewrite $mathbb{E}[Ycdot 1_{{Y>t}}]$. The case where $beta=0$ is an identity. Thanks a lot.
$endgroup$
– Csnake
Dec 30 '18 at 6:02
add a comment |
$begingroup$
Suppose
$lim_{xto infty} x^{alpha+beta}P(|X|>x) = 0$.
Note that
begin{align}
E[|X|^beta cdot 1_{|X|>x}]
&= P(|X|>x)E[|X|^beta | |X| > x] \
&= x^beta P(|X|>x) E[|X/x|^beta | |X/x|>1] \
&le x^beta P(|X|>x) E[|X|^beta], quad text{for large $x$}.
end{align}
Note that for large $x$, $|X/x| < |X|$, and thus
$|X/x|^beta < |X|^beta$.
Assuming $E[|X|^beta] < infty$,
$$
lim_{x to infty} x^alpha E[|X|^beta cdot 1_{|X|>x}] = 0.
$$
answered Dec 30 '18 at 3:53
induction601induction601
1,307314
$endgroup$
$begingroup$
Thanks. But I do not know why the first equation holds. In fact, on the right hand $mathbb{E}[|X|^{beta}| |X|>x]$ is a random variable, but the left hand is a scalar, though a similar formula holds when $mathbb{E}$ changes to $mathbb{P}$.
$endgroup$
– Csnake
Dec 30 '18 at 5:40
add a comment |
$begingroup$
Suppose
$lim_{xto infty} x^{alpha+beta}P(|X|>x) = 0$.
Note that
begin{align}
E[|X|^beta cdot 1_{|X|>x}]
&= P(|X|>x)E[|X|^beta | |X| > x] \
&= x^beta P(|X|>x) E[|X/x|^beta | |X/x|>1] \
&le x^beta P(|X|>x) E[|X|^beta], quad text{for large $x$}.
end{align}
Note that for large $x$, $|X/x| < |X|$, and thus
$|X/x|^beta < |X|^beta$.
Assuming $E[|X|^beta] < infty$,
$$
lim_{x to infty} x^alpha E[|X|^beta cdot 1_{|X|>x}] = 0.
$$
answered Dec 30 '18 at 3:53
induction601induction601
1,307314
$endgroup$
$begingroup$
Thanks. But I do not know why the first equation holds. In fact, on the right hand $mathbb{E}[|X|^{beta}| |X|>x]$ is a random variable, but the left hand is a scalar, though a similar formula holds when $mathbb{E}$ changes to $mathbb{P}$.
$endgroup$
– Csnake
Dec 30 '18 at 5:40
add a comment |
$begingroup$
Suppose
$lim_{xto infty} x^{alpha+beta}P(|X|>x) = 0$.
Note that
begin{align}
E[|X|^beta cdot 1_{|X|>x}]
&= P(|X|>x)E[|X|^beta | |X| > x] \
&= x^beta P(|X|>x) E[|X/x|^beta | |X/x|>1] \
&le x^beta P(|X|>x) E[|X|^beta], quad text{for large $x$}.
end{align}
Note that for large $x$, $|X/x| < |X|$, and thus
$|X/x|^beta < |X|^beta$.
Assuming $E[|X|^beta] < infty$,
$$
lim_{x to infty} x^alpha E[|X|^beta cdot 1_{|X|>x}] = 0.
$$
answered Dec 30 '18 at 3:53
induction601induction601
1,307314
$endgroup$
Suppose
$lim_{xto infty} x^{alpha+beta}P(|X|>x) = 0$.
Note that
begin{align}
E[|X|^beta cdot 1_{|X|>x}]
&= P(|X|>x)E[|X|^beta | |X| > x] \
&= x^beta P(|X|>x) E[|X/x|^beta | |X/x|>1] \
&le x^beta P(|X|>x) E[|X|^beta], quad text{for large $x$}.
end{align}
Note that for large $x$, $|X/x| < |X|$, and thus
$|X/x|^beta < |X|^beta$.
Assuming $E[|X|^beta] < infty$,
$$
lim_{x to infty} x^alpha E[|X|^beta cdot 1_{|X|>x}] = 0.
$$
answered Dec 30 '18 at 3:53
induction601induction601
1,307314
answered Dec 30 '18 at 3:53
induction601induction601
1,307314
answered Dec 30 '18 at 3:53
induction601induction601
1,307314
answered Dec 30 '18 at 3:53
induction601induction601
1,307314
1,307314
$begingroup$
Thanks. But I do not know why the first equation holds. In fact, on the right hand $mathbb{E}[|X|^{beta}| |X|>x]$ is a random variable, but the left hand is a scalar, though a similar formula holds when $mathbb{E}$ changes to $mathbb{P}$.
$endgroup$
– Csnake
Dec 30 '18 at 5:40
add a comment |
$begingroup$
Thanks. But I do not know why the first equation holds. In fact, on the right hand $mathbb{E}[|X|^{beta}| |X|>x]$ is a random variable, but the left hand is a scalar, though a similar formula holds when $mathbb{E}$ changes to $mathbb{P}$.
$endgroup$
– Csnake
Dec 30 '18 at 5:40
$begingroup$
Thanks. But I do not know why the first equation holds. In fact, on the right hand $mathbb{E}[|X|^{beta}| |X|>x]$ is a random variable, but the left hand is a scalar, though a similar formula holds when $mathbb{E}$ changes to $mathbb{P}$.
$endgroup$
– Csnake
Dec 30 '18 at 5:40
$begingroup$
Thanks. But I do not know why the first equation holds. In fact, on the right hand $mathbb{E}[|X|^{beta}| |X|>x]$ is a random variable, but the left hand is a scalar, though a similar formula holds when $mathbb{E}$ changes to $mathbb{P}$.
$endgroup$
– Csnake
Dec 30 '18 at 5:40
add a comment |
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