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Proof for $lim_k to infty P (|X| > k) = 0$ for a random variable X.
Clash Royale CLAN TAG#URR8PPP
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1
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I'm trying to show the following statement.
If $X$ is a random variable, then $lim_k to infty P (|X| > k) = 0$
What I tried:
the above statement is equivalent to the following.
$$
X text : R.V. Rightarrow forall epsilon>0, exists k>0 : Big( P(|X| > k) < epsilon Big)
$$
I defined a sequence of events $a_n = $ which is decreasing.
Since $lim_n to infty a_n = emptyset $, by the continuity of probability, $lim P(a_n) = 0$
However, I now realize that $k>0$ is a real number, so this approach might not be valid.
Is there anyone to help me?
probability random-variables
asked Jul 22 at 10:53
moreblue
1738
add a comment |Â
up vote
1
down vote
favorite
I'm trying to show the following statement.
If $X$ is a random variable, then $lim_k to infty P (|X| > k) = 0$
What I tried:
the above statement is equivalent to the following.
$$
X text : R.V. Rightarrow forall epsilon>0, exists k>0 : Big( P(|X| > k) < epsilon Big)
$$
I defined a sequence of events $a_n = $ which is decreasing.
Since $lim_n to infty a_n = emptyset $, by the continuity of probability, $lim P(a_n) = 0$
However, I now realize that $k>0$ is a real number, so this approach might not be valid.
Is there anyone to help me?
probability random-variables
asked Jul 22 at 10:53
moreblue
1738
Just as a warning to some readers: It's not that uncommon for random variables to take values in the extended real line i.e. to be able to take $pm infty$, in which case this isn't true of course.
– T_M
Jul 23 at 2:56
@T_M Here I take the common definition of 'random variable' as a real valued function : $Omega to mathbbR$
– moreblue
Jul 23 at 3:35
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to show the following statement.
If $X$ is a random variable, then $lim_k to infty P (|X| > k) = 0$
What I tried:
the above statement is equivalent to the following.
$$
X text : R.V. Rightarrow forall epsilon>0, exists k>0 : Big( P(|X| > k) < epsilon Big)
$$
I defined a sequence of events $a_n = $ which is decreasing.
Since $lim_n to infty a_n = emptyset $, by the continuity of probability, $lim P(a_n) = 0$
However, I now realize that $k>0$ is a real number, so this approach might not be valid.
Is there anyone to help me?
probability random-variables
asked Jul 22 at 10:53
moreblue
1738
I'm trying to show the following statement.
If $X$ is a random variable, then $lim_k to infty P (|X| > k) = 0$
What I tried:
the above statement is equivalent to the following.
$$
X text : R.V. Rightarrow forall epsilon>0, exists k>0 : Big( P(|X| > k) < epsilon Big)
$$
I defined a sequence of events $a_n = $ which is decreasing.
Since $lim_n to infty a_n = emptyset $, by the continuity of probability, $lim P(a_n) = 0$
However, I now realize that $k>0$ is a real number, so this approach might not be valid.
Is there anyone to help me?
probability random-variables
asked Jul 22 at 10:53
moreblue
1738
asked Jul 22 at 10:53
moreblue
1738
asked Jul 22 at 10:53
moreblue
1738
asked Jul 22 at 10:53
moreblue
1738
1738
Just as a warning to some readers: It's not that uncommon for random variables to take values in the extended real line i.e. to be able to take $pm infty$, in which case this isn't true of course.
– T_M
Jul 23 at 2:56
@T_M Here I take the common definition of 'random variable' as a real valued function : $Omega to mathbbR$
– moreblue
Jul 23 at 3:35
add a comment |Â
Just as a warning to some readers: It's not that uncommon for random variables to take values in the extended real line i.e. to be able to take $pm infty$, in which case this isn't true of course.
– T_M
Jul 23 at 2:56
@T_M Here I take the common definition of 'random variable' as a real valued function : $Omega to mathbbR$
– moreblue
Jul 23 at 3:35
Just as a warning to some readers: It's not that uncommon for random variables to take values in the extended real line i.e. to be able to take $pm infty$, in which case this isn't true of course.
– T_M
Jul 23 at 2:56
Just as a warning to some readers: It's not that uncommon for random variables to take values in the extended real line i.e. to be able to take $pm infty$, in which case this isn't true of course.
– T_M
Jul 23 at 2:56
@T_M Here I take the common definition of 'random variable' as a real valued function : $Omega to mathbbR$
– moreblue
Jul 23 at 3:35
@T_M Here I take the common definition of 'random variable' as a real valued function : $Omega to mathbbR$
– moreblue
Jul 23 at 3:35
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
Setting $B_n:=<n$ for $n=1,2,dots$ we have: $$bigcup_n=1^inftyB_n=Omega$$and because the sets $B_n$ are disjoint we conclude:$$sum_n=1^inftyP(B_n)=P(Omega)=1$$ or equivalently $$lim_ntoinftysum_k=0^nP(B_k)=1$$
so that $$lim_ntoinftysum_k=n+1^inftyP(B_k)=lim_ntoinfty1-sum_k=0^nP(B_k)=1-1=0$$
Now observe that: $$sum_k=n+1^inftyP(B_k)=P(|X|geq n)$$
because $$bigcup_k=n+1^inftyB_k=$$ and again the sets $B_k$ are disjoint.
answered Jul 22 at 11:11
drhab
86.4k541118
add a comment |Â
up vote
0
down vote
We have
$$
P(|X|>k) = 1-P(|X|leq k) = 1- P(-kleq Xleq k) = 1-P(Xleq k)-P(X<-k)
$$
Let $F(x)$ be the distribution function of $X$. Using elemental properties of this function
$$
lim_kto inftyP(Xleq k) = lim_ktoinfty F(k) = 1
$$
$$
lim_kto inftyP(X<-k) leq lim_kto inftyP(Xleq -k) = lim_kto infty F(-k)= 0
$$
Hence we have the result.
answered Jul 22 at 10:58
Rafael Gonzalez Lopez
637112
How can you verify "by definition"?
– moreblue
Jul 22 at 10:59
I'll do it again
– Rafael Gonzalez Lopez
Jul 22 at 11:00
@moreblue What do you think now?
– Rafael Gonzalez Lopez
Jul 22 at 11:02
Looks fine for me.
– moreblue
Jul 22 at 11:06
add a comment |Â
up vote
0
down vote
If $X$ has a density function $f$, we can use that
$$
P(|X| > k) = int_Bbb Rsetminus[-k,k]f =
int_Bbb Rfchi_Bbb Rsetminus[-k,k]
$$
($chi =$ characteristic function) and apply the Dominated Convergence Theorem.
answered Jul 22 at 11:26
MartÃn-Blas Pérez Pinilla
33.4k42570
1
Two objections: 1) A density might not exist. 2) Advanced means like DCT should not be used to prove elementary questions like this. It might lead easily to circular reasoning.
– drhab
Jul 22 at 16:38
@drhab, (1) True, edited. (2) Also true. But can be worse :-). See math.stackexchange.com/questions/1273862/…. No circularity in this case: the DCT is a general result.
– MartÃn-Blas Pérez Pinilla
Jul 22 at 19:44
add a comment |Â
up vote
0
down vote
If not, then there exists an $varepsilon > 0$ for which $P(|X|>k) > varepsilon$ infinitely often, meaning $P(|X| leq k) < 1-varepsilon$ infinitely often, contradicting the fact that $P(|X| leq k) = F_(k) to 1$ as $k to infty$.
answered Jul 23 at 2:50
Daniel Xiang
1,823414
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Setting $B_n:=<n$ for $n=1,2,dots$ we have: $$bigcup_n=1^inftyB_n=Omega$$and because the sets $B_n$ are disjoint we conclude:$$sum_n=1^inftyP(B_n)=P(Omega)=1$$ or equivalently $$lim_ntoinftysum_k=0^nP(B_k)=1$$
so that $$lim_ntoinftysum_k=n+1^inftyP(B_k)=lim_ntoinfty1-sum_k=0^nP(B_k)=1-1=0$$
Now observe that: $$sum_k=n+1^inftyP(B_k)=P(|X|geq n)$$
because $$bigcup_k=n+1^inftyB_k=$$ and again the sets $B_k$ are disjoint.
answered Jul 22 at 11:11
drhab
86.4k541118
add a comment |Â
up vote
1
down vote
Setting $B_n:=<n$ for $n=1,2,dots$ we have: $$bigcup_n=1^inftyB_n=Omega$$and because the sets $B_n$ are disjoint we conclude:$$sum_n=1^inftyP(B_n)=P(Omega)=1$$ or equivalently $$lim_ntoinftysum_k=0^nP(B_k)=1$$
so that $$lim_ntoinftysum_k=n+1^inftyP(B_k)=lim_ntoinfty1-sum_k=0^nP(B_k)=1-1=0$$
Now observe that: $$sum_k=n+1^inftyP(B_k)=P(|X|geq n)$$
because $$bigcup_k=n+1^inftyB_k=$$ and again the sets $B_k$ are disjoint.
answered Jul 22 at 11:11
drhab
86.4k541118
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Setting $B_n:=<n$ for $n=1,2,dots$ we have: $$bigcup_n=1^inftyB_n=Omega$$and because the sets $B_n$ are disjoint we conclude:$$sum_n=1^inftyP(B_n)=P(Omega)=1$$ or equivalently $$lim_ntoinftysum_k=0^nP(B_k)=1$$
so that $$lim_ntoinftysum_k=n+1^inftyP(B_k)=lim_ntoinfty1-sum_k=0^nP(B_k)=1-1=0$$
Now observe that: $$sum_k=n+1^inftyP(B_k)=P(|X|geq n)$$
because $$bigcup_k=n+1^inftyB_k=$$ and again the sets $B_k$ are disjoint.
answered Jul 22 at 11:11
drhab
86.4k541118
Setting $B_n:=<n$ for $n=1,2,dots$ we have: $$bigcup_n=1^inftyB_n=Omega$$and because the sets $B_n$ are disjoint we conclude:$$sum_n=1^inftyP(B_n)=P(Omega)=1$$ or equivalently $$lim_ntoinftysum_k=0^nP(B_k)=1$$
so that $$lim_ntoinftysum_k=n+1^inftyP(B_k)=lim_ntoinfty1-sum_k=0^nP(B_k)=1-1=0$$
Now observe that: $$sum_k=n+1^inftyP(B_k)=P(|X|geq n)$$
because $$bigcup_k=n+1^inftyB_k=$$ and again the sets $B_k$ are disjoint.
answered Jul 22 at 11:11
drhab
86.4k541118
edited Jul 22 at 11:27
answered Jul 22 at 11:11
drhab
86.4k541118
answered Jul 22 at 11:11
drhab
86.4k541118
answered Jul 22 at 11:11
drhab
86.4k541118
86.4k541118
add a comment |Â
add a comment |Â
up vote
0
down vote
We have
$$
P(|X|>k) = 1-P(|X|leq k) = 1- P(-kleq Xleq k) = 1-P(Xleq k)-P(X<-k)
$$
Let $F(x)$ be the distribution function of $X$. Using elemental properties of this function
$$
lim_kto inftyP(Xleq k) = lim_ktoinfty F(k) = 1
$$
$$
lim_kto inftyP(X<-k) leq lim_kto inftyP(Xleq -k) = lim_kto infty F(-k)= 0
$$
Hence we have the result.
answered Jul 22 at 10:58
Rafael Gonzalez Lopez
637112
How can you verify "by definition"?
– moreblue
Jul 22 at 10:59
I'll do it again
– Rafael Gonzalez Lopez
Jul 22 at 11:00
@moreblue What do you think now?
– Rafael Gonzalez Lopez
Jul 22 at 11:02
Looks fine for me.
– moreblue
Jul 22 at 11:06
add a comment |Â
up vote
0
down vote
We have
$$
P(|X|>k) = 1-P(|X|leq k) = 1- P(-kleq Xleq k) = 1-P(Xleq k)-P(X<-k)
$$
Let $F(x)$ be the distribution function of $X$. Using elemental properties of this function
$$
lim_kto inftyP(Xleq k) = lim_ktoinfty F(k) = 1
$$
$$
lim_kto inftyP(X<-k) leq lim_kto inftyP(Xleq -k) = lim_kto infty F(-k)= 0
$$
Hence we have the result.
answered Jul 22 at 10:58
Rafael Gonzalez Lopez
637112
How can you verify "by definition"?
– moreblue
Jul 22 at 10:59
I'll do it again
– Rafael Gonzalez Lopez
Jul 22 at 11:00
@moreblue What do you think now?
– Rafael Gonzalez Lopez
Jul 22 at 11:02
Looks fine for me.
– moreblue
Jul 22 at 11:06
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have
$$
P(|X|>k) = 1-P(|X|leq k) = 1- P(-kleq Xleq k) = 1-P(Xleq k)-P(X<-k)
$$
Let $F(x)$ be the distribution function of $X$. Using elemental properties of this function
$$
lim_kto inftyP(Xleq k) = lim_ktoinfty F(k) = 1
$$
$$
lim_kto inftyP(X<-k) leq lim_kto inftyP(Xleq -k) = lim_kto infty F(-k)= 0
$$
Hence we have the result.
answered Jul 22 at 10:58
Rafael Gonzalez Lopez
637112
We have
$$
P(|X|>k) = 1-P(|X|leq k) = 1- P(-kleq Xleq k) = 1-P(Xleq k)-P(X<-k)
$$
Let $F(x)$ be the distribution function of $X$. Using elemental properties of this function
$$
lim_kto inftyP(Xleq k) = lim_ktoinfty F(k) = 1
$$
$$
lim_kto inftyP(X<-k) leq lim_kto inftyP(Xleq -k) = lim_kto infty F(-k)= 0
$$
Hence we have the result.
answered Jul 22 at 10:58
Rafael Gonzalez Lopez
637112
edited Jul 22 at 11:02
answered Jul 22 at 10:58
Rafael Gonzalez Lopez
637112
answered Jul 22 at 10:58
Rafael Gonzalez Lopez
637112
answered Jul 22 at 10:58
Rafael Gonzalez Lopez
637112
637112
How can you verify "by definition"?
– moreblue
Jul 22 at 10:59
I'll do it again
– Rafael Gonzalez Lopez
Jul 22 at 11:00
@moreblue What do you think now?
– Rafael Gonzalez Lopez
Jul 22 at 11:02
Looks fine for me.
– moreblue
Jul 22 at 11:06
add a comment |Â
How can you verify "by definition"?
– moreblue
Jul 22 at 10:59
I'll do it again
– Rafael Gonzalez Lopez
Jul 22 at 11:00
@moreblue What do you think now?
– Rafael Gonzalez Lopez
Jul 22 at 11:02
Looks fine for me.
– moreblue
Jul 22 at 11:06
How can you verify "by definition"?
– moreblue
Jul 22 at 10:59
How can you verify "by definition"?
– moreblue
Jul 22 at 10:59
I'll do it again
– Rafael Gonzalez Lopez
Jul 22 at 11:00
I'll do it again
– Rafael Gonzalez Lopez
Jul 22 at 11:00
@moreblue What do you think now?
– Rafael Gonzalez Lopez
Jul 22 at 11:02
@moreblue What do you think now?
– Rafael Gonzalez Lopez
Jul 22 at 11:02
Looks fine for me.
– moreblue
Jul 22 at 11:06
Looks fine for me.
– moreblue
Jul 22 at 11:06
add a comment |Â
up vote
0
down vote
If $X$ has a density function $f$, we can use that
$$
P(|X| > k) = int_Bbb Rsetminus[-k,k]f =
int_Bbb Rfchi_Bbb Rsetminus[-k,k]
$$
($chi =$ characteristic function) and apply the Dominated Convergence Theorem.
answered Jul 22 at 11:26
MartÃn-Blas Pérez Pinilla
33.4k42570
1
Two objections: 1) A density might not exist. 2) Advanced means like DCT should not be used to prove elementary questions like this. It might lead easily to circular reasoning.
– drhab
Jul 22 at 16:38
@drhab, (1) True, edited. (2) Also true. But can be worse :-). See math.stackexchange.com/questions/1273862/…. No circularity in this case: the DCT is a general result.
– MartÃn-Blas Pérez Pinilla
Jul 22 at 19:44
add a comment |Â
up vote
0
down vote
If $X$ has a density function $f$, we can use that
$$
P(|X| > k) = int_Bbb Rsetminus[-k,k]f =
int_Bbb Rfchi_Bbb Rsetminus[-k,k]
$$
($chi =$ characteristic function) and apply the Dominated Convergence Theorem.
answered Jul 22 at 11:26
MartÃn-Blas Pérez Pinilla
33.4k42570
1
Two objections: 1) A density might not exist. 2) Advanced means like DCT should not be used to prove elementary questions like this. It might lead easily to circular reasoning.
– drhab
Jul 22 at 16:38
@drhab, (1) True, edited. (2) Also true. But can be worse :-). See math.stackexchange.com/questions/1273862/…. No circularity in this case: the DCT is a general result.
– MartÃn-Blas Pérez Pinilla
Jul 22 at 19:44
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $X$ has a density function $f$, we can use that
$$
P(|X| > k) = int_Bbb Rsetminus[-k,k]f =
int_Bbb Rfchi_Bbb Rsetminus[-k,k]
$$
($chi =$ characteristic function) and apply the Dominated Convergence Theorem.
answered Jul 22 at 11:26
MartÃn-Blas Pérez Pinilla
33.4k42570
If $X$ has a density function $f$, we can use that
$$
P(|X| > k) = int_Bbb Rsetminus[-k,k]f =
int_Bbb Rfchi_Bbb Rsetminus[-k,k]
$$
($chi =$ characteristic function) and apply the Dominated Convergence Theorem.
answered Jul 22 at 11:26
MartÃn-Blas Pérez Pinilla
33.4k42570
edited Jul 22 at 19:41
answered Jul 22 at 11:26
MartÃn-Blas Pérez Pinilla
33.4k42570
answered Jul 22 at 11:26
MartÃn-Blas Pérez Pinilla
33.4k42570
answered Jul 22 at 11:26
MartÃn-Blas Pérez Pinilla
33.4k42570
33.4k42570
1
Two objections: 1) A density might not exist. 2) Advanced means like DCT should not be used to prove elementary questions like this. It might lead easily to circular reasoning.
– drhab
Jul 22 at 16:38
@drhab, (1) True, edited. (2) Also true. But can be worse :-). See math.stackexchange.com/questions/1273862/…. No circularity in this case: the DCT is a general result.
– MartÃn-Blas Pérez Pinilla
Jul 22 at 19:44
add a comment |Â
1
Two objections: 1) A density might not exist. 2) Advanced means like DCT should not be used to prove elementary questions like this. It might lead easily to circular reasoning.
– drhab
Jul 22 at 16:38
@drhab, (1) True, edited. (2) Also true. But can be worse :-). See math.stackexchange.com/questions/1273862/…. No circularity in this case: the DCT is a general result.
– MartÃn-Blas Pérez Pinilla
Jul 22 at 19:44
1
1
Two objections: 1) A density might not exist. 2) Advanced means like DCT should not be used to prove elementary questions like this. It might lead easily to circular reasoning.
– drhab
Jul 22 at 16:38
Two objections: 1) A density might not exist. 2) Advanced means like DCT should not be used to prove elementary questions like this. It might lead easily to circular reasoning.
– drhab
Jul 22 at 16:38
@drhab, (1) True, edited. (2) Also true. But can be worse :-). See math.stackexchange.com/questions/1273862/…. No circularity in this case: the DCT is a general result.
– MartÃn-Blas Pérez Pinilla
Jul 22 at 19:44
@drhab, (1) True, edited. (2) Also true. But can be worse :-). See math.stackexchange.com/questions/1273862/…. No circularity in this case: the DCT is a general result.
– MartÃn-Blas Pérez Pinilla
Jul 22 at 19:44
add a comment |Â
up vote
0
down vote
If not, then there exists an $varepsilon > 0$ for which $P(|X|>k) > varepsilon$ infinitely often, meaning $P(|X| leq k) < 1-varepsilon$ infinitely often, contradicting the fact that $P(|X| leq k) = F_(k) to 1$ as $k to infty$.
answered Jul 23 at 2:50
Daniel Xiang
1,823414
add a comment |Â
up vote
0
down vote
If not, then there exists an $varepsilon > 0$ for which $P(|X|>k) > varepsilon$ infinitely often, meaning $P(|X| leq k) < 1-varepsilon$ infinitely often, contradicting the fact that $P(|X| leq k) = F_(k) to 1$ as $k to infty$.
answered Jul 23 at 2:50
Daniel Xiang
1,823414
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If not, then there exists an $varepsilon > 0$ for which $P(|X|>k) > varepsilon$ infinitely often, meaning $P(|X| leq k) < 1-varepsilon$ infinitely often, contradicting the fact that $P(|X| leq k) = F_(k) to 1$ as $k to infty$.
answered Jul 23 at 2:50
Daniel Xiang
1,823414
If not, then there exists an $varepsilon > 0$ for which $P(|X|>k) > varepsilon$ infinitely often, meaning $P(|X| leq k) < 1-varepsilon$ infinitely often, contradicting the fact that $P(|X| leq k) = F_(k) to 1$ as $k to infty$.
answered Jul 23 at 2:50
Daniel Xiang
1,823414
answered Jul 23 at 2:50
Daniel Xiang
1,823414
answered Jul 23 at 2:50
Daniel Xiang
1,823414
answered Jul 23 at 2:50
Daniel Xiang
1,823414
1,823414
add a comment |Â
add a comment |Â
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Just as a warning to some readers: It's not that uncommon for random variables to take values in the extended real line i.e. to be able to take $pm infty$, in which case this isn't true of course.
– T_M
Jul 23 at 2:56
@T_M Here I take the common definition of 'random variable' as a real valued function : $Omega to mathbbR$
– moreblue
Jul 23 at 3:35