Replacing continuous variables in a limit with a sequence The Next CEO of Stack OverflowHeine definition of limit of a function at infinity using sequencesProof of limit and limit pointProve that the CDF of a random variable is always right-continuousWhat are the implications of the definition of limiting distribution?Does using Heines definition of functions limit turns the function into a sequence?Proof - Limits of CDFProof; distribution function has limit 1Erroneous argument that every distribution function is left continuous.Proof verification, limit of cumulative distribution functionProving the cdf limit properties in generic caseProof verification. If $x_n$ is a monotone sequence and it has a convergent subsequence $x_n_k$, then $x_n$ is convergent to the same limit.
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Replacing continuous variables in a limit with a sequence
The Next CEO of Stack OverflowHeine definition of limit of a function at infinity using sequencesProof of limit and limit pointProve that the CDF of a random variable is always right-continuousWhat are the implications of the definition of limiting distribution?Does using Heines definition of functions limit turns the function into a sequence?Proof - Limits of CDFProof; distribution function has limit 1Erroneous argument that every distribution function is left continuous.Proof verification, limit of cumulative distribution functionProving the cdf limit properties in generic caseProof verification. If $x_n$ is a monotone sequence and it has a convergent subsequence $x_n_k$, then $x_n$ is convergent to the same limit.
$begingroup$
I have a question regarding the nuts and bolts involved in the proof of the limit of CDFs. The statement is that
Proposition: Let $X$ be a random variable with CDF $F_X(.)$. Then $F_X(.)$ posses the following property.
$$lim_x to infty F_X(x) = 1$$
Proof:
Consider a sequence $x_n$ with $ n in mathbbN$ such that it monotonically increases to $infty$. Then we have
begineqnarray
lim_x to inftyF_X(x) &=& lim_x to infty mathbbP(X leq x) \
&=& lim_n to infty mathbbP(X leq x_n) labeleqnref \
&=& mathbbP left bigcup_n in mathbbN ω : X(ω) ≤ x_n right \
&=& mathbbP(Omega) \
&=& 1.
endeqnarray
My question is regarding the second step where the continuous variable $x$ is replaced by the member of a sequence $x_n$. I feel lack of rigor in this step. To be precise, my questions are
- Why is this step valid?
- The trajectory that $x$ can take while approaching $infty$ are many, while the sequence $x_n$ is assumed to be monotonically increasing. How do we know for sure that this difference in the way to approach infinity will not change the limit?
- Is there a way to make the proof look more rigorous as in is there a rigorous way to substantiate this step of replacing $x$ with $x_n$?
Please help.
limits probability-theory measure-theory probability-distributions probability-limit-theorems
asked Mar 27 at 19:21
TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr
13112
$endgroup$
add a comment |
$begingroup$
I have a question regarding the nuts and bolts involved in the proof of the limit of CDFs. The statement is that
Proposition: Let $X$ be a random variable with CDF $F_X(.)$. Then $F_X(.)$ posses the following property.
$$lim_x to infty F_X(x) = 1$$
Proof:
Consider a sequence $x_n$ with $ n in mathbbN$ such that it monotonically increases to $infty$. Then we have
begineqnarray
lim_x to inftyF_X(x) &=& lim_x to infty mathbbP(X leq x) \
&=& lim_n to infty mathbbP(X leq x_n) labeleqnref \
&=& mathbbP left bigcup_n in mathbbN ω : X(ω) ≤ x_n right \
&=& mathbbP(Omega) \
&=& 1.
endeqnarray
My question is regarding the second step where the continuous variable $x$ is replaced by the member of a sequence $x_n$. I feel lack of rigor in this step. To be precise, my questions are
- Why is this step valid?
- The trajectory that $x$ can take while approaching $infty$ are many, while the sequence $x_n$ is assumed to be monotonically increasing. How do we know for sure that this difference in the way to approach infinity will not change the limit?
- Is there a way to make the proof look more rigorous as in is there a rigorous way to substantiate this step of replacing $x$ with $x_n$?
Please help.
limits probability-theory measure-theory probability-distributions probability-limit-theorems
asked Mar 27 at 19:21
TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr
13112
$endgroup$
1
$begingroup$
math.stackexchange.com/questions/1643588/…
$endgroup$
– d.k.o.
Mar 27 at 20:50
$begingroup$
@d.k.o Thanks for pointing it out!
$endgroup$
– TryingHardToBecomeAGoodPrSlvr
Mar 27 at 21:00
add a comment |
$begingroup$
I have a question regarding the nuts and bolts involved in the proof of the limit of CDFs. The statement is that
Proposition: Let $X$ be a random variable with CDF $F_X(.)$. Then $F_X(.)$ posses the following property.
$$lim_x to infty F_X(x) = 1$$
Proof:
Consider a sequence $x_n$ with $ n in mathbbN$ such that it monotonically increases to $infty$. Then we have
begineqnarray
lim_x to inftyF_X(x) &=& lim_x to infty mathbbP(X leq x) \
&=& lim_n to infty mathbbP(X leq x_n) labeleqnref \
&=& mathbbP left bigcup_n in mathbbN ω : X(ω) ≤ x_n right \
&=& mathbbP(Omega) \
&=& 1.
endeqnarray
My question is regarding the second step where the continuous variable $x$ is replaced by the member of a sequence $x_n$. I feel lack of rigor in this step. To be precise, my questions are
- Why is this step valid?
- The trajectory that $x$ can take while approaching $infty$ are many, while the sequence $x_n$ is assumed to be monotonically increasing. How do we know for sure that this difference in the way to approach infinity will not change the limit?
- Is there a way to make the proof look more rigorous as in is there a rigorous way to substantiate this step of replacing $x$ with $x_n$?
Please help.
limits probability-theory measure-theory probability-distributions probability-limit-theorems
asked Mar 27 at 19:21
TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr
13112
$endgroup$
I have a question regarding the nuts and bolts involved in the proof of the limit of CDFs. The statement is that
Proposition: Let $X$ be a random variable with CDF $F_X(.)$. Then $F_X(.)$ posses the following property.
$$lim_x to infty F_X(x) = 1$$
Proof:
Consider a sequence $x_n$ with $ n in mathbbN$ such that it monotonically increases to $infty$. Then we have
begineqnarray
lim_x to inftyF_X(x) &=& lim_x to infty mathbbP(X leq x) \
&=& lim_n to infty mathbbP(X leq x_n) labeleqnref \
&=& mathbbP left bigcup_n in mathbbN ω : X(ω) ≤ x_n right \
&=& mathbbP(Omega) \
&=& 1.
endeqnarray
My question is regarding the second step where the continuous variable $x$ is replaced by the member of a sequence $x_n$. I feel lack of rigor in this step. To be precise, my questions are
- Why is this step valid?
- The trajectory that $x$ can take while approaching $infty$ are many, while the sequence $x_n$ is assumed to be monotonically increasing. How do we know for sure that this difference in the way to approach infinity will not change the limit?
- Is there a way to make the proof look more rigorous as in is there a rigorous way to substantiate this step of replacing $x$ with $x_n$?
Please help.
limits probability-theory measure-theory probability-distributions probability-limit-theorems
limits probability-theory measure-theory probability-distributions probability-limit-theorems
asked Mar 27 at 19:21
TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr
13112
asked Mar 27 at 19:21
TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr
13112
asked Mar 27 at 19:21
TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr
13112
asked Mar 27 at 19:21
TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr
13112
asked Mar 27 at 19:21
TryingHardToBecomeAGoodPrSlvrTryingHardToBecomeAGoodPrSlvr
13112
13112
1
$begingroup$
math.stackexchange.com/questions/1643588/…
$endgroup$
– d.k.o.
Mar 27 at 20:50
$begingroup$
@d.k.o Thanks for pointing it out!
$endgroup$
– TryingHardToBecomeAGoodPrSlvr
Mar 27 at 21:00
add a comment |
1
$begingroup$
math.stackexchange.com/questions/1643588/…
$endgroup$
– d.k.o.
Mar 27 at 20:50
$begingroup$
@d.k.o Thanks for pointing it out!
$endgroup$
– TryingHardToBecomeAGoodPrSlvr
Mar 27 at 21:00
1
1
$begingroup$
math.stackexchange.com/questions/1643588/…
$endgroup$
– d.k.o.
Mar 27 at 20:50
$begingroup$
math.stackexchange.com/questions/1643588/…
$endgroup$
– d.k.o.
Mar 27 at 20:50
$begingroup$
@d.k.o Thanks for pointing it out!
$endgroup$
– TryingHardToBecomeAGoodPrSlvr
Mar 27 at 21:00
$begingroup$
@d.k.o Thanks for pointing it out!
$endgroup$
– TryingHardToBecomeAGoodPrSlvr
Mar 27 at 21:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This step is valid as long as proving it is obvious/easy/possible. This could be discussed as what is obvious of experimented people may be a full-fledge exercice for beginners, but in any case, it is true.
I understand your point of multiple trajectories for $x$, but take into account that for any of these multiple trajectories, you can extract a monotonically increasing one.
Using the definition of these limits could help to clarify :
$$ lim_x to infty mathbbP(X leq x) = l
iff
forall epsilon >0, exists A mid x > A Rightarrow | mathbbP(X leq x) - l | < epsilon$$
$$ lim_n to infty mathbbP(X leq x_n) = l
iff
forall epsilon >0, exists N mid n > N Rightarrow |mathbbP(X leq x_n) - l | < epsilon $$
So, what you need is to find a way from a $A$ (resp $N$) large enough to a have the nice property, to find a $N$ (resp $A$) large enough to have the other nice property. In order to do so, I would write the definition of : $ lim_n to infty x_n = +infty$.
answered Mar 27 at 20:02
FlorianFlorian
21614
$endgroup$
add a comment |
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$begingroup$
This step is valid as long as proving it is obvious/easy/possible. This could be discussed as what is obvious of experimented people may be a full-fledge exercice for beginners, but in any case, it is true.
I understand your point of multiple trajectories for $x$, but take into account that for any of these multiple trajectories, you can extract a monotonically increasing one.
Using the definition of these limits could help to clarify :
$$ lim_x to infty mathbbP(X leq x) = l
iff
forall epsilon >0, exists A mid x > A Rightarrow | mathbbP(X leq x) - l | < epsilon$$
$$ lim_n to infty mathbbP(X leq x_n) = l
iff
forall epsilon >0, exists N mid n > N Rightarrow |mathbbP(X leq x_n) - l | < epsilon $$
So, what you need is to find a way from a $A$ (resp $N$) large enough to a have the nice property, to find a $N$ (resp $A$) large enough to have the other nice property. In order to do so, I would write the definition of : $ lim_n to infty x_n = +infty$.
answered Mar 27 at 20:02
FlorianFlorian
21614
$endgroup$
add a comment |
$begingroup$
This step is valid as long as proving it is obvious/easy/possible. This could be discussed as what is obvious of experimented people may be a full-fledge exercice for beginners, but in any case, it is true.
I understand your point of multiple trajectories for $x$, but take into account that for any of these multiple trajectories, you can extract a monotonically increasing one.
Using the definition of these limits could help to clarify :
$$ lim_x to infty mathbbP(X leq x) = l
iff
forall epsilon >0, exists A mid x > A Rightarrow | mathbbP(X leq x) - l | < epsilon$$
$$ lim_n to infty mathbbP(X leq x_n) = l
iff
forall epsilon >0, exists N mid n > N Rightarrow |mathbbP(X leq x_n) - l | < epsilon $$
So, what you need is to find a way from a $A$ (resp $N$) large enough to a have the nice property, to find a $N$ (resp $A$) large enough to have the other nice property. In order to do so, I would write the definition of : $ lim_n to infty x_n = +infty$.
answered Mar 27 at 20:02
FlorianFlorian
21614
$endgroup$
add a comment |
$begingroup$
This step is valid as long as proving it is obvious/easy/possible. This could be discussed as what is obvious of experimented people may be a full-fledge exercice for beginners, but in any case, it is true.
I understand your point of multiple trajectories for $x$, but take into account that for any of these multiple trajectories, you can extract a monotonically increasing one.
Using the definition of these limits could help to clarify :
$$ lim_x to infty mathbbP(X leq x) = l
iff
forall epsilon >0, exists A mid x > A Rightarrow | mathbbP(X leq x) - l | < epsilon$$
$$ lim_n to infty mathbbP(X leq x_n) = l
iff
forall epsilon >0, exists N mid n > N Rightarrow |mathbbP(X leq x_n) - l | < epsilon $$
So, what you need is to find a way from a $A$ (resp $N$) large enough to a have the nice property, to find a $N$ (resp $A$) large enough to have the other nice property. In order to do so, I would write the definition of : $ lim_n to infty x_n = +infty$.
answered Mar 27 at 20:02
FlorianFlorian
21614
$endgroup$
This step is valid as long as proving it is obvious/easy/possible. This could be discussed as what is obvious of experimented people may be a full-fledge exercice for beginners, but in any case, it is true.
I understand your point of multiple trajectories for $x$, but take into account that for any of these multiple trajectories, you can extract a monotonically increasing one.
Using the definition of these limits could help to clarify :
$$ lim_x to infty mathbbP(X leq x) = l
iff
forall epsilon >0, exists A mid x > A Rightarrow | mathbbP(X leq x) - l | < epsilon$$
$$ lim_n to infty mathbbP(X leq x_n) = l
iff
forall epsilon >0, exists N mid n > N Rightarrow |mathbbP(X leq x_n) - l | < epsilon $$
So, what you need is to find a way from a $A$ (resp $N$) large enough to a have the nice property, to find a $N$ (resp $A$) large enough to have the other nice property. In order to do so, I would write the definition of : $ lim_n to infty x_n = +infty$.
answered Mar 27 at 20:02
FlorianFlorian
21614
answered Mar 27 at 20:02
FlorianFlorian
21614
answered Mar 27 at 20:02
FlorianFlorian
21614
answered Mar 27 at 20:02
FlorianFlorian
21614
21614
add a comment |
add a comment |
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1
$begingroup$
math.stackexchange.com/questions/1643588/…
$endgroup$
– d.k.o.
Mar 27 at 20:50
$begingroup$
@d.k.o Thanks for pointing it out!
$endgroup$
– TryingHardToBecomeAGoodPrSlvr
Mar 27 at 21:00