What is the relationship between converge(calculus) and converge in probability(statistic) The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)convergence in probability induced by a metricUnderstanding Uniform ContinuityWhat exactly is simple consistency?Formal Definition of Limit and ProofsDefinition of continuityAt what point is the function $f:mathbbRrightarrowmathbbR$ continuous?if $x_0$ is a limit point of $D$, then a function $f:DrightarrowmathbbR$ is continuous at $x_0$ iff $limlimits_xrightarrow x_0f(x)=f(x_0)$Understanding “For every $theta in Theta$” in the definition of the weak consistence of an estimatorRelationship between unit circle and cosine wave?Relation between the concept of Contiguity and Absolutely Continuous Measures$lim$ vs $liminf$ and $limsup$ in the proof convergence in probability implies convergence in distribution

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What is the relationship between converge(calculus) and converge in probability(statistic)



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)convergence in probability induced by a metricUnderstanding Uniform ContinuityWhat exactly is simple consistency?Formal Definition of Limit and ProofsDefinition of continuityAt what point is the function $f:mathbbRrightarrowmathbbR$ continuous?if $x_0$ is a limit point of $D$, then a function $f:DrightarrowmathbbR$ is continuous at $x_0$ iff $limlimits_xrightarrow x_0f(x)=f(x_0)$Understanding “For every $theta in Theta$” in the definition of the weak consistence of an estimatorRelationship between unit circle and cosine wave?Relation between the concept of Contiguity and Absolutely Continuous Measures$lim$ vs $liminf$ and $limsup$ in the proof convergence in probability implies convergence in distribution










0












$begingroup$


I learnt the converge definition in advanced calculus:




For any $epsilon>0$, there is a $N$, such that for all $n>N$, we have $|X_n-x|<epsilon$ then we say $X_n$ converge to $x$.




Then I learnt the consistent estimators in statistics:




For every $epsilon>0$ and every $theta in Theta$, $limlimits_ntoinftyP_theta(|W_n-theta| ge epsilon) = 0$




So my question is that is there any relationship between consistent estimators and converge sequence?



Are the two definition equivalent?



I believe the two definition must have something in common, but I can not find that.



And also , Is there any relationship between converge in distribution[$F_X_nto F_x$] and the definition of continuity in function? [$X_nto x_0, f(X_n) to f(x_0)]$










share|cite|improve this question











$endgroup$











  • $begingroup$
    Convergence of sequences and convergence of functions (one example being estimators) work differently. Both still mean that things get closer and closer, but for numerical sequences they get closer to a single point; for functions many points get closer to other many points at the same time, and there are various ways that those points of a function can get closer and closer.
    $endgroup$
    – AspiringMathematician
    Mar 24 at 16:31











  • $begingroup$
    There are a about 500 (I‘m exaggerating of course) different definitions for the convergence of random variables (I really recommend you to read en.m.wikipedia.org/wiki/Convergence_of_random_variables). Currently you are talking about convergence in probability. Note that the convergence in real analysis deals with numbers, whereas the convergence terms in probability theory deal with random variables. So many of the convergence terms are quite different.
    $endgroup$
    – Maximilian Janisch
    Mar 24 at 16:31










  • $begingroup$
    Have a look at wikipedia. Be aware that random variables are actually functions (not just numbers).
    $endgroup$
    – drhab
    Mar 24 at 16:33










  • $begingroup$
    I see, so basically, the main difference is between function and the single value
    $endgroup$
    – MolinYue
    Mar 25 at 15:30















0












$begingroup$


I learnt the converge definition in advanced calculus:




For any $epsilon>0$, there is a $N$, such that for all $n>N$, we have $|X_n-x|<epsilon$ then we say $X_n$ converge to $x$.




Then I learnt the consistent estimators in statistics:




For every $epsilon>0$ and every $theta in Theta$, $limlimits_ntoinftyP_theta(|W_n-theta| ge epsilon) = 0$




So my question is that is there any relationship between consistent estimators and converge sequence?



Are the two definition equivalent?



I believe the two definition must have something in common, but I can not find that.



And also , Is there any relationship between converge in distribution[$F_X_nto F_x$] and the definition of continuity in function? [$X_nto x_0, f(X_n) to f(x_0)]$










share|cite|improve this question











$endgroup$











  • $begingroup$
    Convergence of sequences and convergence of functions (one example being estimators) work differently. Both still mean that things get closer and closer, but for numerical sequences they get closer to a single point; for functions many points get closer to other many points at the same time, and there are various ways that those points of a function can get closer and closer.
    $endgroup$
    – AspiringMathematician
    Mar 24 at 16:31











  • $begingroup$
    There are a about 500 (I‘m exaggerating of course) different definitions for the convergence of random variables (I really recommend you to read en.m.wikipedia.org/wiki/Convergence_of_random_variables). Currently you are talking about convergence in probability. Note that the convergence in real analysis deals with numbers, whereas the convergence terms in probability theory deal with random variables. So many of the convergence terms are quite different.
    $endgroup$
    – Maximilian Janisch
    Mar 24 at 16:31










  • $begingroup$
    Have a look at wikipedia. Be aware that random variables are actually functions (not just numbers).
    $endgroup$
    – drhab
    Mar 24 at 16:33










  • $begingroup$
    I see, so basically, the main difference is between function and the single value
    $endgroup$
    – MolinYue
    Mar 25 at 15:30













0












0








0





$begingroup$


I learnt the converge definition in advanced calculus:




For any $epsilon>0$, there is a $N$, such that for all $n>N$, we have $|X_n-x|<epsilon$ then we say $X_n$ converge to $x$.




Then I learnt the consistent estimators in statistics:




For every $epsilon>0$ and every $theta in Theta$, $limlimits_ntoinftyP_theta(|W_n-theta| ge epsilon) = 0$




So my question is that is there any relationship between consistent estimators and converge sequence?



Are the two definition equivalent?



I believe the two definition must have something in common, but I can not find that.



And also , Is there any relationship between converge in distribution[$F_X_nto F_x$] and the definition of continuity in function? [$X_nto x_0, f(X_n) to f(x_0)]$










share|cite|improve this question











$endgroup$




I learnt the converge definition in advanced calculus:




For any $epsilon>0$, there is a $N$, such that for all $n>N$, we have $|X_n-x|<epsilon$ then we say $X_n$ converge to $x$.




Then I learnt the consistent estimators in statistics:




For every $epsilon>0$ and every $theta in Theta$, $limlimits_ntoinftyP_theta(|W_n-theta| ge epsilon) = 0$




So my question is that is there any relationship between consistent estimators and converge sequence?



Are the two definition equivalent?



I believe the two definition must have something in common, but I can not find that.



And also , Is there any relationship between converge in distribution[$F_X_nto F_x$] and the definition of continuity in function? [$X_nto x_0, f(X_n) to f(x_0)]$







real-analysis calculus probability probability-theory statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 24 at 16:51









Brian

1,517416




1,517416










asked Mar 24 at 16:08









MolinYueMolinYue

31




31











  • $begingroup$
    Convergence of sequences and convergence of functions (one example being estimators) work differently. Both still mean that things get closer and closer, but for numerical sequences they get closer to a single point; for functions many points get closer to other many points at the same time, and there are various ways that those points of a function can get closer and closer.
    $endgroup$
    – AspiringMathematician
    Mar 24 at 16:31











  • $begingroup$
    There are a about 500 (I‘m exaggerating of course) different definitions for the convergence of random variables (I really recommend you to read en.m.wikipedia.org/wiki/Convergence_of_random_variables). Currently you are talking about convergence in probability. Note that the convergence in real analysis deals with numbers, whereas the convergence terms in probability theory deal with random variables. So many of the convergence terms are quite different.
    $endgroup$
    – Maximilian Janisch
    Mar 24 at 16:31










  • $begingroup$
    Have a look at wikipedia. Be aware that random variables are actually functions (not just numbers).
    $endgroup$
    – drhab
    Mar 24 at 16:33










  • $begingroup$
    I see, so basically, the main difference is between function and the single value
    $endgroup$
    – MolinYue
    Mar 25 at 15:30
















  • $begingroup$
    Convergence of sequences and convergence of functions (one example being estimators) work differently. Both still mean that things get closer and closer, but for numerical sequences they get closer to a single point; for functions many points get closer to other many points at the same time, and there are various ways that those points of a function can get closer and closer.
    $endgroup$
    – AspiringMathematician
    Mar 24 at 16:31











  • $begingroup$
    There are a about 500 (I‘m exaggerating of course) different definitions for the convergence of random variables (I really recommend you to read en.m.wikipedia.org/wiki/Convergence_of_random_variables). Currently you are talking about convergence in probability. Note that the convergence in real analysis deals with numbers, whereas the convergence terms in probability theory deal with random variables. So many of the convergence terms are quite different.
    $endgroup$
    – Maximilian Janisch
    Mar 24 at 16:31










  • $begingroup$
    Have a look at wikipedia. Be aware that random variables are actually functions (not just numbers).
    $endgroup$
    – drhab
    Mar 24 at 16:33










  • $begingroup$
    I see, so basically, the main difference is between function and the single value
    $endgroup$
    – MolinYue
    Mar 25 at 15:30















$begingroup$
Convergence of sequences and convergence of functions (one example being estimators) work differently. Both still mean that things get closer and closer, but for numerical sequences they get closer to a single point; for functions many points get closer to other many points at the same time, and there are various ways that those points of a function can get closer and closer.
$endgroup$
– AspiringMathematician
Mar 24 at 16:31





$begingroup$
Convergence of sequences and convergence of functions (one example being estimators) work differently. Both still mean that things get closer and closer, but for numerical sequences they get closer to a single point; for functions many points get closer to other many points at the same time, and there are various ways that those points of a function can get closer and closer.
$endgroup$
– AspiringMathematician
Mar 24 at 16:31













$begingroup$
There are a about 500 (I‘m exaggerating of course) different definitions for the convergence of random variables (I really recommend you to read en.m.wikipedia.org/wiki/Convergence_of_random_variables). Currently you are talking about convergence in probability. Note that the convergence in real analysis deals with numbers, whereas the convergence terms in probability theory deal with random variables. So many of the convergence terms are quite different.
$endgroup$
– Maximilian Janisch
Mar 24 at 16:31




$begingroup$
There are a about 500 (I‘m exaggerating of course) different definitions for the convergence of random variables (I really recommend you to read en.m.wikipedia.org/wiki/Convergence_of_random_variables). Currently you are talking about convergence in probability. Note that the convergence in real analysis deals with numbers, whereas the convergence terms in probability theory deal with random variables. So many of the convergence terms are quite different.
$endgroup$
– Maximilian Janisch
Mar 24 at 16:31












$begingroup$
Have a look at wikipedia. Be aware that random variables are actually functions (not just numbers).
$endgroup$
– drhab
Mar 24 at 16:33




$begingroup$
Have a look at wikipedia. Be aware that random variables are actually functions (not just numbers).
$endgroup$
– drhab
Mar 24 at 16:33












$begingroup$
I see, so basically, the main difference is between function and the single value
$endgroup$
– MolinYue
Mar 25 at 15:30




$begingroup$
I see, so basically, the main difference is between function and the single value
$endgroup$
– MolinYue
Mar 25 at 15:30










1 Answer
1






active

oldest

votes


















0












$begingroup$

Perhaps this rephrasing will make the parallel more clear:




$X_nto X$ in probability $iff$ for any $epsilon>0$, there is an $N$ so $nge N$ implies $Eleft[fracX_n-Xright]<epsilon.$




For a proof, see convergence in probability induced by a metric.



In other words, if we define $d_P(X,Y)$ to be the funny quantity $Eleft[fracX-Yright]$, then $d_P(X_n,X)$ replaces $|X_n-X|$ in the usual definition of convergence. More generally, in any metric space with a distance function $d$, we have the following notion of convergence:




$x_nto x$ $iff$ for any $epsilon>0$, there is an $N$ so $nge N$ implies $d(x_n,x)<epsilon$.




Your calculus notion of convergence is the one from the metric space $mathbb R$ with distance function $d(x,y)=|x-y|$, while convergence in probability comes $d_P$.



Finally, there is a relationship between convergence and distribution and metric convergence. Let $X_n$ have cdf $F_n$, and let $X$ have cdf $F$.




$X_nto X$ in distribution $iff$ for all $x$ such that $F$ is continuous at $x$, we have $F_n(x)to F(x)$.




Also, convergence in distribution is induced by a metric in the same way convergence in probability is. In this case, the metric is
$$
d_textLevy(X,Y)=infepsilon>0:P(Xle x-epsilon)-epsilon<P(Yle x)le P(Xle x+epsilon)+epsilontext for all xin mathbb R
$$

See https://en.wikipedia.org/wiki/L%C3%A9vy_metric.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you! it seems that I need work harder
    $endgroup$
    – MolinYue
    Mar 27 at 19:28











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Perhaps this rephrasing will make the parallel more clear:




$X_nto X$ in probability $iff$ for any $epsilon>0$, there is an $N$ so $nge N$ implies $Eleft[fracX_n-Xright]<epsilon.$




For a proof, see convergence in probability induced by a metric.



In other words, if we define $d_P(X,Y)$ to be the funny quantity $Eleft[fracX-Yright]$, then $d_P(X_n,X)$ replaces $|X_n-X|$ in the usual definition of convergence. More generally, in any metric space with a distance function $d$, we have the following notion of convergence:




$x_nto x$ $iff$ for any $epsilon>0$, there is an $N$ so $nge N$ implies $d(x_n,x)<epsilon$.




Your calculus notion of convergence is the one from the metric space $mathbb R$ with distance function $d(x,y)=|x-y|$, while convergence in probability comes $d_P$.



Finally, there is a relationship between convergence and distribution and metric convergence. Let $X_n$ have cdf $F_n$, and let $X$ have cdf $F$.




$X_nto X$ in distribution $iff$ for all $x$ such that $F$ is continuous at $x$, we have $F_n(x)to F(x)$.




Also, convergence in distribution is induced by a metric in the same way convergence in probability is. In this case, the metric is
$$
d_textLevy(X,Y)=infepsilon>0:P(Xle x-epsilon)-epsilon<P(Yle x)le P(Xle x+epsilon)+epsilontext for all xin mathbb R
$$

See https://en.wikipedia.org/wiki/L%C3%A9vy_metric.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you! it seems that I need work harder
    $endgroup$
    – MolinYue
    Mar 27 at 19:28















0












$begingroup$

Perhaps this rephrasing will make the parallel more clear:




$X_nto X$ in probability $iff$ for any $epsilon>0$, there is an $N$ so $nge N$ implies $Eleft[fracX_n-Xright]<epsilon.$




For a proof, see convergence in probability induced by a metric.



In other words, if we define $d_P(X,Y)$ to be the funny quantity $Eleft[fracX-Yright]$, then $d_P(X_n,X)$ replaces $|X_n-X|$ in the usual definition of convergence. More generally, in any metric space with a distance function $d$, we have the following notion of convergence:




$x_nto x$ $iff$ for any $epsilon>0$, there is an $N$ so $nge N$ implies $d(x_n,x)<epsilon$.




Your calculus notion of convergence is the one from the metric space $mathbb R$ with distance function $d(x,y)=|x-y|$, while convergence in probability comes $d_P$.



Finally, there is a relationship between convergence and distribution and metric convergence. Let $X_n$ have cdf $F_n$, and let $X$ have cdf $F$.




$X_nto X$ in distribution $iff$ for all $x$ such that $F$ is continuous at $x$, we have $F_n(x)to F(x)$.




Also, convergence in distribution is induced by a metric in the same way convergence in probability is. In this case, the metric is
$$
d_textLevy(X,Y)=infepsilon>0:P(Xle x-epsilon)-epsilon<P(Yle x)le P(Xle x+epsilon)+epsilontext for all xin mathbb R
$$

See https://en.wikipedia.org/wiki/L%C3%A9vy_metric.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you! it seems that I need work harder
    $endgroup$
    – MolinYue
    Mar 27 at 19:28













0












0








0





$begingroup$

Perhaps this rephrasing will make the parallel more clear:




$X_nto X$ in probability $iff$ for any $epsilon>0$, there is an $N$ so $nge N$ implies $Eleft[fracX_n-Xright]<epsilon.$




For a proof, see convergence in probability induced by a metric.



In other words, if we define $d_P(X,Y)$ to be the funny quantity $Eleft[fracX-Yright]$, then $d_P(X_n,X)$ replaces $|X_n-X|$ in the usual definition of convergence. More generally, in any metric space with a distance function $d$, we have the following notion of convergence:




$x_nto x$ $iff$ for any $epsilon>0$, there is an $N$ so $nge N$ implies $d(x_n,x)<epsilon$.




Your calculus notion of convergence is the one from the metric space $mathbb R$ with distance function $d(x,y)=|x-y|$, while convergence in probability comes $d_P$.



Finally, there is a relationship between convergence and distribution and metric convergence. Let $X_n$ have cdf $F_n$, and let $X$ have cdf $F$.




$X_nto X$ in distribution $iff$ for all $x$ such that $F$ is continuous at $x$, we have $F_n(x)to F(x)$.




Also, convergence in distribution is induced by a metric in the same way convergence in probability is. In this case, the metric is
$$
d_textLevy(X,Y)=infepsilon>0:P(Xle x-epsilon)-epsilon<P(Yle x)le P(Xle x+epsilon)+epsilontext for all xin mathbb R
$$

See https://en.wikipedia.org/wiki/L%C3%A9vy_metric.






share|cite|improve this answer









$endgroup$



Perhaps this rephrasing will make the parallel more clear:




$X_nto X$ in probability $iff$ for any $epsilon>0$, there is an $N$ so $nge N$ implies $Eleft[fracX_n-Xright]<epsilon.$




For a proof, see convergence in probability induced by a metric.



In other words, if we define $d_P(X,Y)$ to be the funny quantity $Eleft[fracX-Yright]$, then $d_P(X_n,X)$ replaces $|X_n-X|$ in the usual definition of convergence. More generally, in any metric space with a distance function $d$, we have the following notion of convergence:




$x_nto x$ $iff$ for any $epsilon>0$, there is an $N$ so $nge N$ implies $d(x_n,x)<epsilon$.




Your calculus notion of convergence is the one from the metric space $mathbb R$ with distance function $d(x,y)=|x-y|$, while convergence in probability comes $d_P$.



Finally, there is a relationship between convergence and distribution and metric convergence. Let $X_n$ have cdf $F_n$, and let $X$ have cdf $F$.




$X_nto X$ in distribution $iff$ for all $x$ such that $F$ is continuous at $x$, we have $F_n(x)to F(x)$.




Also, convergence in distribution is induced by a metric in the same way convergence in probability is. In this case, the metric is
$$
d_textLevy(X,Y)=infepsilon>0:P(Xle x-epsilon)-epsilon<P(Yle x)le P(Xle x+epsilon)+epsilontext for all xin mathbb R
$$

See https://en.wikipedia.org/wiki/L%C3%A9vy_metric.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 24 at 20:09









Mike EarnestMike Earnest

27.7k22152




27.7k22152











  • $begingroup$
    Thank you! it seems that I need work harder
    $endgroup$
    – MolinYue
    Mar 27 at 19:28
















  • $begingroup$
    Thank you! it seems that I need work harder
    $endgroup$
    – MolinYue
    Mar 27 at 19:28















$begingroup$
Thank you! it seems that I need work harder
$endgroup$
– MolinYue
Mar 27 at 19:28




$begingroup$
Thank you! it seems that I need work harder
$endgroup$
– MolinYue
Mar 27 at 19:28

















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