If two events co-occur, then is their probability equal?












0












$begingroup$


If two events co-occur, then is their probability equal?



I’m trying to figer out why it is true that if $X_n$ converges to $X$ and $Y_n$ converges to $Y$, then since addition is continuous, $X_n + Y_n$ converges to $X+Y$.



In doing so, I thought it makes sense that since addition maps one-to-one (i.e. is continuous), then
$P[X_n+Y_n $converge to $ X+Y]=P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $.



But I am seein in resources such as this one http://web.math.ku.dk/noter/filer/vidsand12.pdf on page 10, lemma 1.2.10 that one is less than or equal to the other?



$P[X_n+Y_n $converge to $X+Y]ge P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $ .



Why aren’t these two probabilities always equal?






real-analysis probability probability-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
    $endgroup$
    – Shashi
    Jan 9 at 7:08








  • 1




    $begingroup$
    What is the meaning of $P(A)cap P(B)$?
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 7:20






  • 1




    $begingroup$
    Te title also makes no sense.
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 7:21










  • $begingroup$
    I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
    $endgroup$
    – jaja
    Jan 9 at 7:25
















0












$begingroup$


If two events co-occur, then is their probability equal?



I’m trying to figer out why it is true that if $X_n$ converges to $X$ and $Y_n$ converges to $Y$, then since addition is continuous, $X_n + Y_n$ converges to $X+Y$.



In doing so, I thought it makes sense that since addition maps one-to-one (i.e. is continuous), then
$P[X_n+Y_n $converge to $ X+Y]=P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $.



But I am seein in resources such as this one http://web.math.ku.dk/noter/filer/vidsand12.pdf on page 10, lemma 1.2.10 that one is less than or equal to the other?



$P[X_n+Y_n $converge to $X+Y]ge P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $ .



Why aren’t these two probabilities always equal?






real-analysis probability probability-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
    $endgroup$
    – Shashi
    Jan 9 at 7:08








  • 1




    $begingroup$
    What is the meaning of $P(A)cap P(B)$?
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 7:20






  • 1




    $begingroup$
    Te title also makes no sense.
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 7:21










  • $begingroup$
    I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
    $endgroup$
    – jaja
    Jan 9 at 7:25














0












0








0





$begingroup$


If two events co-occur, then is their probability equal?



I’m trying to figer out why it is true that if $X_n$ converges to $X$ and $Y_n$ converges to $Y$, then since addition is continuous, $X_n + Y_n$ converges to $X+Y$.



In doing so, I thought it makes sense that since addition maps one-to-one (i.e. is continuous), then
$P[X_n+Y_n $converge to $ X+Y]=P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $.



But I am seein in resources such as this one http://web.math.ku.dk/noter/filer/vidsand12.pdf on page 10, lemma 1.2.10 that one is less than or equal to the other?



$P[X_n+Y_n $converge to $X+Y]ge P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $ .



Why aren’t these two probabilities always equal?






real-analysis probability probability-theory







share|cite|improve this question











$endgroup$




If two events co-occur, then is their probability equal?



I’m trying to figer out why it is true that if $X_n$ converges to $X$ and $Y_n$ converges to $Y$, then since addition is continuous, $X_n + Y_n$ converges to $X+Y$.



In doing so, I thought it makes sense that since addition maps one-to-one (i.e. is continuous), then
$P[X_n+Y_n $converge to $ X+Y]=P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $.



But I am seein in resources such as this one http://web.math.ku.dk/noter/filer/vidsand12.pdf on page 10, lemma 1.2.10 that one is less than or equal to the other?



$P[X_n+Y_n $converge to $X+Y]ge P[(X_n$ converges to $X ) cap (Y_n $converges to $Y)] =1 $ .



Why aren’t these two probabilities always equal?






real-analysis probability probability-theory




real-analysis probability probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 7:23







jaja

















asked Jan 9 at 6:50









jajajaja

8711




8711








  • 1




    $begingroup$
    You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
    $endgroup$
    – Shashi
    Jan 9 at 7:08








  • 1




    $begingroup$
    What is the meaning of $P(A)cap P(B)$?
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 7:20






  • 1




    $begingroup$
    Te title also makes no sense.
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 7:21










  • $begingroup$
    I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
    $endgroup$
    – jaja
    Jan 9 at 7:25














  • 1




    $begingroup$
    You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
    $endgroup$
    – Shashi
    Jan 9 at 7:08








  • 1




    $begingroup$
    What is the meaning of $P(A)cap P(B)$?
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 7:20






  • 1




    $begingroup$
    Te title also makes no sense.
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 7:21










  • $begingroup$
    I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
    $endgroup$
    – jaja
    Jan 9 at 7:25








1




1




$begingroup$
You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
$endgroup$
– Shashi
Jan 9 at 7:08






$begingroup$
You take the product of the probabilities while in the resource they are speaking about the intersection of two events. You don't know whether these two events are independent. Do you see that?
$endgroup$
– Shashi
Jan 9 at 7:08






1




1




$begingroup$
What is the meaning of $P(A)cap P(B)$?
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:20




$begingroup$
What is the meaning of $P(A)cap P(B)$?
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:20




1




1




$begingroup$
Te title also makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:21




$begingroup$
Te title also makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 7:21












$begingroup$
I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
$endgroup$
– jaja
Jan 9 at 7:25




$begingroup$
I agree actually te title makes no sense. I'm having trouble phrasing it correctly.
$endgroup$
– jaja
Jan 9 at 7:25










1 Answer
1






active

oldest

votes


















2












$begingroup$

What the author probably meant by addition is continuous is that the function $(x, y) mapsto x+y$ is continuous. That let us conclude the convergence of the sum of two convergent random variables.



Indeed, for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges. The converse does not have to be true. Indeed take a divergent $X_n(omega) $ and set $Y_n=-X_n$, then the sum converges but the individual random variables do not.



That means
$${X_n text{ converges to } X } cap {Y_n text{ converges to } Y }subset {X_n +Y_ntext{ converges to } X+Y }$$
The inequality is then a consequence of the subadditivity of the probability measure. Moreover the intersection of two events with probability one has probability one as well. Just look at the complement and see that it is a null set.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
    $endgroup$
    – jaja
    Jan 9 at 7:31








  • 2




    $begingroup$
    @jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
    $endgroup$
    – Shashi
    Jan 9 at 7:35












  • $begingroup$
    Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
    $endgroup$
    – jaja
    Jan 9 at 8:00












  • $begingroup$
    @jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
    $endgroup$
    – Shashi
    Jan 9 at 8:03






  • 1




    $begingroup$
    @jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
    $endgroup$
    – Shashi
    Jan 9 at 17:00














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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

What the author probably meant by addition is continuous is that the function $(x, y) mapsto x+y$ is continuous. That let us conclude the convergence of the sum of two convergent random variables.



Indeed, for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges. The converse does not have to be true. Indeed take a divergent $X_n(omega) $ and set $Y_n=-X_n$, then the sum converges but the individual random variables do not.



That means
$${X_n text{ converges to } X } cap {Y_n text{ converges to } Y }subset {X_n +Y_ntext{ converges to } X+Y }$$
The inequality is then a consequence of the subadditivity of the probability measure. Moreover the intersection of two events with probability one has probability one as well. Just look at the complement and see that it is a null set.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
    $endgroup$
    – jaja
    Jan 9 at 7:31








  • 2




    $begingroup$
    @jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
    $endgroup$
    – Shashi
    Jan 9 at 7:35












  • $begingroup$
    Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
    $endgroup$
    – jaja
    Jan 9 at 8:00












  • $begingroup$
    @jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
    $endgroup$
    – Shashi
    Jan 9 at 8:03






  • 1




    $begingroup$
    @jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
    $endgroup$
    – Shashi
    Jan 9 at 17:00


















2












$begingroup$

What the author probably meant by addition is continuous is that the function $(x, y) mapsto x+y$ is continuous. That let us conclude the convergence of the sum of two convergent random variables.



Indeed, for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges. The converse does not have to be true. Indeed take a divergent $X_n(omega) $ and set $Y_n=-X_n$, then the sum converges but the individual random variables do not.



That means
$${X_n text{ converges to } X } cap {Y_n text{ converges to } Y }subset {X_n +Y_ntext{ converges to } X+Y }$$
The inequality is then a consequence of the subadditivity of the probability measure. Moreover the intersection of two events with probability one has probability one as well. Just look at the complement and see that it is a null set.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
    $endgroup$
    – jaja
    Jan 9 at 7:31








  • 2




    $begingroup$
    @jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
    $endgroup$
    – Shashi
    Jan 9 at 7:35












  • $begingroup$
    Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
    $endgroup$
    – jaja
    Jan 9 at 8:00












  • $begingroup$
    @jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
    $endgroup$
    – Shashi
    Jan 9 at 8:03






  • 1




    $begingroup$
    @jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
    $endgroup$
    – Shashi
    Jan 9 at 17:00
















2












2








2





$begingroup$

What the author probably meant by addition is continuous is that the function $(x, y) mapsto x+y$ is continuous. That let us conclude the convergence of the sum of two convergent random variables.



Indeed, for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges. The converse does not have to be true. Indeed take a divergent $X_n(omega) $ and set $Y_n=-X_n$, then the sum converges but the individual random variables do not.



That means
$${X_n text{ converges to } X } cap {Y_n text{ converges to } Y }subset {X_n +Y_ntext{ converges to } X+Y }$$
The inequality is then a consequence of the subadditivity of the probability measure. Moreover the intersection of two events with probability one has probability one as well. Just look at the complement and see that it is a null set.






share|cite|improve this answer











$endgroup$



What the author probably meant by addition is continuous is that the function $(x, y) mapsto x+y$ is continuous. That let us conclude the convergence of the sum of two convergent random variables.



Indeed, for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges. The converse does not have to be true. Indeed take a divergent $X_n(omega) $ and set $Y_n=-X_n$, then the sum converges but the individual random variables do not.



That means
$${X_n text{ converges to } X } cap {Y_n text{ converges to } Y }subset {X_n +Y_ntext{ converges to } X+Y }$$
The inequality is then a consequence of the subadditivity of the probability measure. Moreover the intersection of two events with probability one has probability one as well. Just look at the complement and see that it is a null set.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 9 at 7:45

























answered Jan 9 at 7:06









ShashiShashi

7,3621628




7,3621628












  • $begingroup$
    What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
    $endgroup$
    – jaja
    Jan 9 at 7:31








  • 2




    $begingroup$
    @jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
    $endgroup$
    – Shashi
    Jan 9 at 7:35












  • $begingroup$
    Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
    $endgroup$
    – jaja
    Jan 9 at 8:00












  • $begingroup$
    @jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
    $endgroup$
    – Shashi
    Jan 9 at 8:03






  • 1




    $begingroup$
    @jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
    $endgroup$
    – Shashi
    Jan 9 at 17:00




















  • $begingroup$
    What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
    $endgroup$
    – jaja
    Jan 9 at 7:31








  • 2




    $begingroup$
    @jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
    $endgroup$
    – Shashi
    Jan 9 at 7:35












  • $begingroup$
    Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
    $endgroup$
    – jaja
    Jan 9 at 8:00












  • $begingroup$
    @jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
    $endgroup$
    – Shashi
    Jan 9 at 8:03






  • 1




    $begingroup$
    @jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
    $endgroup$
    – Shashi
    Jan 9 at 17:00


















$begingroup$
What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
$endgroup$
– jaja
Jan 9 at 7:31






$begingroup$
What does the resource that I cited mean when it says "as addition is continuous"? This statement confuses me since addition is a binary operator, not a function.
$endgroup$
– jaja
Jan 9 at 7:31






2




2




$begingroup$
@jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
$endgroup$
– Shashi
Jan 9 at 7:35






$begingroup$
@jaja oh I see it, they probably mean that the function $f(x, y) =x+y$ is continuous. From there you can indeed conclude that $X_n+Y_n$ converges if $X_n$ and $Y_n$ converges. Is it clear?
$endgroup$
– Shashi
Jan 9 at 7:35














$begingroup$
Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
$endgroup$
– jaja
Jan 9 at 8:00






$begingroup$
Well, I see that part about "addition is continuous" thank you @Sashi. But I do not understand the following: When you say “for any $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges it holds that $X_n(omega) +Y_n(omega) $ converges” then that makes me think that the set containment should actually go the other way: $${X_n +Y_ntext{ converges to } X+Y }subset {X_n text{ converges to } X } cap {Y_n text{ converges to } Y } $$ How does the logic A implies B mean that $Bsubset A$?...what am I missing here?
$endgroup$
– jaja
Jan 9 at 8:00














$begingroup$
@jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
$endgroup$
– Shashi
Jan 9 at 8:03




$begingroup$
@jaja when I say that I take $omega$ such that $X_n(omega) $ and $Y_n(omega) $ converges, I actually take $omegain{X_n text{ converges } } cap{ Y_n text{ converges} } $ and show that $omega in{ X_n+Y_n$ converges $} $. What does that show? The inclusion mentioned in the post, right?
$endgroup$
– Shashi
Jan 9 at 8:03




1




1




$begingroup$
@jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
$endgroup$
– Shashi
Jan 9 at 17:00






$begingroup$
@jaja if you shwo that every $a$ in some set $A$ is also in set $B$ then is not that the same as saying $Asubset B$? That's what I am claiming.
$endgroup$
– Shashi
Jan 9 at 17:00




















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