Difficultly understanding the inclusion exclusion identity because of difficult notation












0












$begingroup$


This is the property: $mathbb{P}(E1cup E2cup ...cup En)= sum_{i}^{n} mathbb{P}(E1) - sum_{i1<i2} mathbb{P}(Ei1Ei2) + ... + -1^{r+1}sum_{i1<i2<...<ir} mathbb{P}(Ei1Ei2...Eir) + -1^{r+1}sum_{1<2<...<n} mathbb{P}(E1E2...En) $



While I understand the property, I don’t understand the summation symbol notation. What does the book mean by $sum_{i1<i2<...<ir}$? Can someone draw out specifically what this sum is so I can understand it? I lost my question privileges due to this question but can someone please let me know what about this question was so bad? I couldn't find the syntax anywhere.






calculus probability







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  • $begingroup$
    I’m not sure why it isn’t showing the whole equation it shows up in my edit
    $endgroup$
    – Person
    Dec 6 '18 at 3:48










  • $begingroup$
    Learn how to write with latex .
    $endgroup$
    – tarit goswami
    Dec 6 '18 at 4:03










  • $begingroup$
    use cup for $cup$ and sum_{i=1}^{n} for $sum_{i=1}^{n}$.
    $endgroup$
    – Nosrati
    Dec 6 '18 at 4:17
















0












$begingroup$


This is the property: $mathbb{P}(E1cup E2cup ...cup En)= sum_{i}^{n} mathbb{P}(E1) - sum_{i1<i2} mathbb{P}(Ei1Ei2) + ... + -1^{r+1}sum_{i1<i2<...<ir} mathbb{P}(Ei1Ei2...Eir) + -1^{r+1}sum_{1<2<...<n} mathbb{P}(E1E2...En) $



While I understand the property, I don’t understand the summation symbol notation. What does the book mean by $sum_{i1<i2<...<ir}$? Can someone draw out specifically what this sum is so I can understand it? I lost my question privileges due to this question but can someone please let me know what about this question was so bad? I couldn't find the syntax anywhere.






calculus probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    I’m not sure why it isn’t showing the whole equation it shows up in my edit
    $endgroup$
    – Person
    Dec 6 '18 at 3:48










  • $begingroup$
    Learn how to write with latex .
    $endgroup$
    – tarit goswami
    Dec 6 '18 at 4:03










  • $begingroup$
    use cup for $cup$ and sum_{i=1}^{n} for $sum_{i=1}^{n}$.
    $endgroup$
    – Nosrati
    Dec 6 '18 at 4:17














0












0








0





$begingroup$


This is the property: $mathbb{P}(E1cup E2cup ...cup En)= sum_{i}^{n} mathbb{P}(E1) - sum_{i1<i2} mathbb{P}(Ei1Ei2) + ... + -1^{r+1}sum_{i1<i2<...<ir} mathbb{P}(Ei1Ei2...Eir) + -1^{r+1}sum_{1<2<...<n} mathbb{P}(E1E2...En) $



While I understand the property, I don’t understand the summation symbol notation. What does the book mean by $sum_{i1<i2<...<ir}$? Can someone draw out specifically what this sum is so I can understand it? I lost my question privileges due to this question but can someone please let me know what about this question was so bad? I couldn't find the syntax anywhere.






calculus probability







share|cite|improve this question











$endgroup$




This is the property: $mathbb{P}(E1cup E2cup ...cup En)= sum_{i}^{n} mathbb{P}(E1) - sum_{i1<i2} mathbb{P}(Ei1Ei2) + ... + -1^{r+1}sum_{i1<i2<...<ir} mathbb{P}(Ei1Ei2...Eir) + -1^{r+1}sum_{1<2<...<n} mathbb{P}(E1E2...En) $



While I understand the property, I don’t understand the summation symbol notation. What does the book mean by $sum_{i1<i2<...<ir}$? Can someone draw out specifically what this sum is so I can understand it? I lost my question privileges due to this question but can someone please let me know what about this question was so bad? I couldn't find the syntax anywhere.






calculus probability




calculus probability






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share|cite|improve this question













share|cite|improve this question




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edited Dec 6 '18 at 17:53







Person

















asked Dec 6 '18 at 3:46









PersonPerson

177




177












  • $begingroup$
    I’m not sure why it isn’t showing the whole equation it shows up in my edit
    $endgroup$
    – Person
    Dec 6 '18 at 3:48










  • $begingroup$
    Learn how to write with latex .
    $endgroup$
    – tarit goswami
    Dec 6 '18 at 4:03










  • $begingroup$
    use cup for $cup$ and sum_{i=1}^{n} for $sum_{i=1}^{n}$.
    $endgroup$
    – Nosrati
    Dec 6 '18 at 4:17


















  • $begingroup$
    I’m not sure why it isn’t showing the whole equation it shows up in my edit
    $endgroup$
    – Person
    Dec 6 '18 at 3:48










  • $begingroup$
    Learn how to write with latex .
    $endgroup$
    – tarit goswami
    Dec 6 '18 at 4:03










  • $begingroup$
    use cup for $cup$ and sum_{i=1}^{n} for $sum_{i=1}^{n}$.
    $endgroup$
    – Nosrati
    Dec 6 '18 at 4:17
















$begingroup$
I’m not sure why it isn’t showing the whole equation it shows up in my edit
$endgroup$
– Person
Dec 6 '18 at 3:48




$begingroup$
I’m not sure why it isn’t showing the whole equation it shows up in my edit
$endgroup$
– Person
Dec 6 '18 at 3:48












$begingroup$
Learn how to write with latex .
$endgroup$
– tarit goswami
Dec 6 '18 at 4:03




$begingroup$
Learn how to write with latex .
$endgroup$
– tarit goswami
Dec 6 '18 at 4:03












$begingroup$
use cup for $cup$ and sum_{i=1}^{n} for $sum_{i=1}^{n}$.
$endgroup$
– Nosrati
Dec 6 '18 at 4:17




$begingroup$
use cup for $cup$ and sum_{i=1}^{n} for $sum_{i=1}^{n}$.
$endgroup$
– Nosrati
Dec 6 '18 at 4:17










2 Answers
2






active

oldest

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0












$begingroup$

Some people would write the sum $sum_{i_1<i_2<cdots<i_r}$ as
$sum_{1 leq i_1<i_2<cdots<i_r leq n}$
in order to emphasize that all the variables $i_1,i_2,ldots,i_r$ have to be between $1$ and $n$ inclusive.



The $r$th sum can be rewritten in more basic summation notation as
$$
sum_{1 leq i_1<i_2<cdots<i_r leq n} mathbb{P}(E_{i_1}E_{i_2}cdots E_{i_r})
= sum_{i_1 = 1}^{n - r + 1}, sum_{i_2 = i_1 + 1}^{n - r + 2},
sum_{i_3 = i_2 + 1}^{n - r + 3} cdots sum_{i_{r-1} = i_{r-2} + 1}^{n - 1},
sum_{i_r = i_{r-1} + 1}^{n} mathbb{P}(E_{i_1}E_{i_2}cdots E_{i_r}).
$$



Each side of the equation ends up with $binom nr$ individual terms in the sum, one for each possible sequence of $r$ increasing integers where none is less than $1$ or greater than $n.$
Each of those sequences picks out a different subset of $r$ of the events in the set ${E_1, E_2, ldots, E_n}.$



Personally, I find the notation on the left is a lot easier to understand.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    There are $r$ distinct terms, where $i_1, ldots, i_r$ are chosen from ${1, ldots, n}$ and satisfies $i_1 < i_2 < ldots, i_r}$.



    For example if $n=4$, $r=3$,



    $$(-1)^{r+1} sum_{i_1 < i_2<i_3}P(E_{i_1}cap E_{i_2}cap E_{i_3})=(-1) left( P(E_1 cap E_2 cap E_3) + P(E_1 cap E_2 cap E_4)+P(E_1 cap E_3 cap E_4)+P(E_2 cap E_3 cap E_3)right)$$



    Using this notation avoid listing the same thing multiple times, for example $P(E_1 cap E_2 cap E_3) = P(E_3 cap E_2 cap E_1)$ and we don't want to list it twice.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

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      0












      $begingroup$

      Some people would write the sum $sum_{i_1<i_2<cdots<i_r}$ as
      $sum_{1 leq i_1<i_2<cdots<i_r leq n}$
      in order to emphasize that all the variables $i_1,i_2,ldots,i_r$ have to be between $1$ and $n$ inclusive.



      The $r$th sum can be rewritten in more basic summation notation as
      $$
      sum_{1 leq i_1<i_2<cdots<i_r leq n} mathbb{P}(E_{i_1}E_{i_2}cdots E_{i_r})
      = sum_{i_1 = 1}^{n - r + 1}, sum_{i_2 = i_1 + 1}^{n - r + 2},
      sum_{i_3 = i_2 + 1}^{n - r + 3} cdots sum_{i_{r-1} = i_{r-2} + 1}^{n - 1},
      sum_{i_r = i_{r-1} + 1}^{n} mathbb{P}(E_{i_1}E_{i_2}cdots E_{i_r}).
      $$



      Each side of the equation ends up with $binom nr$ individual terms in the sum, one for each possible sequence of $r$ increasing integers where none is less than $1$ or greater than $n.$
      Each of those sequences picks out a different subset of $r$ of the events in the set ${E_1, E_2, ldots, E_n}.$



      Personally, I find the notation on the left is a lot easier to understand.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Some people would write the sum $sum_{i_1<i_2<cdots<i_r}$ as
        $sum_{1 leq i_1<i_2<cdots<i_r leq n}$
        in order to emphasize that all the variables $i_1,i_2,ldots,i_r$ have to be between $1$ and $n$ inclusive.



        The $r$th sum can be rewritten in more basic summation notation as
        $$
        sum_{1 leq i_1<i_2<cdots<i_r leq n} mathbb{P}(E_{i_1}E_{i_2}cdots E_{i_r})
        = sum_{i_1 = 1}^{n - r + 1}, sum_{i_2 = i_1 + 1}^{n - r + 2},
        sum_{i_3 = i_2 + 1}^{n - r + 3} cdots sum_{i_{r-1} = i_{r-2} + 1}^{n - 1},
        sum_{i_r = i_{r-1} + 1}^{n} mathbb{P}(E_{i_1}E_{i_2}cdots E_{i_r}).
        $$



        Each side of the equation ends up with $binom nr$ individual terms in the sum, one for each possible sequence of $r$ increasing integers where none is less than $1$ or greater than $n.$
        Each of those sequences picks out a different subset of $r$ of the events in the set ${E_1, E_2, ldots, E_n}.$



        Personally, I find the notation on the left is a lot easier to understand.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Some people would write the sum $sum_{i_1<i_2<cdots<i_r}$ as
          $sum_{1 leq i_1<i_2<cdots<i_r leq n}$
          in order to emphasize that all the variables $i_1,i_2,ldots,i_r$ have to be between $1$ and $n$ inclusive.



          The $r$th sum can be rewritten in more basic summation notation as
          $$
          sum_{1 leq i_1<i_2<cdots<i_r leq n} mathbb{P}(E_{i_1}E_{i_2}cdots E_{i_r})
          = sum_{i_1 = 1}^{n - r + 1}, sum_{i_2 = i_1 + 1}^{n - r + 2},
          sum_{i_3 = i_2 + 1}^{n - r + 3} cdots sum_{i_{r-1} = i_{r-2} + 1}^{n - 1},
          sum_{i_r = i_{r-1} + 1}^{n} mathbb{P}(E_{i_1}E_{i_2}cdots E_{i_r}).
          $$



          Each side of the equation ends up with $binom nr$ individual terms in the sum, one for each possible sequence of $r$ increasing integers where none is less than $1$ or greater than $n.$
          Each of those sequences picks out a different subset of $r$ of the events in the set ${E_1, E_2, ldots, E_n}.$



          Personally, I find the notation on the left is a lot easier to understand.






          share|cite|improve this answer









          $endgroup$



          Some people would write the sum $sum_{i_1<i_2<cdots<i_r}$ as
          $sum_{1 leq i_1<i_2<cdots<i_r leq n}$
          in order to emphasize that all the variables $i_1,i_2,ldots,i_r$ have to be between $1$ and $n$ inclusive.



          The $r$th sum can be rewritten in more basic summation notation as
          $$
          sum_{1 leq i_1<i_2<cdots<i_r leq n} mathbb{P}(E_{i_1}E_{i_2}cdots E_{i_r})
          = sum_{i_1 = 1}^{n - r + 1}, sum_{i_2 = i_1 + 1}^{n - r + 2},
          sum_{i_3 = i_2 + 1}^{n - r + 3} cdots sum_{i_{r-1} = i_{r-2} + 1}^{n - 1},
          sum_{i_r = i_{r-1} + 1}^{n} mathbb{P}(E_{i_1}E_{i_2}cdots E_{i_r}).
          $$



          Each side of the equation ends up with $binom nr$ individual terms in the sum, one for each possible sequence of $r$ increasing integers where none is less than $1$ or greater than $n.$
          Each of those sequences picks out a different subset of $r$ of the events in the set ${E_1, E_2, ldots, E_n}.$



          Personally, I find the notation on the left is a lot easier to understand.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 6:04









          David KDavid K

          53.7k342116




          53.7k342116























              0












              $begingroup$

              There are $r$ distinct terms, where $i_1, ldots, i_r$ are chosen from ${1, ldots, n}$ and satisfies $i_1 < i_2 < ldots, i_r}$.



              For example if $n=4$, $r=3$,



              $$(-1)^{r+1} sum_{i_1 < i_2<i_3}P(E_{i_1}cap E_{i_2}cap E_{i_3})=(-1) left( P(E_1 cap E_2 cap E_3) + P(E_1 cap E_2 cap E_4)+P(E_1 cap E_3 cap E_4)+P(E_2 cap E_3 cap E_3)right)$$



              Using this notation avoid listing the same thing multiple times, for example $P(E_1 cap E_2 cap E_3) = P(E_3 cap E_2 cap E_1)$ and we don't want to list it twice.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                There are $r$ distinct terms, where $i_1, ldots, i_r$ are chosen from ${1, ldots, n}$ and satisfies $i_1 < i_2 < ldots, i_r}$.



                For example if $n=4$, $r=3$,



                $$(-1)^{r+1} sum_{i_1 < i_2<i_3}P(E_{i_1}cap E_{i_2}cap E_{i_3})=(-1) left( P(E_1 cap E_2 cap E_3) + P(E_1 cap E_2 cap E_4)+P(E_1 cap E_3 cap E_4)+P(E_2 cap E_3 cap E_3)right)$$



                Using this notation avoid listing the same thing multiple times, for example $P(E_1 cap E_2 cap E_3) = P(E_3 cap E_2 cap E_1)$ and we don't want to list it twice.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  There are $r$ distinct terms, where $i_1, ldots, i_r$ are chosen from ${1, ldots, n}$ and satisfies $i_1 < i_2 < ldots, i_r}$.



                  For example if $n=4$, $r=3$,



                  $$(-1)^{r+1} sum_{i_1 < i_2<i_3}P(E_{i_1}cap E_{i_2}cap E_{i_3})=(-1) left( P(E_1 cap E_2 cap E_3) + P(E_1 cap E_2 cap E_4)+P(E_1 cap E_3 cap E_4)+P(E_2 cap E_3 cap E_3)right)$$



                  Using this notation avoid listing the same thing multiple times, for example $P(E_1 cap E_2 cap E_3) = P(E_3 cap E_2 cap E_1)$ and we don't want to list it twice.






                  share|cite|improve this answer









                  $endgroup$



                  There are $r$ distinct terms, where $i_1, ldots, i_r$ are chosen from ${1, ldots, n}$ and satisfies $i_1 < i_2 < ldots, i_r}$.



                  For example if $n=4$, $r=3$,



                  $$(-1)^{r+1} sum_{i_1 < i_2<i_3}P(E_{i_1}cap E_{i_2}cap E_{i_3})=(-1) left( P(E_1 cap E_2 cap E_3) + P(E_1 cap E_2 cap E_4)+P(E_1 cap E_3 cap E_4)+P(E_2 cap E_3 cap E_3)right)$$



                  Using this notation avoid listing the same thing multiple times, for example $P(E_1 cap E_2 cap E_3) = P(E_3 cap E_2 cap E_1)$ and we don't want to list it twice.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 5:45









                  Siong Thye GohSiong Thye Goh

                  100k1466117




                  100k1466117






























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