Relation between binomial and negative binomial












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I was reading on negative binomial from a Statistics textbook and came across this portion on probability relation between binomial and negative binomial. $Y$ refers to the number of trials required to get $r$ successes.
Can somebody please explain the relation






self-study binomial negative-binomial







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    enter image description here



    I was reading on negative binomial from a Statistics textbook and came across this portion on probability relation between binomial and negative binomial. $Y$ refers to the number of trials required to get $r$ successes.
    Can somebody please explain the relation






    self-study binomial negative-binomial







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      enter image description here



      I was reading on negative binomial from a Statistics textbook and came across this portion on probability relation between binomial and negative binomial. $Y$ refers to the number of trials required to get $r$ successes.
      Can somebody please explain the relation






      self-study binomial negative-binomial







      share|cite|improve this question









      $endgroup$




      enter image description here



      I was reading on negative binomial from a Statistics textbook and came across this portion on probability relation between binomial and negative binomial. $Y$ refers to the number of trials required to get $r$ successes.
      Can somebody please explain the relation






      self-study binomial negative-binomial




      self-study binomial negative-binomial






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 16 hours ago









      user46697user46697

      367212




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          1 Answer
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          $begingroup$

          Based on binomial distribution, event ${X geq r}$ is the set of outcomes that satisfy "$n$ trials led to $r$ successes or more", which is equivalent to "$r$-th success happened at $n$-th trial or before", which is in turn equivalent to "$n$ trials or less were required to get $r$ successes", and that is it.
          $$begin{align*}
          P{X geq r} &= P{mbox{at least r successes in n trials}}\
          &= P{mbox{r-th success in n-th trial or before}}\
          &= P{mbox{n or fewer trials to get r successes}}\
          &= P{Y leq n}
          end{align*}$$



          The second relation is the complement of first relation that is:
          $$begin{align*}
          P{X geq r} &= P{Y leq n},\
          1 - P{X geq r} &= 1 - P{Y leq n},\
          P{X < r} &= P{Y > n}\
          end{align*}$$



          The second relation means:



          $$P{mbox{less than r successes in n trials}}= P{mbox{more than n trials to get r successes}}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. What about the second relation?
            $endgroup$
            – user46697
            15 hours ago










          • $begingroup$
            In the second relation, is it right to say that the P( less than r successes in n trials) $=$ P( more than n trials to get r successes)
            $endgroup$
            – user46697
            15 hours ago












          • $begingroup$
            (32) is simple the complement of (31)
            $endgroup$
            – Henry
            12 hours ago











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          2












          $begingroup$

          Based on binomial distribution, event ${X geq r}$ is the set of outcomes that satisfy "$n$ trials led to $r$ successes or more", which is equivalent to "$r$-th success happened at $n$-th trial or before", which is in turn equivalent to "$n$ trials or less were required to get $r$ successes", and that is it.
          $$begin{align*}
          P{X geq r} &= P{mbox{at least r successes in n trials}}\
          &= P{mbox{r-th success in n-th trial or before}}\
          &= P{mbox{n or fewer trials to get r successes}}\
          &= P{Y leq n}
          end{align*}$$



          The second relation is the complement of first relation that is:
          $$begin{align*}
          P{X geq r} &= P{Y leq n},\
          1 - P{X geq r} &= 1 - P{Y leq n},\
          P{X < r} &= P{Y > n}\
          end{align*}$$



          The second relation means:



          $$P{mbox{less than r successes in n trials}}= P{mbox{more than n trials to get r successes}}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. What about the second relation?
            $endgroup$
            – user46697
            15 hours ago










          • $begingroup$
            In the second relation, is it right to say that the P( less than r successes in n trials) $=$ P( more than n trials to get r successes)
            $endgroup$
            – user46697
            15 hours ago












          • $begingroup$
            (32) is simple the complement of (31)
            $endgroup$
            – Henry
            12 hours ago
















          2












          $begingroup$

          Based on binomial distribution, event ${X geq r}$ is the set of outcomes that satisfy "$n$ trials led to $r$ successes or more", which is equivalent to "$r$-th success happened at $n$-th trial or before", which is in turn equivalent to "$n$ trials or less were required to get $r$ successes", and that is it.
          $$begin{align*}
          P{X geq r} &= P{mbox{at least r successes in n trials}}\
          &= P{mbox{r-th success in n-th trial or before}}\
          &= P{mbox{n or fewer trials to get r successes}}\
          &= P{Y leq n}
          end{align*}$$



          The second relation is the complement of first relation that is:
          $$begin{align*}
          P{X geq r} &= P{Y leq n},\
          1 - P{X geq r} &= 1 - P{Y leq n},\
          P{X < r} &= P{Y > n}\
          end{align*}$$



          The second relation means:



          $$P{mbox{less than r successes in n trials}}= P{mbox{more than n trials to get r successes}}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. What about the second relation?
            $endgroup$
            – user46697
            15 hours ago










          • $begingroup$
            In the second relation, is it right to say that the P( less than r successes in n trials) $=$ P( more than n trials to get r successes)
            $endgroup$
            – user46697
            15 hours ago












          • $begingroup$
            (32) is simple the complement of (31)
            $endgroup$
            – Henry
            12 hours ago














          2












          2








          2





          $begingroup$

          Based on binomial distribution, event ${X geq r}$ is the set of outcomes that satisfy "$n$ trials led to $r$ successes or more", which is equivalent to "$r$-th success happened at $n$-th trial or before", which is in turn equivalent to "$n$ trials or less were required to get $r$ successes", and that is it.
          $$begin{align*}
          P{X geq r} &= P{mbox{at least r successes in n trials}}\
          &= P{mbox{r-th success in n-th trial or before}}\
          &= P{mbox{n or fewer trials to get r successes}}\
          &= P{Y leq n}
          end{align*}$$



          The second relation is the complement of first relation that is:
          $$begin{align*}
          P{X geq r} &= P{Y leq n},\
          1 - P{X geq r} &= 1 - P{Y leq n},\
          P{X < r} &= P{Y > n}\
          end{align*}$$



          The second relation means:



          $$P{mbox{less than r successes in n trials}}= P{mbox{more than n trials to get r successes}}$$






          share|cite|improve this answer











          $endgroup$



          Based on binomial distribution, event ${X geq r}$ is the set of outcomes that satisfy "$n$ trials led to $r$ successes or more", which is equivalent to "$r$-th success happened at $n$-th trial or before", which is in turn equivalent to "$n$ trials or less were required to get $r$ successes", and that is it.
          $$begin{align*}
          P{X geq r} &= P{mbox{at least r successes in n trials}}\
          &= P{mbox{r-th success in n-th trial or before}}\
          &= P{mbox{n or fewer trials to get r successes}}\
          &= P{Y leq n}
          end{align*}$$



          The second relation is the complement of first relation that is:
          $$begin{align*}
          P{X geq r} &= P{Y leq n},\
          1 - P{X geq r} &= 1 - P{Y leq n},\
          P{X < r} &= P{Y > n}\
          end{align*}$$



          The second relation means:



          $$P{mbox{less than r successes in n trials}}= P{mbox{more than n trials to get r successes}}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 15 hours ago

























          answered 15 hours ago









          EsmailianEsmailian

          25414




          25414












          • $begingroup$
            Thank you. What about the second relation?
            $endgroup$
            – user46697
            15 hours ago










          • $begingroup$
            In the second relation, is it right to say that the P( less than r successes in n trials) $=$ P( more than n trials to get r successes)
            $endgroup$
            – user46697
            15 hours ago












          • $begingroup$
            (32) is simple the complement of (31)
            $endgroup$
            – Henry
            12 hours ago


















          • $begingroup$
            Thank you. What about the second relation?
            $endgroup$
            – user46697
            15 hours ago










          • $begingroup$
            In the second relation, is it right to say that the P( less than r successes in n trials) $=$ P( more than n trials to get r successes)
            $endgroup$
            – user46697
            15 hours ago












          • $begingroup$
            (32) is simple the complement of (31)
            $endgroup$
            – Henry
            12 hours ago
















          $begingroup$
          Thank you. What about the second relation?
          $endgroup$
          – user46697
          15 hours ago




          $begingroup$
          Thank you. What about the second relation?
          $endgroup$
          – user46697
          15 hours ago












          $begingroup$
          In the second relation, is it right to say that the P( less than r successes in n trials) $=$ P( more than n trials to get r successes)
          $endgroup$
          – user46697
          15 hours ago






          $begingroup$
          In the second relation, is it right to say that the P( less than r successes in n trials) $=$ P( more than n trials to get r successes)
          $endgroup$
          – user46697
          15 hours ago














          $begingroup$
          (32) is simple the complement of (31)
          $endgroup$
          – Henry
          12 hours ago




          $begingroup$
          (32) is simple the complement of (31)
          $endgroup$
          – Henry
          12 hours ago


















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