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
asked Mar 16 at 9:09
user46697user46697
438312
$endgroup$
add a comment |
$begingroup$
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
asked Mar 16 at 9:09
user46697user46697
438312
$endgroup$
add a comment |
$begingroup$
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
asked Mar 16 at 9:09
user46697user46697
438312
$endgroup$
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
asked Mar 16 at 9:09
user46697user46697
438312
asked Mar 16 at 9:09
user46697user46697
438312
asked Mar 16 at 9:09
user46697user46697
438312
asked Mar 16 at 9:09
user46697user46697
438312
asked Mar 16 at 9:09
user46697user46697
438312
438312
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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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.
$$beginalign*
PX geq r &= Pmboxat least r successes in n trials\
&= Pmboxr-th success in n-th trial or before\
&= Pmboxn or fewer trials to get r successes\
&= PY leq n
endalign*$$
The second relation is the complement of first relation that is:
$$beginalign*
PX geq r &= PY leq n,\
1 - PX geq r &= 1 - PY leq n,\
PX < r &= PY > n\
endalign*$$
The second relation means:
$$Pmboxless than r successes in n trials= Pmboxmore than n trials to get r successes$$
answered Mar 16 at 10:02
EsmailianEsmailian
42615
$endgroup$
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
Mar 16 at 10:05
$begingroup$
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
Mar 16 at 10:15
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
Mar 16 at 13:33
add a comment |
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1 Answer
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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.
$$beginalign*
PX geq r &= Pmboxat least r successes in n trials\
&= Pmboxr-th success in n-th trial or before\
&= Pmboxn or fewer trials to get r successes\
&= PY leq n
endalign*$$
The second relation is the complement of first relation that is:
$$beginalign*
PX geq r &= PY leq n,\
1 - PX geq r &= 1 - PY leq n,\
PX < r &= PY > n\
endalign*$$
The second relation means:
$$Pmboxless than r successes in n trials= Pmboxmore than n trials to get r successes$$
answered Mar 16 at 10:02
EsmailianEsmailian
42615
$endgroup$
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
Mar 16 at 10:05
$begingroup$
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
Mar 16 at 10:15
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
Mar 16 at 13:33
add a comment |
$begingroup$
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.
$$beginalign*
PX geq r &= Pmboxat least r successes in n trials\
&= Pmboxr-th success in n-th trial or before\
&= Pmboxn or fewer trials to get r successes\
&= PY leq n
endalign*$$
The second relation is the complement of first relation that is:
$$beginalign*
PX geq r &= PY leq n,\
1 - PX geq r &= 1 - PY leq n,\
PX < r &= PY > n\
endalign*$$
The second relation means:
$$Pmboxless than r successes in n trials= Pmboxmore than n trials to get r successes$$
answered Mar 16 at 10:02
EsmailianEsmailian
42615
$endgroup$
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
Mar 16 at 10:05
$begingroup$
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
Mar 16 at 10:15
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
Mar 16 at 13:33
add a comment |
$begingroup$
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.
$$beginalign*
PX geq r &= Pmboxat least r successes in n trials\
&= Pmboxr-th success in n-th trial or before\
&= Pmboxn or fewer trials to get r successes\
&= PY leq n
endalign*$$
The second relation is the complement of first relation that is:
$$beginalign*
PX geq r &= PY leq n,\
1 - PX geq r &= 1 - PY leq n,\
PX < r &= PY > n\
endalign*$$
The second relation means:
$$Pmboxless than r successes in n trials= Pmboxmore than n trials to get r successes$$
answered Mar 16 at 10:02
EsmailianEsmailian
42615
$endgroup$
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.
$$beginalign*
PX geq r &= Pmboxat least r successes in n trials\
&= Pmboxr-th success in n-th trial or before\
&= Pmboxn or fewer trials to get r successes\
&= PY leq n
endalign*$$
The second relation is the complement of first relation that is:
$$beginalign*
PX geq r &= PY leq n,\
1 - PX geq r &= 1 - PY leq n,\
PX < r &= PY > n\
endalign*$$
The second relation means:
$$Pmboxless than r successes in n trials= Pmboxmore than n trials to get r successes$$
answered Mar 16 at 10:02
EsmailianEsmailian
42615
edited Mar 16 at 10:19
answered Mar 16 at 10:02
EsmailianEsmailian
42615
answered Mar 16 at 10:02
EsmailianEsmailian
42615
answered Mar 16 at 10:02
EsmailianEsmailian
42615
42615
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
Mar 16 at 10:05
$begingroup$
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
Mar 16 at 10:15
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
Mar 16 at 13:33
add a comment |
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
Mar 16 at 10:05
$begingroup$
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
Mar 16 at 10:15
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
Mar 16 at 13:33
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
Mar 16 at 10:05
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
Mar 16 at 10:05
$begingroup$
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
Mar 16 at 10:15
$begingroup$
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
Mar 16 at 10:15
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
Mar 16 at 13:33
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
Mar 16 at 13:33
add a comment |
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