Relation between binomial and negative binomial
$begingroup$
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 yesterday
user46697user46697
367212
$endgroup$
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 yesterday
user46697user46697
367212
$endgroup$
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 yesterday
user46697user46697
367212
$endgroup$
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 yesterday
user46697user46697
367212
asked yesterday
user46697user46697
367212
asked yesterday
user46697user46697
367212
asked yesterday
user46697user46697
367212
asked yesterday
user46697user46697
367212
367212
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:
$$P{mbox{less than r successes in n trials}}= P{mbox{more than n trials to get r successes}}$$
answered yesterday
EsmailianEsmailian
26914
$endgroup$
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
yesterday
$begingroup$
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
yesterday
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
yesterday
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "65"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f397830%2frelation-between-binomial-and-negative-binomial%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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:
$$P{mbox{less than r successes in n trials}}= P{mbox{more than n trials to get r successes}}$$
answered yesterday
EsmailianEsmailian
26914
$endgroup$
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
yesterday
$begingroup$
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
yesterday
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
yesterday
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.
$$begin{align*}
P{X geq r} &= P{mbox{at least r successes in n trials}}\
&= P{mbox{r-th success in n-th trial or before}}\
&= P{mbox{n or fewer trials to get r successes}}\
&= P{Y leq n}
end{align*}$$
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:
$$P{mbox{less than r successes in n trials}}= P{mbox{more than n trials to get r successes}}$$
answered yesterday
EsmailianEsmailian
26914
$endgroup$
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
yesterday
$begingroup$
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
yesterday
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
yesterday
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.
$$begin{align*}
P{X geq r} &= P{mbox{at least r successes in n trials}}\
&= P{mbox{r-th success in n-th trial or before}}\
&= P{mbox{n or fewer trials to get r successes}}\
&= P{Y leq n}
end{align*}$$
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:
$$P{mbox{less than r successes in n trials}}= P{mbox{more than n trials to get r successes}}$$
answered yesterday
EsmailianEsmailian
26914
$endgroup$
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:
$$P{mbox{less than r successes in n trials}}= P{mbox{more than n trials to get r successes}}$$
answered yesterday
EsmailianEsmailian
26914
edited yesterday
answered yesterday
EsmailianEsmailian
26914
answered yesterday
EsmailianEsmailian
26914
answered yesterday
EsmailianEsmailian
26914
26914
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
yesterday
$begingroup$
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
yesterday
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
yesterday
add a comment |
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
yesterday
$begingroup$
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
yesterday
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
yesterday
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
yesterday
$begingroup$
Thank you. What about the second relation?
$endgroup$
– user46697
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
yesterday
$begingroup$
(32) is simple the complement of (31)
$endgroup$
– Henry
yesterday
add a comment |
Thanks for contributing an answer to Cross Validated!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f397830%2frelation-between-binomial-and-negative-binomial%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
