Mixing
How is this equality established? (binomial/factorials)
$begingroup$
In Rick Durrett's book, in a proof for the asymptotic behaviour of Poisson to normal, he uses the following identity:
$$frac{n!n^m}{(n+m)!} = left(prod_{k=1}^m 1 + k/n right)^{-1}$$
I'm just wondering how this established? As Durrett states it without proof, and I am not sure how to show this.
probability self-learning
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,071115
$endgroup$
add a comment |
$begingroup$
In Rick Durrett's book, in a proof for the asymptotic behaviour of Poisson to normal, he uses the following identity:
$$frac{n!n^m}{(n+m)!} = left(prod_{k=1}^m 1 + k/n right)^{-1}$$
I'm just wondering how this established? As Durrett states it without proof, and I am not sure how to show this.
probability self-learning
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,071115
$endgroup$
$begingroup$
You have $prod_{k = 1}^m frac{n}{n + k}$, from which the $n^m$ term is obvious. Then write out what the quotient of two factorials is.
$endgroup$
– T. Bongers
Dec 4 '18 at 1:08
$begingroup$
Thanks! I will give it another go
$endgroup$
– Xiaomi
Dec 4 '18 at 1:20
$begingroup$
I noticed an error in my answer so I deleted it. The correct answer is now posted. If it helps give it a big ole' check.
$endgroup$
– clathratus
Jan 26 at 22:35
add a comment |
$begingroup$
In Rick Durrett's book, in a proof for the asymptotic behaviour of Poisson to normal, he uses the following identity:
$$frac{n!n^m}{(n+m)!} = left(prod_{k=1}^m 1 + k/n right)^{-1}$$
I'm just wondering how this established? As Durrett states it without proof, and I am not sure how to show this.
probability self-learning
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,071115
$endgroup$
In Rick Durrett's book, in a proof for the asymptotic behaviour of Poisson to normal, he uses the following identity:
$$frac{n!n^m}{(n+m)!} = left(prod_{k=1}^m 1 + k/n right)^{-1}$$
I'm just wondering how this established? As Durrett states it without proof, and I am not sure how to show this.
probability self-learning
probability self-learning
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,071115
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,071115
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,071115
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,071115
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,071115
1,071115
$begingroup$
You have $prod_{k = 1}^m frac{n}{n + k}$, from which the $n^m$ term is obvious. Then write out what the quotient of two factorials is.
$endgroup$
– T. Bongers
Dec 4 '18 at 1:08
$begingroup$
Thanks! I will give it another go
$endgroup$
– Xiaomi
Dec 4 '18 at 1:20
$begingroup$
I noticed an error in my answer so I deleted it. The correct answer is now posted. If it helps give it a big ole' check.
$endgroup$
– clathratus
Jan 26 at 22:35
add a comment |
$begingroup$
You have $prod_{k = 1}^m frac{n}{n + k}$, from which the $n^m$ term is obvious. Then write out what the quotient of two factorials is.
$endgroup$
– T. Bongers
Dec 4 '18 at 1:08
$begingroup$
Thanks! I will give it another go
$endgroup$
– Xiaomi
Dec 4 '18 at 1:20
$begingroup$
I noticed an error in my answer so I deleted it. The correct answer is now posted. If it helps give it a big ole' check.
$endgroup$
– clathratus
Jan 26 at 22:35
$begingroup$
You have $prod_{k = 1}^m frac{n}{n + k}$, from which the $n^m$ term is obvious. Then write out what the quotient of two factorials is.
$endgroup$
– T. Bongers
Dec 4 '18 at 1:08
$begingroup$
You have $prod_{k = 1}^m frac{n}{n + k}$, from which the $n^m$ term is obvious. Then write out what the quotient of two factorials is.
$endgroup$
– T. Bongers
Dec 4 '18 at 1:08
$begingroup$
Thanks! I will give it another go
$endgroup$
– Xiaomi
Dec 4 '18 at 1:20
$begingroup$
Thanks! I will give it another go
$endgroup$
– Xiaomi
Dec 4 '18 at 1:20
$begingroup$
I noticed an error in my answer so I deleted it. The correct answer is now posted. If it helps give it a big ole' check.
$endgroup$
– clathratus
Jan 26 at 22:35
$begingroup$
I noticed an error in my answer so I deleted it. The correct answer is now posted. If it helps give it a big ole' check.
$endgroup$
– clathratus
Jan 26 at 22:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $n,minBbb N_0$, $m>n$ we define
$$begin{align}
P(n,m)=&prod_{k=1}^mfrac1{1+k/n}\
=&prod_{k=1}^mfrac{n}{n+k}\
=&n^mprod_{k=1}^mfrac{1}{n+k}\
end{align}$$
Then note that
$$begin{align}
prod_{k=1}^{m}frac1{n+k}=&frac{n!}{n!}prod_{k=1}^{m}frac1{n+k}\
=&frac{n!}{n!prod_{k=1}^{m}(n+k)}\
=&frac{n!}{(prod_{r=1}^{n}r)(prod_{k=n+1}^{n+m}k)}\
=&frac{n!}{prod_{r=1}^{n+m}r}\
=&frac{n!}{(n+m)!}\
end{align}$$
So
$$P(n,m)=frac{n!n^m}{(n+m)!}$$
answered Jan 26 at 22:11
clathratusclathratus
5,1821338
$endgroup$
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: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
},
noCode: 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%2fmath.stackexchange.com%2fquestions%2f3024966%2fhow-is-this-equality-established-binomial-factorials%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
$begingroup$
For $n,minBbb N_0$, $m>n$ we define
$$begin{align}
P(n,m)=&prod_{k=1}^mfrac1{1+k/n}\
=&prod_{k=1}^mfrac{n}{n+k}\
=&n^mprod_{k=1}^mfrac{1}{n+k}\
end{align}$$
Then note that
$$begin{align}
prod_{k=1}^{m}frac1{n+k}=&frac{n!}{n!}prod_{k=1}^{m}frac1{n+k}\
=&frac{n!}{n!prod_{k=1}^{m}(n+k)}\
=&frac{n!}{(prod_{r=1}^{n}r)(prod_{k=n+1}^{n+m}k)}\
=&frac{n!}{prod_{r=1}^{n+m}r}\
=&frac{n!}{(n+m)!}\
end{align}$$
So
$$P(n,m)=frac{n!n^m}{(n+m)!}$$
answered Jan 26 at 22:11
clathratusclathratus
5,1821338
$endgroup$
add a comment |
$begingroup$
For $n,minBbb N_0$, $m>n$ we define
$$begin{align}
P(n,m)=&prod_{k=1}^mfrac1{1+k/n}\
=&prod_{k=1}^mfrac{n}{n+k}\
=&n^mprod_{k=1}^mfrac{1}{n+k}\
end{align}$$
Then note that
$$begin{align}
prod_{k=1}^{m}frac1{n+k}=&frac{n!}{n!}prod_{k=1}^{m}frac1{n+k}\
=&frac{n!}{n!prod_{k=1}^{m}(n+k)}\
=&frac{n!}{(prod_{r=1}^{n}r)(prod_{k=n+1}^{n+m}k)}\
=&frac{n!}{prod_{r=1}^{n+m}r}\
=&frac{n!}{(n+m)!}\
end{align}$$
So
$$P(n,m)=frac{n!n^m}{(n+m)!}$$
answered Jan 26 at 22:11
clathratusclathratus
5,1821338
$endgroup$
add a comment |
$begingroup$
For $n,minBbb N_0$, $m>n$ we define
$$begin{align}
P(n,m)=&prod_{k=1}^mfrac1{1+k/n}\
=&prod_{k=1}^mfrac{n}{n+k}\
=&n^mprod_{k=1}^mfrac{1}{n+k}\
end{align}$$
Then note that
$$begin{align}
prod_{k=1}^{m}frac1{n+k}=&frac{n!}{n!}prod_{k=1}^{m}frac1{n+k}\
=&frac{n!}{n!prod_{k=1}^{m}(n+k)}\
=&frac{n!}{(prod_{r=1}^{n}r)(prod_{k=n+1}^{n+m}k)}\
=&frac{n!}{prod_{r=1}^{n+m}r}\
=&frac{n!}{(n+m)!}\
end{align}$$
So
$$P(n,m)=frac{n!n^m}{(n+m)!}$$
answered Jan 26 at 22:11
clathratusclathratus
5,1821338
$endgroup$
For $n,minBbb N_0$, $m>n$ we define
$$begin{align}
P(n,m)=&prod_{k=1}^mfrac1{1+k/n}\
=&prod_{k=1}^mfrac{n}{n+k}\
=&n^mprod_{k=1}^mfrac{1}{n+k}\
end{align}$$
Then note that
$$begin{align}
prod_{k=1}^{m}frac1{n+k}=&frac{n!}{n!}prod_{k=1}^{m}frac1{n+k}\
=&frac{n!}{n!prod_{k=1}^{m}(n+k)}\
=&frac{n!}{(prod_{r=1}^{n}r)(prod_{k=n+1}^{n+m}k)}\
=&frac{n!}{prod_{r=1}^{n+m}r}\
=&frac{n!}{(n+m)!}\
end{align}$$
So
$$P(n,m)=frac{n!n^m}{(n+m)!}$$
answered Jan 26 at 22:11
clathratusclathratus
5,1821338
edited Jan 26 at 22:33
answered Jan 26 at 22:11
clathratusclathratus
5,1821338
answered Jan 26 at 22:11
clathratusclathratus
5,1821338
answered Jan 26 at 22:11
clathratusclathratus
5,1821338
5,1821338
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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%2fmath.stackexchange.com%2fquestions%2f3024966%2fhow-is-this-equality-established-binomial-factorials%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
$begingroup$
You have $prod_{k = 1}^m frac{n}{n + k}$, from which the $n^m$ term is obvious. Then write out what the quotient of two factorials is.
$endgroup$
– T. Bongers
Dec 4 '18 at 1:08
$begingroup$
Thanks! I will give it another go
$endgroup$
– Xiaomi
Dec 4 '18 at 1:20
$begingroup$
I noticed an error in my answer so I deleted it. The correct answer is now posted. If it helps give it a big ole' check.
$endgroup$
– clathratus
Jan 26 at 22:35