How is this equality established? (binomial/factorials)
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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
1,050115
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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
1,050115
$endgroup$
$begingroup$
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
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
1,050115
$endgroup$
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
1,050115
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,050115
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,050115
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,050115
asked Dec 4 '18 at 1:02
XiaomiXiaomi
1,050115
1,050115
$begingroup$
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
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
$begingroup$
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
add a comment |
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$begingroup$
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