Question

If T~BIN(n, p), then find the expectation:

E[T(T-1)...(T-k+1)]

Answer 1

Basically, this is trying to compute the sum
\[\sum_{t=0}^n t(t-1)(t-2)\cdots (t-k+1) {n \choose t} p^t (1-p)^{n-t} \]
I tried doing this:
\[\sum_{t=k}^n {n \choose t} p^t (1-p)^{n-t} \]since the sum is 0 at \(t=0, 1, \ldots, k-1\)\[=\sum_{t=0}^{n-k} {n \choose t+k} p^{t+k} (1-p)^{n-(t+k)} \]\[=p^k\sum_{t=0}^{n-k} {n \choose t+k} p^{t} (1-p)^{(n-k)-t} \]

I feel like I need to change \({n \choose t+k}\) to \({n-k \choose t}\) somehow so that I get a Binomial(n-k, p) and then that would sum to 1..but I am having trouble doing so.

Answer 2

Wait a minute, what is k?

Answer 3

Just some constant integer you have to deal with?

Answer 4

What you are getting is something similar to the CDF for the binomial distribution. I don't think there is any simplifying it past that.

Answer 5

Yes k is just some integer with values 1, 2, ...

Answer 6

On second though, when \(t \geq k\) the \(t(t-1)...(t-k+1)\) will not necessarily be \(1\).

Answer 7

Recall the derivation the expectation of \(E(T)\):
\[\begin{align*}E(T)&=\sum_{t=0}^nt\binom nt p^t (1-p)^{n-t}\\
&=\sum_{t=0}^nt\frac{n!}{t!(n-t)!} p^t (1-p)^{n-t}\\
&=0+\sum_{t=1}^nt\frac{n!}{t!(n-t)!} p^t (1-p)^{n-t}\\
&=np\sum_{t=1}^{n-1}\frac{(n-1)!}{(t-1)!(n-t)!} p^{t-1} (1-p)^{n-t}\\
&=np\sum_{t=1}^{n-1}\frac{(n-1)!}{(t-1)!((n-1)-(t-1))!} p^{t-1} (1-p)^{(n-1)-(t-1)}\\
&=np\sum_{t=0}^{n-1}\frac{(n-1)!}{t!((n-1)-t)!} p^{t-1} (1-p)^{(n-1)-t}\\
&=np\color{red}{\sum_{t=0}^{n-1}\binom{n-1}t p^{t-1} (1-p)^{(n-1)-t}}\\
&=np
\end{align*}\]
The red part is 1, since that's the pdf of \(\text{Bin}(n-1,p)\).

No doubt finding the expectation of \(E\bigg[T(T-1)\cdots(T-(k-1))\bigg]\) involves a similar process.

Answer 8

When \(t=k\) you have \(t(t−1)...(t−k+1) = k(k-1)\dots (1) = k!\)
When \(t=k+1\) you get \(\frac{(k+1)!}{1!}\).
Whne \(t=k+c\) you get \((n+c)!/c!\)

Answer 9

whoops, should say \((k+c)!/c!\)

Answer 10

And \(c\) is just \(t-k\). So that simplifies to \(t(t−1)...(t−k+1) = t!/(t-k)!\).
This is permutations formula, and is just \(k! \binom tk\).

Answer 11

Hopefully, someone can check if I'm right:
\[E\bigg[T(T-1)\cdots(T-(k-1))\bigg]\\
~~~~=\sum_{t=0}^nt(t-1)\cdots(t-(k-1))\binom nt p^t (1-p)^{n-t}\\
~~~~=\sum_{t=0}^n t (t-1)\cdots(t-(k-1))\frac{n!}{t!(n-t)!} p^t (1-p)^{n-t}\\
~~~~=0+\sum_{t=k}^n t (t-1)\cdots(t-(k-1))\frac{n!}{t!(n-t)!} p^t (1-p)^{n-t}\\
~~~~=n(n-1)\cdots(n-(k-1))~p^{k-2}\\
~~~~~~~~\times \sum_{t=k}^n t \frac{(n-(k-2))!}{(t-(k-2))!(n-(k-2)-(t-(k-2)))!} p^{t-(k-2)} (1-p)^{n-(k-2)-(t-(k-2))}\\
~~~~=n(n-1)\cdots(n-(k-1))~p^{k-2}\\
~~~~=p^{k-2}\frac{n!}{(n-k)!}\]
As @wio mentioned, the fraction is equivalent to \({}_nP_k\).