Writing the first-order loss function of the negative binomial distribution in terms of the CDF and the PMF.

Question

So the negative binomial distribution uses two parameters $r$ and $p$, where $r \gt 0$ and $0 \lt p \lt 1$. We assume $r$ is an integer in this question.

The probability mass function is defined as:

$$f(s) = \dbinom{s+r-1}{r-1} (1-p)^rp^s$$

and the cumulative distribution as:

$$F(S) =\sum_{s=0}^S f(s) $$

The mean and variance are respectively $\frac{rp}{1-p}$ and $\frac{rp}{(1-p)^2}$.

Now I got the following first-order loss function, defined as:

$$L(q) =\sum_{s=q+1}^\infty (s-q) f(s) $$

According to a book, I can write this in terms of $F(q)$ and $f(q)$ like this:

$$L(q)=-(q-r\beta)(1-F(q))+(q+r)\beta f(q)$$

, where $\beta = \frac{p}{1-p}$.

But I don't see how they derived this...

I tried to split up $L(q)$ into:

$$L(q) =\sum_{s=q+1}^\infty s f(s) - \sum_{s=q+1}^\infty q f(s) $$

, where the right term is $ -q(1-F(q)) $ and the left term can be split up in:

$$\sum_{s=q+1}^\infty s f(s) = \sum_{s=0}^\infty s f(s) - \sum_{s=0}^q s f(s) $$

$$= \frac{rp}{1-p} - \sum_{s=0}^q \frac{r(r+1)..(r+s-1)}{(s-1)!}(1-p)^rp^s $$

But I don't see how I can go further...

Any hints or solutions are welcome.

Answer 1

We can easily prove the claim by induction on $q$ (not sure if this is the solution you were looking at). For $q=0$ we get $$L(0)=\sum_{s=0}^\infty sf(s)=r\beta=-(0-r\beta)(1-f(0))+(0+r\beta)f(0)$$ Then if we assume that the thesis is true for $q$ we obtain $$L(q+1)=\sum_{s=q+2}^\infty (s-q-1)f(s)=\sum_{s=q+1}^\infty (s-q)f(s)-(q+1-q)f(q+1)-\sum_{s=q+2}^\infty f(s)=-(q-r\beta)(1-F(q))+(q+r)\beta f(q)-f(q+1)-\sum_{s=q+2}^\infty f(s)=-(q-r\beta)(1-F(q))+(q+r)\beta f(q)-(1-F(q))=-(q+1-r\beta)(1-F(q))+(q+r)\beta f(q)=-(q+1-r\beta)(1-F(q+1))-(q+1-r\beta)f(q+1)+(q+r)\beta f(q)$$ So for the inductive step we are left to prove that $$(q+1-r\beta)f(q+1)+(q+r)\beta f(q)=(q+1+r)\beta f(q+1)$$ that is equivalent to $$f(q+1)=\frac{\beta}{1+\beta}\frac{q+r}{q+1}f(q)$$ which follows immediately from the definition of $f$, noticed that $p=\frac{\beta}{1+\beta}$.