I am trying to evaluate the following summation:
$\hspace{2in}f=\sum_{z=0}^{N-1}Q(z)\phi(z)$
Here, $\hspace{1in}Q(z)=\left(\begin{array}{c}N-1 \\ z\end{array}\right)\left(\frac{ q}{N}\right)^{z}\left(1-\frac{ q}{N}\right)^{N-1-z}$
$\hspace{1.5in}\phi({z})$ is CDF of gaussian RV ~$N(0,a+z)$
Is there a way of evaluating $f$ analytically?
Any help is appreciated.
If it is impossible, is there a good approximation that can be helpful. I can assume $N$ is a large positive integer.
Here's a heuristic approach that can easily be done strict.
Claim: If $N \to \infty$ then $$f \approx \sum_{z=\max([q]-10, 0)}^{[q]+10} \phi(z) e^{-q}\frac{q^z}{z!} \in (0,1)$$ for $q \le 10$ and error may be estimated easily. In case when $q \ge 10$ we have $f \approx \sum_{q - \sqrt{10q} < z < q+\sqrt{10q}} \phi(z) e^{-q}\frac{q^z}{z!} \in (0,1)$.
Corollary 1. If $q - \sqrt{10q} > 2 + 2\sqrt{a+1}$ and $ q > 10$ then $f \approx 1$.
As $q$ is fixed then number of terms in sums is fixed and not bigger than $2\max(10, \sqrt{10q}) + 1$.
Proof of the claim. We suppose that $N >> a$ and $N >>q$. Consider the case $q \le 10$.
As we have $z \ge 0$ and $N(0,a+z)$ exists we will suppose $a > 0$ (case $a=0$ is close to case $a=0.001$).
Put $\Phi(u) = \int_{-\infty}^u \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-a)^2}{2}} dx$, then $\phi(z) = \Phi(\frac{z}{\sqrt{z+a}})$. For $z\ge 0$ function $\frac{z}{\sqrt{z+a}}$ is non-decreasing, positive and tends to 1 as $z \to +\infty$. Hence $\phi(k)$, $k = 0, 1, 2, \ldots$ is non-decreasing, positive and tends to 1 as $z \to +\infty$.
We have $$f = \sum_{z=\max([q]-10, 0)}^{[q]+10} \phi(z) Q(z) + \sum_{z \in A} \phi(z) Q(z) $$ where $A = \{ z \ge 0: (z < \max([q]-10, 0)) \vee (z > [q]+10)\}$ But $$\sum_{z \in A} \phi(z) Q(z) \le \sum_{z \ge 0, |z-q| \ge 9} \phi(z) Q(z) \le \sum_{z \ge 0, |z-q| \ge 9} Q(z). $$
Put $\xi \sim Bin(\frac{q}{N}, N-1)$. Hence $$\sum_{z \in A} \phi(z) Q(z) \le P(|\xi - q| \ge 9) = P(|\xi - E\xi - \frac{q}N| \ge 9) \le $$ $$P(|\xi - E\xi| \ge 9 - \frac{q}N) \approx P(Pois(q) \ge 9) \le \frac{D Pois(q)}{81} \approx 0 $$ as $q \le 10$.
Hence $$f \approx \sum_{z=\max([q]-10, 0)}^{[q]+10} \phi(z) Q(z)$$.
If $N \to \infty$ then $Q(k) \to P(Posi(q)=k) = e^{-q}\frac{q^k}{k!}$ by Poisson limit theorem. When we write $Q(k) \approx e^{-q}\frac{q^k}{k!}$ and error may be estimated easily from Stirling's formula.
Thus $$f \approx \sum_{z=\max([q]-10, 0)}^{[q]+10} \phi(z) e^{-q}\frac{q^z}{z!}$$.
Case $q \ge 10$ is similar. Instead of $A = \{ z \ge 0: (z < \max([q]-10, 0)) \vee (z > [q]+10)\}$ and $$\sum_{z \in A} \phi(z) Q(z) \le P(|\xi - q| \ge 9) \le P(|\xi - E\xi| \ge 9 - \frac{q}N) \approx P(Pois(q) \ge 9) \le \frac{D Pois(q)}{81} \approx 0 $$ we will have $A = \{ z \ge 0: (z > q + \sqrt{10q}) \vee z< q - \sqrt{10q}\}$ and $$\sum_{z \in A} \phi(z) Q(z) \le P(|\xi - q| \ge \sqrt{10q}-1) \approx P(|\xi - E\xi| \ge \sqrt{10q} ) \approx P(Pois(q) \ge \sqrt{10q}) \le \frac{D Pois(q)}{10q} \approx 0 $$.
Corollary2. In case when $q \ge 0$ is arbitrary we have $f \approx \sum_{q - \max(\sqrt{10q},10) <z < q+\max(\sqrt{10q},10)} \phi(z) e^{-q}\frac{q^z}{z!} \in (0,1)$.
Proof of the corollary 1: $\Phi(z) \approx 1$ for $z \ge 2$, thus $\phi(z) \approx 1$ for $\frac{z}{\sqrt{z+a}} \ge 2$. As we have $\frac{z}{\sqrt{z+a}} \ge 2$ iff $z \ge 2 + 2 \sqrt{a+1}$ then $\phi(z) \approx 1$ for all $z \ge 2 + 2 \sqrt{a+1}$ .
If $q - \sqrt{10q} > 2 + 2\sqrt{a+1}$ and $ q > 10$ then $$f \approx\sum_{q - \sqrt{10q} < z < q+\sqrt{10q}}\phi(z) e^{-q}\frac{q^z}{z!} \approx \sum_{q - \sqrt{10q} < z < q+\sqrt{10q}} e^{-q}\frac{q^z}{z!} = P(q - \sqrt{10q} < Pois(q) < q +\sqrt{10q})$$.
But $$P(q - \sqrt{10q} < Pois(q) < q +\sqrt{10q}) \approx 1$$ by Chebyshev's inequality.
