I need to solve an optimization problem that involves an expected value like 
$$F(n,x) = \sum\_{k=0}^n \binom{n}{k} p^k(1 - p)^{n - k} f(k,x).$$


Here $f(k,x)$ is actually a probability coming from a normal distribution. And what I need, in particular, is to solve a first order condition that involves the derivative of $F$ with respect to $x$. Numerically it is OK, but I was wondering how I could obtain a nice analytical form. I tried several things, like replacing $f(k,x)$ with other normal-like distributions, but none worked. 


I need this to show some asymptotic properties, so I could use a solution that considers the case when $n\rightarrow\infty$. Also, approximations are equally welcome. 


Update upon the first comment: 


I can simplify $f$ in the following way.
$$f(k,x)= \Pr[y\geq Ak+B(n-k)+Cx] \text{ where } y\sim N(n,\sigma).$$


And I want to solve


$$\max qF(n,x)+(1-q)G(n,x)$$


with respect to $x$ where $q\in(0,1)$ and
$$G(n,x) = \sum\_{k=0}^n \binom{n}{k} p^k(1 - p)^{n - k} g(k,x)$$
with 
$$g(k,x)=\Pr[y'\leq Ak+B(n-k)+Cx] \text{ where } y'\sim N(-n,\sigma).$$


Upon a comment: $p\in(0,1)$ and $A,B,C\in\mathbb R\_{++}$. 



Expected value (probability) maximization with binomial distribution
The maximum converges to $1$ as $n\to\infty$. Indeed, one can write 
$$F(n,x)=P(X\_n<w)\quad\text{and}\quad G(n,x)=P(X\_n>w-d), 
$$
where $X\_n:=\dfrac{(a-b)Y\_n+Z}n$, $Y\_n$ and $Z$ are independent random variables (r.v.'s), $Y\_n$ has the binomial distribution with parameters $(n,p)$, $Z$ is standard normal, $a:=A/\sigma$, $b:=B/\sigma$, $w:=-b\_1-cx/n$, $b\_1:=(B-1)/\sigma$, $c:=C/\sigma$, $d:=2/\sigma>0$. So, the maximum of $qF(n,x)+(1-q)G(n,x)$ in $x\in\mathbb{R}$ equals $m\_n:=\max\_{w\in\mathbb{R}} [q P(X\_n<w)+(1-q)P(X\_n>w-d)]$. Note that, by the law of large numbers, $X\_n\to\mu:=(a-b)p$ as $n\to\infty$ (say, in probability). So, taking $w=\mu+d/2$, one has $P(X\_n<w)\to1$ and $P(X\_n>w-d)\to1$. On the other hand, $m\_n\le q+(1-q)=1$. Thus, $m\_n\to1$ as $n\to\infty$. 


The signs of $A,B,C$ are irrelevant for this conclusion, as long as $A,B,C$ are fixed and $C\ne0$. The assumptions on the r.v.'s $y$ and $y'$, including their normality, are also mostly irrelevant; for instance, one can replace $y$ and $y'$ by any r.v.'s $V\_n$ and $W\_n$ such that for some real $\delta>0$ one has $P(V\_n-W\_n\ge n\delta)\to0$ as $n\to\infty$. 


**Addendum:** **juror**: As I understood, you were able to rewrite $F(n,x)$ as $P(AY\_n+B(n-Y\_n)+Cx+Z\sigma<n)$. Then you can further rewrite this as 
$$P((A-B)Y\_n+Z\sigma<-[(B-1)n+Cx])=P\Big(\frac{(a-b)Y\_n+Z}n<-\,\frac{(B-1)+Cx/n}\sigma\Big)
=P(X\_n<w), 
$$
as was stated previously (recall the definitions of $a,b,X\_n,w$). $G(n,x)$ is dealt with quite similarly. 


As for the law of large numbers (LLN), here are the details. Note that $Y\_n$ equals $T\_1+\dots+T\_n$ in distribution, where $T\_1,\dots,T\_n$ are independent identically distributed Bernoulli r.v.'s, with $ET\_i=p$ for all $i$. Hence, by the LLN, $Y\_n/n\to p$ (in distribution) as $n\to\infty$. It is also clear that $Z/n\to0$. Hence, by the Slutsky theorem, $X\_n\to(a-b)p$. 


**Addendum 2, on $\text{arg max}$:** Recall that any distribution function is monotonic. So, one can see that, for each $\delta\in(0,1)$, the mentioned convergences $P(X\_n<w)\to1$ and $P(X\_n>w-d)\to1$ are uniform in $w$ such that $|w-(\mu+d/2)|\le(1-\delta)d/2$. Similarly, for each real $\delta>0$, the convergence $P(X\_n<w)\to0$ is uniform in $w$ such that $w-(\mu+d/2)\le-(1+\delta)d/2$ and the convergence $P(X\_n>w-d)\to0$ is uniform in $w$ such that $w-(\mu+d/2)\ge(1+\delta)d/2$. It follows that, for each real $\delta>0$ and all large enough $n$, the set 
$\text{arg max}\_{w\in\mathbb{R}} [q P(X\_n<w)+(1-q)P(X\_n>w-d)]$ is contained in the interval $\{w\in\mathbb{R}\colon|w-(\mu+d/2)|<(1+\delta)d/2\}$. Substituting here $w=-b\_1-cx/n$, we conclude that, for each real $\delta>0$ and all large enough $n$, the set 
$$\text{arg max}\_{x\in\mathbb{R}} [qF(n,x)+(1-q)G(n,x)]
=\{x\in\mathbb{R}\colon qF(n,x)+(1-q)G(n,x)=m\_n\}$$ 
is contained in the interval $$\{x\in\mathbb{R}\colon|cx/n+b\_1+\mu+d/2|<(1+\delta)d/2\};$$ 
recall that 
$$m\_n:=\max\_{w\in\mathbb{R}} [q P(X\_n<w)+(1-q)P(X\_n>w-d)]=\max\_{x\in\mathbb{R}} [qF(n,x)+(1-q)G(n,x)].$$ 
Similarly, for each real $\varepsilon>0$, each $\delta\in(0,1)$, and all large enough $n$, the set $\{x\in\mathbb{R}\colon qF(n,x)+(1-q)G(n,x)>m\_n-\varepsilon\}$ of all $\varepsilon$-maximizers contains the set $\{x\in\mathbb{R}\colon qF(n,x)+(1-q)G(n,x)>1-\varepsilon\}$, which in turn contains the interval 
$$\{x\in\mathbb{R}\colon|cx/n+b\_1+\mu+d/2|\le(1-\delta)d/2\}. 
$$


