Asymptotic Approximation to Number of Knapsack Solutions Is there an asymptotic approximation to the fraction of sets satisfying a knapsack feasibility constraint?

More precisely, imagine I have a large number $n$ of items with bounded weights $X_1,...,X_n \in [0,1]$. For a given bound $Y$, I want to compute the fraction of sets $S \subset \{1,...,n\}$ which satisfy the feasibility constraint $\sum_{i \in S} X_i \leq Y$. That is, I want to compute:

$$N(X_1,...,X_n,Y) =  \frac{\# \{S: \sum_{i \in S} X_i \leq Y\}}{2^n}$$

I want to understand how the ``partial derivatives'' of $N$ behave when $X$ and $Y$ change. That is, I want to compute some analogue of $\frac{\partial N}{\partial X_i}$ or $\frac{\partial N}{\partial Y}$. As defined, the function $N$'s partial derivatives are not very interesting: the function is constant, except at regions of discontinuity where the number of feasible sets changes, and at those points the derivatives do not exist.

More concretely, I want to know if there's a function $F(X_1,...,X_n,Y,n)$ such that 


F is differentiable in $X_1,...,X_n,Y$
$\lim_{n \to \infty} \frac{F(X_1,...,X_n,Y, n)}{N(X_1,...,X_n,Y)} = 1$


I know that, when $X_1,...,X_n = 1$, then $N(X_1,...,X_n,Y)=\sum_{k=0}^{\lfloor Y \rfloor} \frac{\binom{n}{k}}{2^n}$, which converges to the normal CDF $\Phi(Y; \frac{n}{2}, \frac{n}{4})$. But I don't know if this kind of approximation generalizes to arbitrary bounded $X_i$.
The answer is no, assuming we interpret Condition 2 as follows:


For every infinite sequence $X_1,X_2,\ldots$ with each $X_i$ in $[0,1]$,
$$\lim_{n \to \infty} \frac{F(X_1,...,X_n,Y, n)}{N(X_1,...,X_n,Y,n)} = 1.$$




Lemma 1. No function $F$ that satisfies the above condition is continuous (much less differentiable) in every $X_i$.

Proof. Fix any candidate function $F$ that is continuous in each $X_i$.  Fix $Y=1$. To complete the proof, we show there must be a sequence $X_1', X_2', \ldots$ such that
$$\lim\inf_{n\rightarrow\infty} \frac{F(X_1',\ldots,X_n',1,n)}{N(X_1',\ldots,X_n',1,n)} = 0$$ 
or
$$\lim\sup_{n\rightarrow\infty} \frac{F(X_1',\ldots,X_n',1,n)}{N(X_1',\ldots,X_n',1,n)} = \infty.$$

Start by fixing an arbitrarily large index $n_1$.  To ease notation, for $z\in[0,1]$, define
$$\begin{align}
 F_1(z) &amp;{} = F(z, z, \ldots, z, 1, n_1), \text{ and } \\
 N_1(z) &amp; {}= N(z, z, \ldots, z, 1, n_1).
\end{align}$$
(That is, the restrictions of $F$ and $N$ to $X_1=X_2=\cdots=X_{n_1}=z$, $Y=1$, and $n=n_1$.)

Fix $\epsilon_1&gt;0$ such that 
$$F_1(1/2) - F_1(1/2+\epsilon_1) \le 1/2^{n_1}.$$
(This is possible because $F_1$ is continuous, as $F$ is continuous in each $X_i$.)

Meanwhile, by inspection
$$
\begin{aligned}
N_1(1/2+\epsilon_1) &amp;{} = (1 + n_1)/2^{n_1} &amp;&amp;{} = \Theta(n_1/2^{n_1}), &amp; \text{ while} \\
N_1(1/2) &amp;{} = \Big(1 + n_1 + {n_1 \choose 2}\Big)/2^{n_1} &amp;&amp;{} = \Theta(n_1^2/2^{n_1}).
\end{aligned}
$$

Case 1.1. First consider the case that $F_1(1/2+\epsilon_1) \le n_1^{3/2}/2^{n_1}$.

In this case, define $X'_1 = X'_2 = \cdots = X'_{n_1} = 1/2$.
Then
$$\frac{F(X'_1, \ldots, X'_{n_1}, 1, n_1)}{N(X'_1,\ldots, X'_{n_1}, 1, n_1)} = 
\frac{F_1(1/2)}{N_1(1/2)} 
\le 
\frac{F_1(1/2+\epsilon_1)+1/2^{n_1}}{N_1(1/2)} 
= O(1/\sqrt{n_1}).$$

Case 1.2. Otherwise $F_1(1/2+\epsilon_1) &gt; n_1^{3/2}/2^{n_1}$.

In this case, define $X'_1 = X'_2 = \cdots = X'_{n_1} = 1/2 + \epsilon_1$.
Then $$\frac{F(X'_1, \ldots, X'_{n_1}, 1, n_1)}{N(X'_1,\ldots, X'_{n_1}, 1, n_1)} = \frac{F_1(1/2+\epsilon_1)}{N_1(1/2+\epsilon_1)} = \Omega(\sqrt{n_1}).$$

So, at index $n=n_1$, the ratio in question is either $O(1/\sqrt n)$ or $\Omega(\sqrt n)$.



To continue, choose index $n_2 \ge n_1^2$,
and set the values of $X'_{n_1+1}, X'_{n_1+2}, \ldots, X'_{n_2}$ as follows. Define
$$\begin{align}
 F_2(z) &amp;{} = F(X'_1, X'_2, \ldots, X'_{n_1}, z, z, \ldots, z, 1, n_2), \text{ and } \\
 N_2(z) &amp; {}= N(X'_1, X'_2, \ldots, X'_{n_1}, z, z, \ldots, z, 1, n_2).
\end{align}$$
(Above, $z$ occurs $n_2-n_1$ times in each parameter list.)
Fix $\epsilon_2&gt;0$ such that 
$$F_2(1/2) - F_2(1/2+\epsilon_2) \le 1/2^{n_2}.$$
(This is possible because $F_2$ is continuous, as $F$ is continuous in each $X_i$.)

By inspection, using that $n_2 \ge n_1^2$,
$$
\begin{aligned}
N_2(1/2+\epsilon_2) &amp;{} = O((n_1^2 + n_2)/2^{n_2}) &amp;&amp;{} = O(n_2/2^{n_2}), &amp; \text{ while} \\
N_2(1/2) &amp;{} \ge {n_2 \choose 2}/2^{n_2} &amp;&amp;{} = \Theta(n_2^2/2^{n_2}).
\end{aligned}
$$

Case 2.1. First consider the case that $F_2(1/2+\epsilon_2) \le n_2^{3/2}/2^{n_2}$.

In this case, define $X'_{n_1+1} = X'_{n_1+2} = \cdots = X'_{n_2} = 1/2$.
Then
$$\frac{F(X'_1, \ldots, X'_{n_2}, 1, n_2)}{N(X'_1,\ldots, X'_{n_2}, 1, n_2)} = 
\frac{F_2(1/2)}{F_1(1/2)} 
\le 
\frac{F_2(1/2+\epsilon_2)+1/2^{n_2}}{N_2(1/2)} 
= O(1/\sqrt {n_2}).$$

Case 2.2. Otherwise $F_2(1/2+\epsilon_2) &gt; n_2^{3/2}$.

In this case, define $X'_{n_1+1} = X'_{n_1+2} = \cdots = X'_{n_2} = 1/2 + \epsilon_2$.
Then $$\frac{F(X'_1, \ldots, X'_{n_2}, 1, n_2)}{N(X'_1,\ldots, X'_{n_2}, 1, n_2)} = \frac{F_2(1/2+\epsilon_2)}{N_2(1/2+\epsilon_2)} = \Omega(\sqrt{n_2}).$$

So, at index $n=n_2$, the ratio in question is either $O(1/\sqrt n)$ or $\Omega(\sqrt n)$.



Continuing in this way, we can set $X'_1, X'_2, \ldots$ so that there is an infinite sequence of indices $1\le n_1 &lt; n_2 &lt; n_3 &lt; \cdots$
such that, at each index $n=n_i$, the ratio in question is either $O(1/\sqrt n)$ or $\Omega(\sqrt n)$.

Then either there are infinitely many indices $n$ such that the ratio is $O(1/\sqrt n)$, in which case
$$\lim\inf_{n\rightarrow\infty} \frac{F(X_1',\ldots,X_n',1,n)}{N(X_1',\ldots,X_n',1, n)} = 0,$$ 
or there are infinitely many indices $n$ such that the ratio is $\Omega(\sqrt n)$, in which case
$$\lim\sup_{n\rightarrow\infty} \frac{F(X_1',\ldots,X_n',1,n)}{N(X_1',\ldots,X_n',1, n)} = \infty.$$
$~\Box$