Let $X_1, \cdots, X_N$ be a random sample (that is, it is i.i.d) from a population with density $f(x)$. Consider the following uniform kernel density estimate of $f(x)$: $$\widehat{f}(x) = \frac{1}{2Nh} \sum_{i=1}^{N} 1(x-h\le X_i \le x+h)$$ Show that $2Nh\widehat{f}(x) \sim \text{Binomial}(N, p_h(x))$ for $p_h(x) = h \int 1(-1 \le u \le 1) f(x+uh) du$.
What I've done so far is this: Rearrange so we get $$2Nh\widehat{f}(x) = \sum_{i=1}^{N} 1(x-h\le X_i \le x+h)$$
Now clearly $1(x-h\le X_i \le x+h)$ is a Bernoulli random variable with parameter $p = \int_{x-h}^{x+h} f(u) du$. Since the $X_i$'s are i.i.d, the sum of $N$ independent Bernoulli random variables is Binomial, so $$2Nh\widehat{f}(x) \sim \text{Binomial}(N, p)$$ But, how can I show that $p = p_h(x)$?
It's a change of variables.
\begin{align} \int_{x-h}^{x+h} f(t) \mathop{dt} =h\int_{-1}^1 f(x+uh) \mathop{du} \end{align} by the substitution $u = \frac{t-x}{h}$.