Suppose we have $X_1, \dots, X_n$ independent random variables with distribution $N(0, 1)$. The density function is:
\begin{equation} \phi(x) = \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^2}{2}} \end{equation}
Suppose we have a fixed value $x > 0$. I want to compute the expected value of: \begin{equation} Y(x) = \min_{i \in [1, n]\; \phi(X_i) > \phi(x)} \left| x - X_i \right| \end{equation} which represents the distance to the nearest neighbor of higher density. $Y$ is a non-negative random variable. For any $t \in [0, 2x]$, the probability that $Y$ exceeds $t$ equals the probability that any $X_i$, such that $\phi(X_i) > \phi(x)$, lies inside $\left[-x, +x-t \right]$, hence: \begin{align} \mathbb{P}(Y(x) < t) &= 1 - \prod_{i \in [1, n]\; \phi(X_i) > \phi(x)}\mathbb{P}(X_i \in [-x, +x-t] | X_i \in [-x, +x])\\ &= 1 - \left( \frac{1}{A} \int_{-x}^{+x-t} \phi(u)\, du \right)^{nA} \end{align} where $A = \int_{-x}^{+x} \phi(u)\,du$. The previous line gives the CDF of $Y(x)$. By differentiating this expression with respect to $t$ with the help of the fundamental theorem of calculus, I am able to recover the PDF of $Y(x)$: \begin{equation} f_Y(t)=n\cdot\phi(x-t)\cdot\left(\frac{1}{A}\int_{-x}^{+x-t}\phi(u)\,du\right)^{nA-1}\cdot1_{t\in[0, 2x]} \end{equation} Now I want to compute the mean of this PDF, $\mu_Y(x) = \int_0^{2x}tf_Y(t)\,dt$, and ideally find a relation between $\mu_Y(x)$ and $\phi(x)$. However I don't know how to integrate this PDF because of the term which looks like the Gauss error function.
How can I estimate this mean?
Thanks.