, 2023

Solution:

Our job is to arrange the nine numbers in a way that maximizes the spikiness. Let an element be a peak if it is higher than its neighbor(s) and a valley if it is lower than its neighbor(s). It is not hard to show that an optimal arrangement has every element either a peak or a valley (if you have some number that is neither, just move it to the end to increase spikiness). Since $9$ is odd, there are two possibilities: the end points are either both peaks or both valleys.

Sort the numbers from least to greatest: $x_{1}, \ldots, x_{9}$. If we arrange them in such a way that it starts and ends with peaks, the factor of $x_{i}$ added to the final result will be $[-2,-2,-2,-2,1,1,2,2,2]$, respectively. If we choose the other way (starting and ending with valleys), we get $[-2,-2,-2,-1,-1,2,2,2,2]$. Notice that both cases have a base value of $[-2,-2,-2,-1,0,1,2,2,2]$, but then we add on $\max \left(x_{6}-x_{5}, x_{5}-x_{4}\right)$.

Since the expected value of $x_{i}$ is $\frac{i}{10}$, our answer is $-\frac{2}{10}(1+2+3)-\frac{4}{10}+\frac{6}{10}+\frac{2}{10}(7+8+9)+\mathbb{E}\left(\max \left(x_{6}-x_{5}, x_{5}-x_{4}\right)\right)$. This last term actually has value $\frac{3}{4} \mathbb{E}\left(x_{6}-x_{4}\right)=\frac{3}{4} \cdot \frac{2}{10}$. This is because if we fix all values except $x_{5}$, then $x_{5}$ is uniformly distributed in $\left[x_{4}, x_{6}\right]$. Geometric probability tells us that the distance from $x_{5}$ to its farthest neighbor is $\frac{3}{4}$ to total distance between its two neighbors $\left(x_{6}-x_{4}\right)$. We add this all up to get $\frac{79}{20}$.