What Conditions Make the Expectancy of Max{n-q, q-1} Equal to 3n/4?

[evinda]

Hello! (Wave)

When is the expectancy of $max \{n-q,q-1\}$ , where $n$ is a fixed number and $q$ is in $[0,n]$ , so $\max\{n-q,q-1 \}$ is in $[\frac{n}{2},n]$, equal to $\frac{\frac{n}{2}+n}{2}=\frac{3n}{4}$? (Thinking)

 

[chisigma]

Hello! (Wave)

When is the expectancy of $max \{n-q,q-1\}$ , where $n$ is a fixed number and $q$ is in $[0,n]$ , so $\max\{n-q,q-1 \}$ is in $[\frac{n}{2},n]$, equal to $\frac{\frac{n}{2}+n}{2}=\frac{3n}{4}$? (Thinking)

Let suppose n odd [the case n even is quite similar...], then...

$\displaystyle \max \{n - q, q - 1\} =\begin{cases}n - q &\text{if}\ q \le \frac{n-1}{2}\\ q - 1 &\text{if}\ q\ge \frac{n+1}{2}\end{cases}\ (1)$

If we call M the expected value of $\max \{n - q, q - 1\}$ then is...

$\displaystyle M = \frac{1}{n+1}\ \{\sum_{i=0}^{\frac{n-1}{2}} (n-i) + \sum_{i =\frac{n+1}{2}}^{n} (i-1)\} = \frac{n\ (3\ n - 1)}{4\ (n + 1)}\ (2)$

Kind regards

$\chi$ $\sigma$