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

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The discussion focuses on determining the conditions under which the expected value of max{n-q, q-1} equals 3n/4, with n being a fixed number and q ranging from 0 to n. It establishes that for odd n, the expression for max{n-q, q-1} can be simplified based on the value of q. The expected value M is calculated using a summation approach, leading to the formula M = (n(3n - 1))/(4(n + 1)). The discussion emphasizes the relationship between the values of n and q in achieving the desired expectancy. The mathematical exploration seeks to clarify the conditions necessary for this equality to hold.
evinda
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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)
 
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evinda said:
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

$\chi$ $\sigma$
 
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