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418531 ภาคต้น 2552/โจทย์ปัญหาความน่าจะเป็น II/เฉลยข้อ 7 - ประวัติรุ่นแก้ไข
2026-04-11T18:30:56Z
ประวัติรุ่นแก้ไขของหน้านี้ในวิกิ
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Cardcaptor: หน้าที่ถูกสร้างด้วย 'สำหรับ <math>i = 1, 2, \ldots, n\,</math> ให้ <math>X_i = |a_i - i|\,</math> จาก law of total expectation เร…'
2009-08-03T09:22:43Z
<p>หน้าที่ถูกสร้างด้วย 'สำหรับ <math>i = 1, 2, \ldots, n\,</math> ให้ <math>X_i = |a_i - i|\,</math> จาก law of total expectation เร…'</p>
<p><b>หน้าใหม่</b></p><div>สำหรับ <math>i = 1, 2, \ldots, n\,</math> ให้ <math>X_i = |a_i - i|\,</math> จาก law of total expectation เราได้ว่า<br />
<table cellpadding="5"><br />
<tr><br />
<td align="right"><math>E[X_i] \,</math></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\Pr(a_i < i)E[X_i\ |\ a_i < i] + \Pr(a_i = i) E[X_i\ |\ a_i = i] + \Pr(a_i > i) E[X_i\ |\ a_i > i] \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\Pr(a_i < i)E[i - a_i \ |\ a_i < i] + \Pr(a_i = i) \cdot 0 + \Pr(a_i > i) E[a_i - i\ |\ a_i > i] \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\frac{i-1}{n} E[i - a_i \ |\ a_i < i] + \frac{n-i}{n} E[a_i - i\ |\ a_i > i] \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\frac{i-1}{n} \big( i - E[a_i\ |\ a_i < i] \big) + \frac{n-i}{n}\big( E[a_i \ |\ a_i > i] - i \big) \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\frac{i-1}{n} \bigg( i - \frac{i}{2} \bigg) + \frac{n-i}{n}\bigg( \frac{n+i+1}{2} - i \bigg) \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\frac{i-1}{n} \cdot \frac{i}{2} + \frac{n-i}{n} \cdot \frac{n-i+1}{2} \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\frac{1}{2n} [(i-1)i + (n-i)(n-i+1)] \,</math></td><br />
</tr><br />
</table><br />
<br />
ฉะนั้น<br />
<table cellpadding="5"><br />
<tr><br />
<td align="right"><math>E\bigg[ \sum_{i=1}^{n} |a_i - i| \bigg] \,</math></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\sum_{i=1}^{n} E[ X_i ] \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\sum_{i=1}^{n} \frac{1}{2n} [(i-1)i + (n-i)(n-i+1)] \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\frac{1}{2n} \bigg( \sum_{i=1}^n (i-1)i + \sum_{i=1}^n (n-i)(n-i+1) \bigg)\,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\frac{1}{2n} \bigg( \sum_{i=1}^n (i-1)i + \sum_{i=1}^n (i-1)i \bigg)\,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\frac{1}{n} \sum_{i=1}^n (i-1)i \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\frac{1}{n} \bigg( \sum_{i=1}^n i^2 - \sum_{i=1}^n i \bigg) \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\frac{1}{n} \bigg( \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} \bigg) \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>(n+1) \bigg( \frac{2n+1}{6} - \frac{1}{2} \bigg) \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>(n+1) \frac{2n-2}{6} \,</math></td><br />
</tr><br />
<tr><br />
<td align="right"></td><br />
<td align="center"><math>= \,</math></td><br />
<td align="left"><math>\frac{n^2 - 1}{3} \,</math></td><br />
</tr><br />
</table></div>
Cardcaptor