1991 Paper 2 Q15

Year: 1991
Paper: 2
Question Number: 15

Course: LFM Stats And Pure
Section: Uniform Distribution

Difficulty: 1600.0 Banger: 1484.0

Problem

Integers \(n_{1},n_{2},\ldots,n_{r}\) (possibly the same) are chosen independently at random from the integers \(1,2,3,\ldots,m\). Show that the probability that \(\left|n_{1}-n_{2}\right|=k\), where \(1\leqslant k\leqslant m-1\), is \(2(m-k)/m^{2}\) and show that the expectation of \(\left|n_{1}-n_{2}\right|\) is \((m^{2}-1)/(3m)\). Verify, for the case \(m=2\), the result that the expection of \(\left|n_{1}-n_{2}\right|+\left|n_{2}-n_{3}\right|\) is \(2(m^{2}-1)/(3m).\) Write down the expectation, for general \(m\), of \[ \left|n_{1}-n_{2}\right|+\left|n_{2}-n_{3}\right|+\cdots+\left|n_{r-1}-n_{r}\right|. \] Desks in an examination hall are placed a distance \(d\) apart in straight lines. Each invigilator looks after one line of \(m\) desks. When called by a candidate, the invigilator walks to that candidate's desk, and stays there until called again. He or she is equally likely to be called by any of the \(m\) candidates in the line but candidates never call simultaneously or while the invigilator is attending to another call. At the beginning of the examination the invigilator stands by the first desk. Show that the expected distance walked by the invigilator in dealing with \(N+1\) calls is \[ \frac{d(m-1)}{6m}[2N(m+1)+3m]. \]

No solution available for this problem.

Rating Information

Difficulty Rating: 1600.0

Difficulty Comparisons: 0

Banger Rating: 1484.0

Banger Comparisons: 1

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Problem source
Integers $n_{1},n_{2},\ldots,n_{r}$ (possibly the same) are chosen
independently at random from the integers $1,2,3,\ldots,m$. Show
that the probability that $\left|n_{1}-n_{2}\right|=k$, where $1\leqslant k\leqslant m-1$,
is $2(m-k)/m^{2}$ and show that the expectation of $\left|n_{1}-n_{2}\right|$
is $(m^{2}-1)/(3m)$. Verify, for the case $m=2$, the result that
the expection of $\left|n_{1}-n_{2}\right|+\left|n_{2}-n_{3}\right|$
is $2(m^{2}-1)/(3m).$ Write down the expectation, for general $m$,
of 
\[
\left|n_{1}-n_{2}\right|+\left|n_{2}-n_{3}\right|+\cdots+\left|n_{r-1}-n_{r}\right|.
\]
Desks in an examination hall are placed a distance $d$ apart in straight
lines. Each invigilator looks after one line of $m$ desks. When called
by a candidate, the invigilator walks to that candidate's desk, and
stays there until called again. He or she is equally likely to be
called by any of the $m$ candidates in the line but candidates never
call simultaneously or while the invigilator is attending to another
call. At the beginning of the examination the invigilator stands by
the first desk. Show that the expected distance walked by the invigilator
in dealing with $N+1$ calls is 
\[
\frac{d(m-1)}{6m}[2N(m+1)+3m].
\]