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There are n distinct points marked on the ring, each of which is either blue or red with equal probabilities independently of each other. These n points divide the ring into n arcs. If an arc has both endpoints red, the arc is also considered red. If N is the number of red arcs, compute E[N] and var(N).

I do not understand where to start this problem, I understand one will need to use indicators but how would one go about this? Also, The condition that if both end points are red then the arc is, is a bit confusing, how will this be utilized in solving the problem?

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Explanation of problem: There are $n$ arcs. They are all "short," just joining two consecutive points on the circle. We colour an arc red iff both of its endpoints are red. We are only interested in arcs, the colours of the points are irrelevant except insofar as they determine whether an arc will be coloured red or not. Or to put it another way, we want to count the situations in which we have two consecutive red points.

So for example the pattern BRRRB gives us $2$ red arcs. For further information about how indicator functions can be used, please read the first four lines only of the solution outline.

Outline of solution: Call the points $P_1,P_2,\dots,P_n$, with the understanding that $P_{n+1}=P_1$.
Suppose they are arranged counterclockwise in that order around the circle.

Let random variable $X_i$ be $1$ if $P_i$ is red and $P_{i+1}$ are red. Otherwise, let $X_i=0$.

Then the number of $N$ of red arcs is $X_1+X_2+\cdots+X_n$.

The expectation of this sum is the sum of the expectations. It should be easy for you to find $\Pr(X_i=1)$, and hence $E(X_i)$.

The variance is more complicated. We use the fact that $\text{Var}(N)=E(N^2)-(E(N))^2$.

So we need $E(N^2)$. Expand $(X_1+X_2+\cdots +X_n)^2$. By the linearity of expectation, we need $E(X_i^2)$ and $E(X_iX_j)$. The first is no problem, since $X_i^2=X_i$. For the second, there is a need to distinguish between the cases where $P_i$ and $P_j$ are neighbours, and the cases where they are not.

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@Olivia, For Variance, you may expand it using covariance, and note that $Cov (V_i V_j) = 0 $ if $j\neq i \pm 1$, and calculate it otherwise. – Calvin Lin Feb 7 '13 at 2:43
the direct expansion isn't hard. $Cov (V_i V_{i+1} ) = E[V_i V_{i+1}] - E[V_i]E[V_{i+1}] = \frac {1}{8} - \frac {1}{16} = \frac {1}{16}$. So the variance is $\sum Var (V_i) - 2\sum Cov(V_i V_j) = \frac {3n}{16} - \frac {2n}{16} = \frac {n}{16}$. – Calvin Lin Feb 7 '13 at 2:45
@CalvinLin: The procedure I suggested for variance uses only very basic machinery. Yours is smoother if the OP has some background. – André Nicolas Feb 7 '13 at 2:50
The expectation of this sum is the sum of the expectations. It should be easy for you to find Pr(Xi=1), and hence E(Xi). This phrase,while helpful, does not help me, I am confused as to how to find out the probability, they give me I have n arcs each equally likely to be blue or red, so 1/2 chance blue or red right? The confusion is what the whole 2 ends being red condition comes in. otherwise I understand what you had up til there. Also can you be a little more explicit. I know this is quite trivial but I need to understand the reasoning behind the approach. – Olivia Irving Feb 7 '13 at 2:56
@AndréNicolas I thought that using of Indicator Variables would indicate knowledge of Expectation / Variance formulas, but apparently no necessarily. – Calvin Lin Feb 7 '13 at 3:03

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