A die is thrown $n+2$ times. After each throw a '$+$' is recorded for $4$, $5$, or $6$ and '$-$' for $1$,$2$, or $3$, the signs forming an ordered sequence. To each, except the first and the last sign, is attached a characteristic random variable which takes the value $1$ if both the neighboring signs differ from the one between them and $0$ otherwise. If $X_1, X_2, \ldots , X_n$ are the characteristic random variables, find the mean and variance of $X = \sum_{i=1}^n X_i$

Problematic part:

$$ V(X) = V(X_1+X_2+\cdots+X_n) = \sum_{i=1}^n V(X_i) +2\sum_{i=1}^n\sum_{j=1, j>i}^n Cov(X_i,X_j) $$

Calculating the variance of $X$ requires calculating the covariance of two arbitrarily chosen random variables $X_i$ and $X_j$.

The formula then used is $Cov(X_i,X_j) = E(X_iX_j) - E(X_i)E(X_j)$

Which brings us to the essence of my problem -- How to find $E(X_iX_j)$?

It is certain that $X_iX_j$ can take only two values, namely, $0$ and $1$. Therefore $E(X_iX_j) = 1.P(X_iX_j=1) + 0.P(X_iX_j=0)= P(X_iX_j=1)$

But at this point I'm not sure how to compute the probability. The book I'm using says the probability is $1/8$, but I can't seem to wrap my head around the reasoning. An intuitive explanation would be highly appreciated!


Assuming the die is thrown independently, $X_i$ and $X_j$ are independent if $j \geq i + 3$. So in that case the convariance is $0$. So you only have to calculate $E[X_iX_{i+1}] = E[X_1X_2]$ and $E[X_iX_{i+2}]=E[X_1X_3]$ which should be easy by just looking at all the possible outcomes.

  • $\begingroup$ Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion. $\endgroup$ – s0ulr3aper07 Nov 29 '18 at 9:30

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