$N$ men are sitting at the table. Each of them tosses the dice. Let $A$ be the random variable representing number of people, who got rolled number that is equal to both neighbour's . Find $\mathbb{E}X$ and $\mathbb{D}X$. ($\mathbb{D}$ - stands for dispersion/variance here).
I managed to calculate $\mathbb{E}X$, still haven't found $\mathbb{D}X$. May you help me with $\mathbb{D}X$?
I've been thinking about using somehow the following property:
It is known that $\mathbb{D}X=\mathbb{E}X^2-(\mathbb{E}X)^2$ While knowing what $\mathbb{E}X$ is equal to we may find $(\mathbb{E}X)^2$ easily, in order to find $\mathbb{E}X^2$ I want to use following property:
for $B= \varphi (A): \mathbb{E} B= \sum_{k} \varphi (x_k) \cdot P(ξ = x_k) $
You can calculate $\ \mathbb{E}\left(X^2\right)\ $ by using the same indicators $\ A_i\ $ you used to find $\ \mathbb{E}\left(X\right)\ $. If $\ j\ge i+3\ $, then $\ A_i\ $ and $\ A_j\ $ are independent, so $\ \mathbb{E}\left(A_iA_j\right)= \mathbb{E}\left(A_i\right) \mathbb{E}\left(A_j\right)=\left(\frac{11}{36}\right)^2\ $, and \begin{align} X^2&=\left(\sum_{i=1}^NA_i\right)^2\\ &= \sum_{i=1}^N \sum_{j=1}^N A_iA_j\\ &= \sum_{i=1}^NA_i +2 \sum_{i=1}^{N-1}\sum_{j=i+1}^NA_iA_j\\ &= \sum_{i=1}^NA_i+2\sum_{i=1}^{N-2}A_i\left(A_{i+1}+A_{i+2} \right) + 2\sum_{i=1}^{N-3}\sum_{j=i+3}^NA_iA_j+2A_{N-1}A_N\ . \end{align} So \begin{align} \mathbb{E}\left(X^2\right)&=(N-2) \mathbb{E}\left(A_i\right)+(N-3)(N-2) \mathbb{E}\left(A_i\right)^2\\ &\hspace{2em}+(2N-3)\mathbb{E}\left(A_iA_{i+1}\right)+ 2(N-2)\mathbb{E}\left(A_iA_{i+2}\right)\ . \end{align} Can you calculate $\ \mathbb{E}\left(A_iA_{i+1}\right)\ $ and $\ \mathbb{E}\left(A_iA_{i+2}\right)\ $?