There are $10$ empty boxes numbered $1, 2, \dots , 10$ placed sequentially on the circumference of a circle.We perform $100$ independent trials. At each trial, one box is selected with probability $\dfrac{1}{10}$ and one ball is placed in each of the two neighbouring boxes of the selected one.

Define $X_k$ to be the number of balls in the $k$-th box at the end of $100$ trials.

Find $E[X_k]$ for $1 \le k \le 10$.

Possibly, this is a repost..But I have a another take on this problem. Suppose, I define the event $E_k:$ as the $k^{\text{th}}$ box is chosen in a particular step. So, $P(E_k)=\frac{1}{10}$ and define $Y_{k,m}: $the number of balls in the $k^{\text{th}}$ box after $m$ trials.

So, here $X_k=Y_{k,100}$ Now, $P(Y_{k,m}=n)=P(Y_{k,m}=n-1;E_{k-1})P(E_{k-1})+P(Y_{k,m}=n-1;E_{k+1})P(E_{k+1})$ So, we derive at the recursion $P(Y_{k,m}=n)=\frac{1}{5}[P(Y_{k,m}=n-1)]$ Iterating $n-2$ times, we have $P(Y_{k,m}=n)=(\frac{1}{5})^{n-1}[P(Y_{k,m}=1)] =(\frac{1}{5})^{n-1}[P(E_{k+1} \cup E_{k-1})]=(\frac{1}{5})^{n}$

Now, this is independent of $k,m$ So, can I say from here that $X_k \sim \text{Binomal}(100,\frac{1}{5})?$


You are correct in stating that $X_k\sim\mathsf{Binom}(100,\frac15)$.

You can write $X_k=\sum_{i=1}^{100} X_{k,i}$ where $X_{k,i}$ takes value $1$ if a ball is placed in the $k$-th box at trial $i$. The $X_{k,i}$ are iid with Bernoulli distribution with parameter $\frac15$.

But to find $\mathsf EX_k$ you actually do not need to know the distribution of $X_k$.

By every trial there are $2$ balls placed so after $100$ trials there are $200$ balls present so that:$$X_1+\cdots+X_{10}=200$$

Then by linearity of independence we get:$$\mathsf EX_1+\cdots+\mathsf EX_{10}=200$$

Then by symmetry we find that for a fixed $k\in\{1,\dots,10\}$ we have: $$10\mathsf EX_k=200\text{ and consequently }\mathsf EX_k=20$$


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