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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.

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

(b) Find $\text{Cov} (X_k,X_5)$ for $1 \le k \le 10$.

Trial:I think $X_k\sim \text{Bin}(n=100,p=1/5)$ and $E(X_k)=np=20,V(X_k)=npq=16$.Am I right? Lastly how I find $\text{Cov}(X_k,X_5)$.

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The general situation you have here is described by a multinomial distribution. The Wikipedia article about the multinomial distribution explains:

For n independent trials each of which leads to a success for exactly one of k categories, with each category having a given fixed success probability [...]

In your case, two balls are put in the boxes in each trial. You have already recognized that this leads to a success probability of 1/5 for each of the boxes. So your answer for (a) is correct.

In order to answer question (b), recall the formula for the covariance for $i\neq j$:

$$ Cov[X_{i}, X_{j}] = E[X_{i}\cdot X_{j}] - E[X_{i}]\cdot E[X_{j}] $$

If only one ball was put in a box in each trials, we could set $E[X_{i}\cdot X_{j}]=0$ because $X_{i}$ and $X_{j}$ could not be 1 (i.e. receive a ball) in the same trial. It is easily seen that this assumption is true whenever $j\neq i \pm 1 \text{ mod } 10$. For these cases, the covariance is $-np_{i}p_{j} = -100\cdot (1/5)\cdot (1/5) = -4$.

For the case that $j=i \pm 1 \text{ mod } 10$, the term $E[X_{i}\cdot X_{j}]\neq 0$ because both boxes receive a ball at the same trial. From the Wikipedia article about the binomial distribution we get

$$ Cov[X_{i}, X_{j}] = n\cdot (p_{b}-p_{i}p_{j}) $$

where $p_{b}$ denotes the probability that both boxes receive a ball. In this case, $p_{b}=1/10$. So finally, whenever $j=i \pm 1 \text{ mod } 10$ the covariance is $100\cdot (\frac{1}{10} - \frac{1}{5}\cdot \frac{1}{5})=6$.

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    $\begingroup$ By "$j = \pm 2i$" do you perhaps mean $j = i \pm 1 \mod 10$? $\endgroup$
    – whuber
    Apr 14 '13 at 19:05
  • $\begingroup$ Dear whuber. Exactly! I wasn't rigorous enough. Thanks for clarifying. $\endgroup$ Apr 14 '13 at 21:16
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This is my first post on StackExchange, and I hope it is helpful. COOLSerdash gave a very complete answer already which I mostly agree with, but I will try to add a different perspective on the problem.

As you stated, $X_{k} \sim Binomial(n=100,p=\frac{1}{5}) \equiv 100 \times Bernoulli(\frac{1}{5})$

$\text{Cov}(X_5,X_k) = \mathbb{E}[(X_5-\mathbb{E}[X_5])(X_k-\mathbb{E}[X_k])]$

This leads to the expression $\mathbb{E}[X_5X_k]-\mathbb{E}[X_5]\mathbb{E}[X_k] = \mathbb{E}[X_5X_k]-100\times(\mathbb{E}(Bernoulli(\frac{1}{5}))$ $= \mathbb{E}[X_5X_k]-100\times(\frac{1}{25})=\mathbb{E}[X_5X_k]-4$.

Therefore, we have three conditions:

$k = 5 ; Cov(X_5,X_5) = Var(X_5) = Var(X_k) = 16$

$k \neq 3,7; Cov(X_5,X_k) = \mathbb{E}[X_5X_k]-4 = 0$

Consider the events $A := ${box $5$ gets a ball} , $B$ {box $k$ gets a ball} $\neq 3,7$, then $ \mathbb{P}(A,B) = \mathbb{P}(A)\mathbb{P}(B)$ due to independence.

$\therefore \mathbb{E}[X_5X_k]=100\times \mathbb{E}[A]\mathbb{E}[B]=100\times(\frac{1}{5}\times \frac{1}{5}) = 4$

$k = 3,7; Cov(X_5,X_k) = \mathbb{E}[X_5X_k]-4 = 6$ (as COOLSerdash stated)

With the defined events $A$ and $B$ for $k = 3,7$, $ \mathbb{P}(A,B) = \frac{1}{10}$ since the probability $A \cap B$ is the probability that box 4 (if $k=3$) or box 6 (if $k=7$) gets a ball.

Hope this helps!

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