Due on Sep 16 (20 points)

1. In line with the topic discussed in class, consider a stick that is one foot long. If this stick is randomly broken into three separate pieces, what is the expected length of the longest piece among them? Please refrain from merely guessing the answer; instead, include a detailed explanation of the steps you take to arrive at your conclusion.

2. (Markov chain) Recall that for five random variables, say \(X_1,X_2,X_3,X_4,X_5\), forms a Markov chain \(X_1 \rightarrow X_2 \rightarrow X_3 \rightarrow X_4 \rightarrow X_5 \) if \(X_5 \bot X^3 |X_4\) (i.e., \(p(x_5|x_4,x^3)=p(x_5|x_4))\), \(X_4 \bot X^2 | X_3\), and \(X_3 \bot X_1|X_2\). We will show in the following that the above conditional independence conditions can be summarized and are equivalent to the joint distribution equation \(p(x^5)=p(x_1) p(x_2|x_1) p(x_3|x_2) p(x_4|x_3) p(x_5|x_4)\).
   Remark: This proof can be generalized to a chain of any length using mathematical induction.

   1. (From Conditional Independence to Joint Distribution) Show that the conditional independence conditions lead to the joint distribution equation (Hint: add one \(x\) at a time. That is, show \(p(x^3) = p(x_1) p(x_2|x_1) p(x_3|x_2)\) with the conditional independence conditions, and then \(p(x^4) = p(x_1) p(x_2|x_1) p(x_3|x_2) p(x_4|x_3)\), and so on.)

   2. (From Joint Distribution to Conditional Independence) Show that the joint distribution equation leads to the conditional independence conditions (Hint: use marginalization)

   3. (Symmetry in the Markov Chain) Show that \(p(x^5) = p(x_1) p(x_2|x_1) p(x_3|x_2) p(x_4|x_3) p(x_5|x_4)\) leads to \(p(x^5)=p(x_5) p(x_4|x_5) p(x_3|x_4) p(x_2|x_3) p(x_1|x _2)\). Consequently, the Markov chain definition is symmetric. That is, \(X_1 \rightarrow X_2 \rightarrow X_3 \rightarrow X_4 \rightarrow X_5 \equiv X_5 \rightarrow X_4 \rightarrow X_3 \rightarrow X_2 \rightarrow X_1\). (Hint: if you are stuck, try to show each of the conditional independence conditions are satisfied.)