HW1 (Due 9/18)
Compute the mean and variance of \(X\) when
\(X\) is continuous and uniformly distributed between 0 and 2. That is, \(p_X(x)=\begin{cases}0.5, &\mbox{if } 0\le x\le 2\\0, & \mbox{otherwise}\end{cases}\)
\(X\) is discrete and \(Pr(X=1)=0.4, Pr(X=2)=0.2, Pr(X=3)=0.1, Pr(X=4)=0.3\)
(Markov chain) Let's 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., \(p(x_5x_4,x^3)=p(x_5x_4))\), \(X_4 \bot X^2  X_3\), and \(X_3 \bot X_1X_2\). We will show in the following that the above conditional independence conditions can be summarized and are equivalent to
\(p(x^5)=p(x_1) p(x_2x_1) p(x_3x_2) p(x_4x_3) p(x_5x_4)\). (Remark: note that the proof below can be generalized to chain of any length with mathematical induction.)
Show that the conditional independence conditions lead to the joint distribution equation (Hint: add one \(x\) at a time. That is, show \(p(x^3) = p(x_1) p(x_2x_1) p(x_3x_2)\) with the conditional independence conditions, and then \(p(x^4) = p(x_1) p(x_2x_1) p(x_3x_2) p(x_4x_3)\), and so on.)
Show that the joint distribution equation leads to the conditional independence conditions (Hint: use marginalization)
Show that \(p(x^5) = p(x_1) p(x_2x_1) p(x_3x_2) p(x_4x_3) p(x_5x_4)\) leads to \(p(x^5)=p(x_5) p(x_4x_5) p(x_3x_4) p(x_2x_3) p(x_1x _2)\). 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.)
