# HW1 (Due 9/18)

1. Compute the mean and variance of $$X$$ when

1. $$X$$ is continuous and uniformly distributed between 0 and 2. That is, $$p_X(x)=\begin{cases}0.5, &\mbox{if } 0\le x\le 2\\0, & \mbox{otherwise}\end{cases}$$

2. $$X$$ is discrete and $$Pr(X=1)=0.4, Pr(X=2)=0.2, Pr(X=3)=0.1, Pr(X=4)=0.3$$

2. (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., $$p(x_5|x_4,x^3)=p(x_5|x_4))$$, $$X_4 \bot X^2 | X_3$$, 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 $$p(x^5)=p(x_1) p(x_2|x_1) p(x_3|x_2) p(x_4|x_3) p(x_5|x_4)$$. (Remark: note that the proof below can be generalized to chain of any length with mathematical induction.)

1. 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. Show that the joint distribution equation leads to the conditional independence conditions (Hint: use marginalization)

3. 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.)