# Independence: Discrete Random Variables

I am taking my first course in Measure Theory and I am struggling with the following problem:

Consider the following random variables $$X_1,\dots,X_k$$ for which countable subsets $$A_1,\dots,A_k$$ are defined such that:

$$P(X_j \in A_j)=1\ \ \forall j$$

Show that $$X_1,\dots,X_k$$ are independent iff for all $$a_j \in A_j$$

$$P(X_j=a_j,\forall j)=\prod_{i=1}^{k}P(X_j=a_j)$$

I know that the independence of random variables can be written as:

$$P(X_j\leq x_j,\forall j)=\prod_{i=1}^{k}P(X_j \leq x_j)$$ for $$x_j \in \mathbb{R}$$

However, I am having trouble taking this into a discrete universe. Thanks in advance for any advice.

• Here $P(X_j=a_j, \forall j)=P(X_1=a_1, X_2=a_2,\dots X_k=a_k)$, is it? Jun 16 '17 at 17:39
• Simply decompose $P(\forall j,X_j\leqslant x_j)$ into $$\sum_{(a_j)}P(\forall j,X_j=a_j)$$ for the suitable collection of $k$tuples $(a_j)$, use the independence on each of these terms to deduce that $$P(\forall j,X_j=a_j)=\prod_jP(X_j=a_j)$$ and conclude.
– Did
Jun 16 '17 at 18:14
• @Did someone downvoted my solution, if you help me to identify what is wrong, it will be very helpful to me. I need to know! Jun 17 '17 at 12:48

## 1 Answer

Let $Y_j=1_{(X_j=a_j)},j=1(1)k$, then $E(Y_j)=P(X_j=a_j)$. Note that since $X_1, X_2, \dots ,X_k$ are independent then $Y_1, Y_2, \dots , Y_k$ are too.

Now $E(Y_1Y_2\dots Y_k)=E(Y_1)E(Y_2)\dots E(Y_k)$ gives the proof of "only if" part.

Let $B_{ij}=\{X_i=a_j\}$, $a_j\in A_i$. From the expression given, we can conclude that the $\sigma$-algebra generated by $A_j$ are also independent. Then $X_1, X_2, \dots ,X_k$ are independent.

• Who ever downvoted please identify what is wrong, I need to know. Jun 17 '17 at 12:45