2
$\begingroup$

Let $x_1,x_2,x_3$ be (real and continuous) random variables.

  1. If $x_1,x_2,x_3$ are mutually independent, then $x_1 + x_2$ and $x_2 + x_3$ are independent?
  2. If $x_1,x_2,x_3$ are mutually independent, then $x_1 x_2$ and $x_2 x_3$ are independent?
  3. What If $x_1,x_2,x_3$ are pairwise independent?

In general, when I prove things like these, what kind of technique do I need to use?

$\endgroup$
1
  • $\begingroup$ The answer for all three questions is: not nesessarily. In fact that the asserted independence is false except in very special cases. For example, in 1) x1+x2 and x2+x3 are independent only when x2 is a constant random variable. The best way to handle questions like these is to use characteristic functions. $\endgroup$ Commented Jan 20, 2017 at 6:08

1 Answer 1

1
$\begingroup$

(1) False. Let $X_1$, $X_2$, $X_3$ be mutually independent normal$(0,1)$ variables. Compute $$ E(X_1+X_2)(X_2+X_3) = E(X_1X_2 + X_1X_3+X_2^2 + X_2X_3) = 0 + 0 + 1 + 0=1. $$ This is different from $$E(X_1+X_2)E(X_2+X_3)=0,$$ and therefore $X_1+X_2$ and $X_2+X_3$ are not independent.

(2) False. Take the $X$'s from (1) and define $Y_i := \exp(X_i)$ . Then the $Y$'s are mutually independent. But $Y_1Y_2$ and $Y_2Y_3$ are not independent; if they were, then $\log(Y_1Y_2):=X_1+X_2$ and $\log(Y_2Y_3):=X_2+X_3$ would be independent.

(3) False. Find normal$(0,1)$ variables $X_1$, $X_2$, $X_3$ that are pairwise independent. For example, see two of the answers to this question. Next, note that the calculation in (1) requires only pairwise independence.

These types of assertions are typically false. One strategy is to experiment with expectations, as done in step (1); what would independence imply about expectations? Another strategy is to check if the assertion holds with discrete variables; if it doesn't, then it likely won't hold for continuous ones. Also it's good to visualize what variables look like when they are independent, and when they are pairwise independent. (Refer to the answers to that other question again.)

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.