1
$\begingroup$

Suppose that there exists a random vector $\eta\equiv (\eta_1, \eta_2, \eta_3)$ continuously distributed on $\mathbb{R}^3$ and with full support. Can we always find a vector $\epsilon\equiv (\epsilon_1,\epsilon_2,\epsilon_3)$ continuously distributed on $\mathbb{R}^3$ and with full support such that $$ \eta_1\equiv \epsilon_1-\epsilon_3\\ \eta_2\equiv \epsilon_1-\epsilon_2\\ \eta_3\equiv \epsilon_2-\epsilon_3\\ $$

If yes, how can we show it? If not, can you provide a counterexample?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

You have $\begin{bmatrix}1 & 0 & -1 \\ 1 & -1 & 0 \\ 0 & 1 & -1\end{bmatrix} \epsilon = \eta$, but this matrix is not invertible, so the vectors of the form $(\epsilon_1-\epsilon_3, \epsilon_1-\epsilon_2, \epsilon_2 - \epsilon_3)$ cannot span all of $\mathbb{R}^3$.

Improve this answer
$\endgroup$
4
$\begingroup$

Hint:

Consider $$(\epsilon_1-\epsilon_3) - (\epsilon_1-\epsilon_2) - (\epsilon_2-\epsilon_3)$$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks. I'm not understanding it, sorry. Could you help more? Is it a proof for "yes", or a counterexample for "no"? $\endgroup$
    – Star
    Commented Oct 5, 2020 at 18:35
  • 2
    $\begingroup$ @user3285148 It is a hint to help you find the answer. What happens when you simplify my expression? Now suppose $\eta_1 -\eta_2-\eta_3 \not = 0$ such as $\eta_1=10, \eta_2=2,\eta_3=3$. Can you see why there is no possible $\epsilon_1,\epsilon_2,\epsilon_3$ satisfying your requirements? $\endgroup$
    – Henry
    Commented Oct 5, 2020 at 18:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.