2
$\begingroup$

Consider the following problem: $$z_1 = a_1 x_1$$ $$z_2 = a_2 x_2$$ where $a_1, a_2$ are i.i.d. (regardless of their distribution; in the actual case study it is a symmetric Bernoulli distribution with equiprobable symbols (+1, -1)) and independent w.r.t. $x_1, x_2$.

$x_1$ and $x_2$ are dependent regardless of their distribution (which is continuous, in particular it could be a sequence of two correlated Gaussian RVs with zero mean and a generic covariance matrix $\Sigma_x$).

Question: Are the product variables $z_1$ and $z_2$ dependent? What is the PDF of $S=z_1 + z_2$?

My take: since $S$ is a function of $a_1, a_2, x_1, x_2$ and $x_1, x_2$ are dependent, the joint PDF $f_{z_1, z_2}(\zeta_1, \zeta_2) = f_{a_1, a_2, x_1, x_2} (\alpha_1, \alpha_2, \xi_1, \xi_2)$ is not in general the product of the marginals, so $z_1$ and $z_2$ are dependent.

$\endgroup$
0
$\begingroup$

If $x_1=x_2$, then the conditional distribution of $z_2$ given $z_1$ is concentrated at $\pm z_1$, whereas the unconditional distribution of $z_2$ is continuous. Hence $z1$ and $z2$ are not independent.

To compute the PDF of $S=z_1+z_2$ use the decomposition $$ \mathbb{P}(S\leq s) = \frac{1}{4}\sum\int\int f_{x_1,x_2}(\xi_1,\xi_2)\mathbf{1}_{\{\pm \xi_1\pm \xi_2\leq s\}}\mathbb{d}^2(\xi_1,\xi_2), $$ where $f_{x_1,x_2}$ is the joint PDF of $x_1,x_2$ and the sum is over $\{\pm\}^2$.

The generalization to other distributions of $a_i$ should be obvious.

$\endgroup$
  • $\begingroup$ Thank you, I was thinking of something like: $f(z_1, z_2) = f(a_1 x_1, a_2 x_2 | a_1, a_2) f(a_1) f(a_2)$ ($a_1,a_2$ independent) which should lead to the same result, i.e. the PDF: $$f_{z_1, z_2} = \frac{1}{4} (f(-x_1, -x_2 | a_1 = -1, a_2 = -1) + f(x_1, -x_2 | a_1 = 1, a_2 = -1) + f(x_1, x_2 | a_1 = 1, a_2 = 1) + f(-x_1, x_2 | a_1 = -1, a_2 = 1))$$ which clearly depends on $f(x_1, x_2)$ which is not the product of the marginals since they are not independent in general. What if they were? Could we claim $z_1$ and $z_2$ are independent? $\endgroup$ – V.C. Mar 28 '13 at 18:11
  • $\begingroup$ If $x_1$ and $x_2$ are independent, then so are $z_1$ and $z_2$, yes. $\endgroup$ – Eckhard Mar 28 '13 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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