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.


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.

  • $\begingroup$ 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

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