Show that a particular process is white noise Given $0< p < 1$ and

*

*$T_t \overset{i.i.d.}{\sim} t _5 $, Student's-t distribution with 5 degrees of freedom;

*$B_t \overset{i.i.d.}{\sim} B(1,p)$, Bernoulli distrution.

Define:
$$\epsilon_t = B_t T_t, \, \forall\, t $$
I want to show that $\{\epsilon_t\}_{t \in \mathbb{Z}}$ is a white noise process.
First, my lecture notes does not suppose any thing about the independence or dependence between $T_t$ and $B_t$. There is a possibility that my reading notes are considering $T_t$ and $B_t$ independent, but not written. This would be a fault.
So, is it possible to show that $\{\epsilon_t\}_{t \in \mathbb{Z}}$ is white noise not assuming the independence of $T_t$ and $B_t$?
If they are independent, then $E(\epsilon_t) = E(T_t B_t) = E(T_t)E(B_t) = 0$, because $E(T_t)=0$. But if $T_t$ and $B_t$ are dependent?
How about the other properties?

*

*$E(\epsilon_t^2) = \sigma^2 < \infty\,\, \forall t $;

*$E(\epsilon_t \epsilon_s) = 0, \,\,\forall s \neq t.$
Some help?
 A: The independent case has been considered in the comments, so here I focus on the dependent case.
If $T_t$ and $B_t$ are not independent I think the statement is false.
Let's consider $t$ fixed and show that two variables $T$, $B$ can have the right marginals, but $E[BT] \ne 0$, therefore we cannot have white noise.
We take a generic joint $p(b,t)$. Than the conditions on the marginals are:

*

*$p(0,t)+p(1,t)=f^{stud}(t)$

*$\int dt p(1,t)=p$

*$\int dt p(0,t)=1-p$
It is easy to check that if we have $g(t)$ such that:

*

*$0\le g(t)\le f^{stud}(t)$ ;

*$\int dt g(t)=p$ :

,than a solution to the conditions is $p(1,t)=g(t)$ and $p(0,t)=f^{stud}(t)-g(t)$, so we can parametrize in some sense all solutions.
Now we can verify that:
$E[BT]=\int g(t)tdt$
And in general this does not vanish under the given conditions.
EDIT: we can check for compatibility what happens in the independent case. In this case $p(0,t)$ is proportional to $p(1,t)$ and hence, by condition [1], to $f^{stud}(t)$. This leads that $E[BT]$, written in terms of $g(t)$, is proportional to the expectation of the student distribution, and therefore vanishes. Of course in this case we can proceed in a much more direct way.
