I am stuck with the following question.
Suppose $T \perp C |Z$, does it follow that $T \perp C $?
Suppose $T \perp C $, does it follow that $T \perp C |Z$?
I think both don't follow, but I don't know how to show it
Thanks in advance
I am stuck with the following question.
Suppose $T \perp C |Z$, does it follow that $T \perp C $?
Suppose $T \perp C $, does it follow that $T \perp C |Z$?
I think both don't follow, but I don't know how to show it
Thanks in advance
For your first, consider the following possibilities (with equal probability)
T C Z 1 1 1 1 0 1 0 1 1 0 0 1 2 2 0 2 3 0 3 2 0 3 3 0
For your second, consider the following possibilities (with equal probability)
T C Z 1 1 1 0 0 1 1 0 0 0 1 0
Two (related) classes of examples one can try to play with in such cases: 1. Random variables $X_k=\varphi(Y_k,T)$ for a given function $\varphi$ and some independent random variables $(Y_k)$ and $T$. 2. Products of independent random signs.
For example, case 1. yields independent random variables $(X_k)$ conditionally on $T$ and one can easily imagine situations where the random variables $(X_k)$ are not (unconditionally) independent. (Hint: try $T$ very very very large with positive probability.)
Likewise, independent Bernoulli $\pm1$ random variables $X$ and $Y$ are, well... independent and, conditionally on their product $Z=XY$, $X$ and $Y$ are not independent since $Y=XZ$ is a (non degenerate) deterministic function of $X$ and $Z$.