Let $X,\ Y,\ Z >0\ $ random variables such that $X=YZ$ where $Y$ and $Z$ are not independent. If I have the following two relationships: $$\mathbb{E}[X|Y=c]=\mathbb{E}[Y|Y=c]\ \mathbb{E}[Z|Y=c],\ \forall c\ge0 $$ and $$\mathbb{E}[X|Y\ge c]=\mathbb{E}[Y|Y\ge c]\ \mathbb{E}[Z|Y\ge c],\ \forall c\ge0 \label{1}$$ Does the first imply the second or vice versa or no implication is present?
For the first, $$\begin{align} \mathbb{E}[X|Y\ge c]&={1 \over \mathbb{P}(Y\ge c)}\int_0^{\infty}\int_c^{\infty}xf_{X|Y}f_Y\ dy\ dx\\ &={1 \over \mathbb{P}(Y\ge c)}\int_c^{\infty}\mathbb{E}[X|Y=y]f_Y\ dy\\ &={1 \over \mathbb{P}(Y\ge c)}\int_c^{\infty}\mathbb{E}[Y|Y=y]\ \mathbb{E}[Z|Y=y]f_Y\ dy\\ &\not={1 \over \mathbb{P}(Y\ge c)^2}\int_c^{\infty}\mathbb{E}[Y|Y=y]\ f_Y\ dy\ \int_c^{\infty}\mathbb{E}[Z|Y=y]f_Y\ dy\\ \end{align}$$ For the second implication I get: $${\mathbb{P}(Y\ge c)}\int_c^{\infty}\mathbb{E}[X|Y=y]f_Y\ dy\\=\int_c^{\infty}\mathbb{E}[Y|Y=y]\ f_Y\ dy\ \int_c^{\infty}\mathbb{E}[Z|Y=y]f_Y\ dy\\$$ Which gives: $$\int_c^{\infty}\int_c^{\infty}\mathbb{E}[X|Y=y]f_Y\ f_Y\ dy\ dl\\=\int_c^{\infty}\int_c^{\infty}\mathbb{E}[Y|Y=y]\ \mathbb{E}[Z|Y=l]f_Y\ f_Y\ dy\ dl\\$$ Which implies $\mathbb{E}[X|Y=c]=\mathbb{E}[Y|Y=c]\ \mathbb{E}[Z|Y=c],\ \forall c\ge0 $.
Is my reasoning correct?