I'm reading on some branching process theory in Harris' Theory of Branching Processes and encountered an inequality which looks simple but is eluding me.
The full version is a bit complicated to explain (it is equation 6.1 in Harris, if you have it), but it essentially boils down to:
$$P(X=x|Y=y) \geq P(X=x|Z=z) \cdot P(Z=z|Y=y)$$
I sense that this is super trivial, but I'm still shaky on whether this is true in general, and if so, how to put it in a measure-theoretic context (i.e. shown rigorously using conditional expectations). Any hints?
EDIT:
Here are more details about the specific case. We have a multitype Galton-Watson branching process $\{Z_n\}_{n \geq 0}$ on $k$ types (i.e. each $Z_n$ is a $k$-dimensional vector). Denote the standard basis vectors $\{e_i\}$, i.e. $e_i$ is the vector with 1 in the i-th position and 0's elsewhere.
Denote by $S$ the set of types $i$ such that: $$ P(Z_n = 0 | Z_0 = e_i) = 0, \quad n = 1, 2, \ldots $$
In other words, $S$ is the set of types such that starting with 1 individual of that type in the branching process leads to the branching process having 0 probability of going extinct.
Suppose there are $r>0$ of these types and number them $1, 2, \ldots, r$. Let $z$ be a $k$-dimensional, non-negative integer-valued vector and suppose that $z^1 + \ldots + z^r = 0$. That is, assume that $z$ is a vector with 0's in all of the positions corresponding to types in $S$. Then:
$$ P(Z_{n+1} = 0 | Z_0 = e_i) \geq P(Z_{n+1} = 0 | Z_1 = z) \cdot P(Z_1 = z | Z_0 = e_i) $$
In the case where $X$ and $Y$ are conditionally independent given $Z$, we have \begin{align*} P(X=x \mid Y=y) &\ge P(X=x, Z=z \mid Y=y)\\ &= P(X=x \mid Z=z,Y=y) P(Z=z \mid Y=y)\\ &= P(X=x \mid Z=z) P(Z=z \mid Y=y). \end{align*}
I'm not sure if anything in general holds though...