Suppose $p(X \mid Y) = 1$, how do I show that $p(X \mid Y, Z) = 1$?

Question

Suppose $X, Y, Z$ are subsets of an event space and I know that $p(X \mid Y) = 1$, how do I show that $p(X \mid Y, Z) = 1$, in this case $Z$ can be any subset I guess. Here $p(X \mid Y, Z)$ refers to the probability that $X$ occur given both $Y$ and $Z$ occur. This statement makes sense intuitively, but how would I go about proving it from the basic probability rules, ie definition of conditional probability.

Answer 1

Firstly, convince yourself that: \begin{equation} Y\subseteq X \iff X\cap Y = Y \tag{1} \end{equation} (hint: assume $Y\subseteq X$ and show inclusion both ways, i.e. that $X\cap Y \subseteq Y$ and $X\cap Y \supseteq Y$).

By assumption, and using the definition of conditional probability we have: $$ P(X|Y) = \frac{P(X\cap Y)}{P(Y)} =1 $$ This implies that $P(X\cap Y) = P(Y)$. By equation 1, this is true iff $Y\subseteq X$.

By the definition of conditional probability (and assuming $P(Y \cap Z) \neq 0$): $$ \begin{align} P(X|Y \cap Z)&=\frac{P(X\cap Y \cap Z)}{P(Y \cap Z)} \\ &= \frac{P(Y \cap Z)}{P(Y \cap Z)} \\ &= 1 \end{align} $$ where the second equality comes from substituting $X\cap Y = Y$ as justified in equation 1.