Confused by Probability Notation and Solution

Question

Suppose $P(Z=0)=P(Z=1)=1/2$ and $Y$ ~ $N(0,1)$. $Y$ and $Z$ are independent, and $X=YZ$

The question is to find the law of $X$.

I know the Law is the function $P_X$ from $B$ to $[0,1]$, where $B$ are the borel sets.

The Solution Manual has:

$P(X\in B)$ = $P(X \in B|Z=0)P(Z=0)+P(X \in B|Z=1)P(Z=1)$

$=P(0 \in B)\frac{1}{2} + P(Y \in B)\frac{1}{2}$

$=\frac{\delta_0 + \mu_N}{2}(B)$

I don't quite understand how to interpret $P(0 \in B)$...because $B$ is what it is, it is not chosen randomly. $0$ is either in $B$ or it isn't.

Secondly, I don't know if $\delta_0$ and $\mu_N$ are supposed to be standard notation, but I don't know what they are referring to.

Answer 1

You're right, $0$ is either in $B$ or it's not. Therefore, $P(X\in B)$ equals either $0$ or $1$, which is represented by $\delta_0$.

The $\delta_0$ is the Kronecker delta, which takes value either $0$ or $1$. In this case, $\delta_0=1$ if $0\in B$ and $\delta_0=0$ otherwise.

I would normally take $\mu_N$ to be the mean of $N$, but I'm not positive if it's playing that role here. Will think on it and update.

EDIT: Based on my interpretation of the problem, $\mu_N(B)$ is representing $P(Y\in B)$. Since $Y$~$N(0,1)$ this is the same as saying "What's the area under the normal distribution for the range $B$?"

This is not notation I normally use, but maybe it's standard somewhere. Check your book a little more carefully and see if it mentions it elsewhere (potentially when introducing $N$).

Kronecker delta reading: http://en.wikipedia.org/wiki/Kronecker_delta

Answer 2

The $\mu_N$ is supposed to be the measure induced by $Y$. The subscript $N$ is presumably to denote that it is from an $N(0,1)$ variable. It's somewhat sloppy/nonstandard notation.

Duncan has already cleared up the point about what $\delta_0$ means.

$P(0 \in B)$ might be easier to understand when written out formally. It means $P \left ( \left \{ \omega \in \Omega : 0 \in B \right \} \right )$. Since the condition in this comprehension doesn't depend on $\omega$, this set is either $\Omega$ (if $0 \in B$) or $\emptyset$ (if $0 \notin B$). In the first case the probability is 1, in the second case it is 0.