Normal distribution -Why $\Phi(a,b)=\Phi(b)-\Phi(a)$? I am reading "Probability" by Pitman and in the section that talks about normal distribution the book sais that $\Phi(a,b)=\Phi(b)-\Phi(a)$.
Is this a definition or a theorem ? if it's a theorem why is it true ?
 A: If you replace $\Phi(a,b)$ and $\Phi(a),\Phi(a)$ with the integrals that define them, you get something like
$$\int_a^b \circ~dx = \int_{-\infty}^b \circ~dx - \int_{-\infty}^a \circ~dx,$$
which is elementary calculus.
A: By definition, $\Phi(z)$ is the probability that $Z\le z$, where $Z$ is standard normal. And $\Phi(a,b)$ (for $a\le b$) is the probability that $a\le Z\le b$. But
$$P(a\le Z\le b)=P(Z\le b)-P(Z<a).$$
The probability that $Z$ is exactly $a$ is $0$, so $P(Z<a)=P(Z\le a)$, and the result follows.
More informally, $\Phi(a)$ is the area under the standard normal curve, from $-\infty$ to $a$, and $\Phi(a,b)$ is the area under the curve, from $a$ to $b$. Now the area from $-\infty$ to $a$, plus the area from $a$ to $b$, is the area from $-\infty$ to $b$. In symbols,
$$\Phi(a)+\Phi(a,b)=\Phi(b),$$
and therefore $\Phi(a,b)=\Phi(b)-\Phi(a)$. 
The fact that $\Phi(a,b)=\Phi(b)-\Phi(a)$ is a theorem. However, it is intuitively clear from the intended meaning of $\Phi(a,b)$ and $\Phi(z)$. 
A: Suppose $b\le a$.  Then
$$
\Pr(X \le a) = \Pr(X \le b\text{ or }b<X\le a) = \Pr(X\le b) + \Pr(b<X\le a).\tag{1}
$$
This holds because of $E$ and $F$ are mutually exclusive events then $\Pr(E\text{ or }F)=\Pr(E)+\Pr(F)$.
Subtract $\Pr(X\le b)$ from both sides of $(1)$ and it says:
$$
\Pr(X \le a) - \Pr(X\le b) = \Pr(b<X\le a).
$$
This holds regardless of whether the distribution of $X$ is discrete or continuous or a mixture of the two or something else.  But when it's continuous the expression $\Pr(b<X\le a)$ becomes insensitive to the difference between "${<}$" and "$\le$".
