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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 ?

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3 Answers 3

up vote 2 down vote accepted

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 $\Phi(z)$.

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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.

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Thank you. This explanation was concise and elegant. –  000 Apr 12 '12 at 3:07

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 "$\le$".

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