Normal distribution (why $\Phi(0,1)=0.5\Phi(-1,1)$)

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I am reading "Probability" by Pitman and in the section that talks about normal distribution states: $\Phi(0,1)=0.5\Phi(-1,1)$.

Can someone explain this equality ?

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1 Answer

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Let $\Phi(b)$ be the cumulative distribution function of the normalized normal distribution. That is, $\Phi(b)=P(X<b)$. Then it's convenient to introduce $\Phi(a,b)=\Phi(b)-\Phi(b)$, which expresses the probability $P(a<X<b)$. (Could also be written as $P(a\le X<b)$, which is the same probability because $P(X=a)=0$.) Since $X$ is symmetric, $$P(0<X<1)=P(-1<X<0) \\ = \frac12 (P(0<X<1)+P(-1<X<0)) \\ =\frac12 P(-1<X<1)$$

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