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

Can someone explain this equality ?

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    Yes. **Symmetry**.2012-04-11
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    If $\Phi(a,b)$ is the probability that a standard normal random variable lies between $a$ and $b$, then this follows from the fact that the normal distribution is symmetric about its mean, in this case zero. So $\Phi(-1,0) = \Phi(0,1)$ and $\Phi(-1,1) = \Phi(-1,0) - \Phi(0,1) = 2\Phi(0,1)$.2012-04-11
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    @RahulNarain - if $\Phi(-1,0)=\Phi(0,1)$ then $\Phi(-1,0)-\Phi(0,1)=0$...Why $\Phi(-1,1)=\Phi(-1,0)-\Phi(0,1)=2\Phi(0,1)$ ?2012-04-11
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    Oops, I meant $\Phi(-1,1) = \Phi(-1,0) + \Phi(0,1) = 2\Phi(0,1)$, with a plus sign. You know that $\Phi(a,b) + \Phi(b,c) = \Phi(a,c)$, right?2012-04-11
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    @RahulNarain - I think so...it depends if I understand right - is $\Phi(a,b)=\Phi(b)-\Phi(a)$ a definition or a theorem ? if it's the definition then it's ok..2012-04-11
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    I don't know, check your book. But how does it matter?2012-04-11
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    If it's the definition then I understand it, if it's a theorem then I don't understand the theorom hence I don't rally understand this statement..2012-04-11

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