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Math Help - Abs Value & Normal Distribution

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    Abs Value & Normal Distribution

    I'm pretty sure that this is correct?

     P(|z|=1) = P(z>1)+P(z<-1) = 2\times P(z>1)

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    Quote Originally Posted by pickslides View Post
    I'm pretty sure that this is correct?

     P(|z|=1) = P(z>1)+P(z<-1) = 2\times P(z>1)

    The probability of what you've posted is equal to zero:

     P(|z|=1) = P(z=1)+P(z=-1) = 0 + 0 = 0

    since you're dealing with a continuous variable.

    What is correct is that

     P(|z| {\color{red}>} 1) = P(z>1)+P(z<-1) = 2\times P(z>1)
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    MHF Contributor matheagle's Avatar
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    And we're assuming this is a st normal?
    (sorry didn't notice the title)
    You only need this to be a distribution that is symmetric about 0.
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    Quote Originally Posted by mr fantastic View Post

    What is correct is that

     P(|z| {\color{red}>} 1) = P(z>1)+P(z<-1) = 2\times P(z>1)
    \Rightarrow P(|z|< 1) = P(z<1)-P(z<-1) = 2\times P(0<z<1) ?

    and P(|z|< 1) + P(|z|> 1) = 1 or am I just spiralling out of control now?
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    Quote Originally Posted by matheagle View Post
    And we're assuming this is a st normal?
    (sorry didn't notice the title)
    You only need this to be a distribution that is symmetric about 0.
    Yep, going with all these assumptions, just never bumped into a problem with the abs wraped around the Z before.
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    Moo
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    Quote Originally Posted by pickslides View Post
    \Rightarrow P(|z|< 1) = P(z<1)-P(z<-1) = 2\times P(0<z<1) ?

    and P(|z|< 1) + P(|z|> 1) = 1 or am I just spiralling out of control now?
    No, these are false...

    What you did in a first time was right, except for the red part corrected by Mr F :

    \mathbb{P}(|Z|{\color{red}>}1)=\mathbb{P}(Z<-1)+\mathbb{P}(Z>1)

    Why so ? Because |Z|>1 means "Z<-1 OR Z>1"
    Since the event "Z<-1" and the event "Z>1" cannot happen at the same time, \mathbb{P}(|Z|>1)=\mathbb{P}(\{Z<-1\} \cup \{Z>1\})=\mathbb{P}(Z<-1)+\mathbb{P}(Z>1)

    (this is the formula of P(A u B)=P(A)+P(B) when A and B are disjoint)
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    Quote Originally Posted by pickslides View Post
    \Rightarrow P(|z|< 1) = P(z<1)-P(z<-1) = 2\times P(0<z<1) ?

    and P(|z|< 1) + P(|z|> 1) = 1 or am I just spiralling out of control now?
    Quote Originally Posted by Moo View Post
    No, these are false...

    [snip]
    No moo, they're true
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    Thanks Mr F, I think you have helped to consolidate my understanding.
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