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Math Help - A function of random variables

  1. #1
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    A function of random variables

    Hi, I'm studying for an exam and I'm studying this problem:

    Let  X_1, X_2,...,X_n be i.i.d. to  N( \mu , 1) with  \mu unknown.

    And suppose that  Y_i = I(X_i<0) \ \ \ \ \forall i = 1,...,n

    The solution says that then Y_i = Bernulli(p) where p=Pr(X<0)= \Phi (- \mu ) where  \Phi ( \dot ) is the standard nomral cdf.

    But I don't understand the last part when the distribution can be written as the standard nomral cdf.

    Please help, thanks!!!
    Last edited by tttcomrader; September 17th 2012 at 12:04 PM.
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  2. #2
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    Re: A function of random variables

    Hey tttcomrader.

    Basically the reason is that if you have a normal random variable X with mean mu and variance v then [X - mu]/SQRT(v) ~ N(0,1). Now your variance is 1 which means that P(X < 0) = P((X - mu)/1 < (0-mu)/1) = P(Z < -mu) = phi(-mu).
    Thanks from tttcomrader
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