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Math Help - A question regarding Bernoulli distribution with a Normal distribution variable

  1. #1
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    A question regarding Bernoulli distribution with a Normal distribution variable

    This is from an old PhD exam:

    Given X_1, X_2, \ldots , X_n are i.i.d. to  N( \mu , 1 )

    And define Y_i = I \{ X_i < 0 \}

    Then Y_1, Y_2, \ldots , Y_n are i.i.d. to Bernoulli(p) where p = Pr (X<0) = \Phi (- \mu )

    Now, my question really is... Why is Pr (X<0) = \Phi (- \mu ) ?

    Thank you!
    Last edited by tttcomrader; July 29th 2013 at 09:40 AM.
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  2. #2
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    Re: A question regarding Bernoulli distribution with a Normal distribution variable

    This is obvious from a picture. If f(x) is the the p.d.f. of X and ϕ(x) is the p.d.f. of the standard normal distribution, then f(x) is ϕ(x) shifted μ to the right. Therefore, the area under f(x) up to 0 is the same as the area under ϕ(x) up to -μ. More formally, f(x) = ϕ(x - μ), so

    F(x)=\int_{-\infty}^x f(t)\,dt= \int_{-\infty}^x \phi(t-\mu)\,dt= \int_{-\infty}^{x-\mu} \phi(s)\,ds= \Phi(x-\mu)

    In general, if F(x) is a c.d.f. of a normal distribution with mean μ and deviation σ, then F(x)=\Phi((x-\mu)/\sigma).
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