# Thread: A question regarding Bernoulli distribution with a Normal distribution variable

1. ## 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!

2. ## 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 $F(x)=\Phi((x-\mu)/\sigma)$.