A function of random variables

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

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

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

The solution says that then $\displaystyle Y_i = Bernulli(p) $ where $\displaystyle p=Pr(X<0)= \Phi (- \mu ) $ where $\displaystyle \Phi ( \dot )$ 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!!!

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).