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).
Hi, I'm studying for an exam and I'm studying this problem:
Let be i.i.d. to with unknown.
And suppose that
The solution says that then where where 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!!!
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).