Let X ~ Normal Standar find X given X > 0

Question

Let $X ~ N (0,1)$ Find the conditional PDF of $X$ given $X> 0$

I understand that this would be given by $ \frac{f(x,x>0)}{f(x>0)}$, but I really do not know how to start, if you can give me ideas I appreciate.

Answer 1

Let $g$ be the density of $X$. ($g(x)=\frac 1 {\sqrt {2\pi}} e^{-x^{2}/2})$. $P\{X\leq x |X>0\}=\frac {\int_0^{x} g(t)dt} {\int_0^{\infty} g(t)dt}=2\int_0^{x} g(t)dt$ for $x>0$ and $0$ for $x<0$. Hence the density function is $2g(x)$for $x>0$, $0$ for $x<0$.

Answer 2

Let $Y$ denote the R.V. $X|X>0$. By definition, the support of $Y$ is $[0, \infty)$.

To compute the PDF, let's compute the CDF first.

$$P(Y>y) = P(X>y|X>0) = \frac{P(X>y, X>0)}{P(X>0)}$$.

Since $y \geq 0$, the numerator is simply $P(X>y)$. Since $X \sim N(0,1)$, the denominator is ${1 \over 2}$. Thus,

$$P(Y>y) = 2P(X>y)$$

Now you can see that $f_Y(y) = \sqrt{{2 \over \pi}}e^{-y^2/2}$ for $y \geq 0$ and $0$ otherwise.