Conditional pdfs on Standard Normal Distribution

Question

Consider $Z \sim N[0,1]$. The standard normal probability density function is given by \begin{equation} \phi(x)=\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \end{equation} We define \begin{equation} \Phi(x)=\int^z_{-\infty} \phi(x)dx \end{equation} which is the probability that $Z<z$.

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I am asked to compute the probability density of Z conditional on Z>0.

I know this translates to \begin{equation} P(Z=z|Z>0)=\frac{P(Z=z \cap Z>0)}{P(Z>0)} \end{equation} I know that I compute the numerator from the given relation for $\Phi(x)$ but I am having trouble with the numerator. How can I simplify the numerator $P(Z=z \cap Z>0)$?

Answer 1

I would firstly compute $\begin{equation} \text{P(Z<z|Z>0)}\end{equation}$ to obtain the cdf.

$$\begin{equation} \text{P(Z<z|Z>0)}\end{equation}=\frac{P(Z<z\cap Z>0)}{P(Z>0)}=\frac{P(0<Z<z) }{P(Z>0)}$$

• $P(0<Z<z)=P(Z<z)-P(z<0)=\Phi(z)-\frac12$
• $P(Z>0)=\frac12$

$\Rightarrow \frac{P(0<Z<z) }{P(Z>0)}= 2\Phi(z)-1$, where $\Phi(z)$ is the cdf of the standard normal distribution.

Let $A$ be the event that $Z>0$. Then the cdf is

$$F_{Z|A}(z)=\begin{cases} 0, \ z<0 \\ 2\cdot \Phi(z)-1, \ 0 \leq z<\infty \end{cases}$$

Now differentiate to obtain the pdf.

Answer 2

$P(Z=z|Z\gt 0)=\sqrt{\frac{2}{\pi}}e^{-\frac{z^2}{2}}$ for $z\gt 0$ and $=0$ otherwise. Here P is the probability density function, not the distribution function.