The distribution function of X+

Question

Let's assume that we have a symmetric density function around $y$ axis, i.e. $\mathbb{P}(X\leq 0)=\mathbb{P}(X\geq 0)=\Phi(0)=0.5$ where $\Phi(\cdot)$ is the distribution function of $X$.

$$y^+=\max(y,0)$$

Is this approach correct?

Let $x\geq 0$ \begin{eqnarray} &&\mathbb{P}(X^+\leq x)=\mathbb{P}(\max(X,0)\leq x)=\\ &&\mathbb{P}(X< 0)\times \mathbb{P}(\max(X,0)\leq x \vert X< 0)+\mathbb{P}(X\geq 0)\times \mathbb{P}(\max(X,0)\leq x \vert X \geq 0)=\\ &&\frac{1}{2}\times \mathbb{P}(\max(X,0)\leq x \vert X< 0)+\frac{1}{2}\times \mathbb{P}(\max(X,0)\leq x \vert X\geq 0)=\\ && \frac{1}{2}+\frac{1}{2}\Phi(x) \end{eqnarray} Does $X^+|X$ has the same distribution as $X$?

Can we infer that \begin{eqnarray} \Phi_{X^+}(z)=\frac{1}{2}+\frac{1}{2}\Phi(z),&& \mbox{ if }z\geq 0\\ &&\\ \Phi_{X^+}(z)=0,&& \mbox{ if }z< 0 \end{eqnarray}

Answer 1

You need to be more careful in calculating the second part:

$$ \mathbb{P}(\max(X,0)\leq x|X \geq 0) = \frac {\mathbb{P}(X\leq x, X \geq 0) }{\mathbb{P}(X \geq 0)} = 2\left(\Phi(x) - \frac {1} {2}\right)$$

and thus you will obtain the result agree with your empirical finding.

Note that $X^+$ is a (measurable) function of $X$, so $X^+ \in \sigma(X)$ and thus $X^+|X$ has the same distribution as $X^+$