Let $\mathbf{\Pi}$ be a homogeneous Poisson point process on the real number line with intensity $\lambda$. Let $r^{+}$ denote the distance from the origin to the closest point of $\mathbf{\Pi}$ on the +'ve real axis and let $r^{-}$ denote the distance from the origin to the closest point of $\mathbf{\Pi}$ on the -'ve real axis.
I want to find the distribution of these functions of the randoms variables associated with the process, namely:
$$\mathbf{P}(r^{+} \leq x), $$
$$\mathbf{P}(r^{-} \leq x),$$
$$\mathbf{P}(\textrm{min}(r^{-},r^{+}) \leq x).$$
I know that the probability of points generated by a Poisson point process defined on the real line with intensity $\lambda \gt 0$ in a region $[a,b]$ where $a \leq b$ is given by:
$$\mathbf{P} \{N(a,b] = n\} = \frac{[\lambda(b-a)]^n}{n!}e^{-\lambda(b-a)}.$$
My solution to find $\mathbf{P}(r^{+} \leq x)$ is to consider a region $[0,r^+]$ where no points exist, i.e $N=0,$ which gives:
$$\mathbf{P}(r^{+} \leq x) =1- e^{-\lambda r^{+}}$$
Similarly,
$$\mathbf{P}(r^{-} \leq x) = 1- e^{-\lambda r^{-}} $$
$$\mathbf{P}(\textrm{min}(r^{-},r^{+}) \leq x) =(1-\lambda e^{-2\lambda x})^2. $$
Could anyone comment on the validity of my method to tackle this?
There is some confusion in your post; $r^+$ is a random variable, and $x$ is a fixed number, so the formula for $P(r^+\le x)$ cannot involve $r^+$. It will be some deterministic function of $x$.
In particular, $r^+\le x$ if and only if there is some arrival in the interval $[0,x]$, so $$ P(r^+\le x)=1-e^{-\lambda x} $$ The corresponding probability for $r^-$ is the same. Finally, since $r^+$ and $r^-$ are independent, \begin{align} P(\min(r^+,r^-)\le x) &=P(r^+\le x\cup r^-\le x) \\&=P(r^+\le x)+P(r^-\le x)-P(r^+\le x\cap r^-\le x) \\&=P(r^+\le x)+P(r^-\le x)-P(r^+\le x)P(r^-\le x) \\&=(1-e^{-\lambda x})+(1-e^{-\lambda x})-(1-e^{-\lambda x})^2 \\&=1-e^{-2\lambda x} \end{align} Another way to arrive at this; we see that $\min(r^+,r^-)>x$ if and only if there are no arrivals in $[-x,x]$.