I am stuck in obtaining the probability density of $Z_t = X_t + Y_t$ where $X_t = |W_t|, Y_t = |B_t|$, $W_t$ and $B_t$ are independent Brownian motions.
I am aware that the probability density of the absolute value of a Brownian motion $X_t = |W_t|, t \geq 0$ is given by $$p_t(x) = \frac{2}{\sqrt {2\pi t}}e^{-\frac{x^2}{2t}},\quad x \geq 0 $$
My approach is to use a previously proven lemma that the probability of a sum of random variables is the convolution lemma, that goes
Let $X$, $Y$ be 2 positive independent random variables whose probability density function $f$ and $g$ respectively. Denote $h$ as the probability density of $X+Y$ then $$h(x) = \int_{0}^{x}f(x-\tau)g(\tau)\,d\tau$$
Thus, in the context of my problem, I have \begin{align} h(x) &= \int_{0}^{x} \frac{2}{\sqrt{2\pi t}}e^{-\frac{(x-\tau)^2}{2t}}\frac{2}{\sqrt{2\pi t}}e^{-\frac{{\tau}^2}{2t}}\, d\tau \\ &= \frac{2}{\pi t} \int_{0}^{x} e^{-\frac{x^2 - 2x\tau + 2{\tau}^2}{2t}}\,d\tau \\ &= \cdots \\ &= \frac{2}{\pi t} e^{\frac{-x^2}{4t}}\int_{0}^{x} e^{-\frac{1}{t}(\tau - \frac{x}{2})^2} \, d\tau \\ &= \frac{2}{\sqrt{\pi t}} e^{\frac{-x^2}{4t}} \frac{1}{\sqrt{\pi t}} \int_{0}^{x} e^{-\frac{1}{t}(\tau - \frac{x}{2})^2} \, d\tau \end{align}
and then I am stuck.
The given answer is $$h(x) = \frac{2}{\sqrt{\pi t}}e^{-\frac{x^2}{4t}} \left[ 2N\left( \frac{x}{\sqrt{2t}} \right) - 1\right]$$. However, I do not know what $2N\left( \frac{x}{\sqrt{2t}} \right)$ is, but I included it in case it is helpful. I suspect it is either the distribution function or is the cumulative distribution.
Thanks!
The function $N$ is most certainly the distribution function of a standard normal random variable $B_1$, so that $N(x)=\int_{-\infty}^x \frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy$ and $N(0)=1/2$. By symmetry, $$\begin{split} \int_{0}^{x}\frac{1}{\sqrt{\pi t}}e^{-(y-x/2)^2/t}dy&= \mathbb{P}(0<B_{t/2}+x/2<x)= 2\mathbb{P}(0<B_{t/2}<x/2)\\ &=2\mathbb{P}(0<B_1\sqrt{t/2}<x/2)=2\mathbb{P}(0<B_1<x/\sqrt{2t})=2N(x/\sqrt{2t})-1. \end{split}$$ So you can connect your answer to the one you were after.