PDF of the sum of two dependent normal random variables

$Y=X\mathbb{1}_{W=0}-X\mathbb{1}_{W=1}$ , so $Z=X+Y=X+X\mathbb{1}_{W=0}-X\mathbb{1}_{W=1}=X(1+\mathbb{1}_{W=0}-\mathbb{1}_{W=1})$ . Since $W\sim \operatorname{Bernoulli}{(\frac{1}{2})}$ , $1=\mathbb{1}_{W=0}+\mathbb{1}_{W=1}$ , hence $Z=X(\mathbb{1}_{W=0}+\mathbb{1}_{W=1}+\mathbb{1}_{W=0}-\mathbb{1}_{W=1})=2X\mathbb{1}_{W=0}=\Phi(X,W)$ , where $\Phi(x,w)=2x\mathbb{1}_{w=0}$ . Now, let $\Psi:\mathbb{R}\rightarrow \mathbb{R}$ be a suitably general (ie positive or bounded) measurable function. $\mathbb{E}[\Psi(Z)]=\mathbb{E}[\Psi\circ\Phi(X,W)]$ . Using the joint density of $(X,W)$ , given that $X$ and $W$ are independent, we compute this as $\mathbb{E}[\Psi(Z)]=\iint_{\mathbb{R^2}}\Psi\circ\Phi(x,w)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\frac{1}{2}(\delta_{0}(w)+\delta_{1}(w))dxdw$ , where $\delta_{a}$ is the Dirac suppported at $a$ . So, $\mathbb{E}[\Psi(Z)]=\iint_{\mathbb{R^2}}\Psi(2x\mathbb{1}_{w=0})\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\frac{1}{2}(\delta_{0}(w)+\delta_{1}(w))dxdw$ $=\frac{1}{2}(\int_{\mathbb{R}}\Psi(2x)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx+\int_{\mathbb{R}}\Psi(0)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx)$ $=\frac{1}{2}(\int_{\mathbb{R}}\Psi(2x)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx+\Psi(0))$ , by integrating with respect to $w$ . Now we write $\Psi(0)=\int_{\mathbb{R}}\Psi(x)\delta_{0}(x)dx$ and $\int_{\mathbb{R}}\Psi(2x)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=\int_{\mathbb{R}}\Psi(x)\frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{8}}dx$ by the change of variables $2x\leftrightarrow x$ . Finally, summing these up, we obtain that $\mathbb{E}[\Psi(Z)]=\frac{1}{2}(\int_{\mathbb{R}}\Psi(x)(\frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{8}}+\delta_{0}(x))dx)=\int_{\mathbb{R}}\Psi(x)(\frac{1}{4\sqrt{2\pi}}e^{-\frac{x^2}{8}} +\frac{1}{2}\delta_{0}(x))dx$ . Since $\Psi$ was arbitrary, this shows that the PDF of $Z$ , $f_{Z}$ , is given by the expression $f_{Z}(z)=\frac{1}{4\sqrt{2\pi}}e^{-\frac{z^2}{8}} +\frac{1}{2}\delta_{0}(z)$ which was one of the choices, by reconciling the notations $\delta_{0}(z)=\delta z$

Note: since there is Dirac gymnastics involved, strictly speaking, $f_{Z}$ is not a PDF.

Note that by symmetry of $\mathcal{N}(0,1)$ around zero, $-X$ is also $\mathcal{N}(0,1)$ . In particular, we can write

$F_Y(y)=P(Y\leq y)$

$=P(Y\leq y|W=0)P(W=0)+P(Y\leq y|W=1)P(W=1)$

$=\frac12P(X\leq y|W=0)+\frac12P(-X\leq y|W=1)$

$=\frac12 P(X\leq y)+\frac12P(-X\leq y)$ since $X$ and $W$ are independent.

$=\frac12\Phi(y) +\frac12\Phi(y)$

$=\Phi(y)$

Thus $Y\sim \mathcal{N}(0,1)$ . Now note that,

$Z=X+Y= \left \{ {\text{2X, with probability 0.5} \atop \text{0, with probability 0.5} }\right.$

Thus $Z$ is a mixed random variable and its PDF is given by

$f_Z(z)=\frac12 \delta_0(z) +\frac12$ (PDF of 2X at z)

$f_Z(z)=\frac12 \delta_0(z) +\frac12$ (PDF of a N(0,4) at z)

$f_Z(z)=\frac12 \delta(z) +\frac{1}{4\sqrt{2\pi}}e^{-\frac{z^2}{8}}$