Let $X$ and $Y$ be independent $\mathcal N(0,1)$ random variables. Let $Z\sim\mathcal U(0,1)$ be independent of $X$ and $Y$. What is the distribution of $U=ZX+(1-Z)Y$?
Clearly, $[U\mid Z=z]=zX+(1-z)Y\sim\mathcal N(0,z^2+(1-z)^2)$, using the reproductive property of Normal distribution. So by the total probability theorem, density of $U$ is given by
$$f_U(u)=\int_0^1 f_{U\mid Z=z}(u|z)\cdot1\,\mathrm{d}z$$
$$=\int_0^1\frac{1}{\sqrt{z^2+(1-z)^2}}\phi\left(\frac{u}{\sqrt{z^2+(1-z)^2}}\right)\,\mathrm{d}z$$
[where $\phi$ is the pdf of $\mathcal N(0,1)$ distribution]. But seeing that $U$ has a mixed distribution, it may not have a density in which case the above integral would not make sense. In any case, I don't think this integral can be evaluated. So I should look for the CDF of $U$.
$\displaystyle F_U(u)=\Pr(U\leqslant u)=\int_0^1\Pr(zX+(1-z)Y\leqslant u\mid Z=z)\cdot1\,\mathrm{d}z$
$\displaystyle\qquad\qquad\qquad\qquad\quad=\int_0^1\Pr(zX+(1-z)Y\leqslant u)\,\mathrm{d}z$
$\displaystyle\qquad\qquad\qquad\qquad\quad=\int_0^1\Phi\left(\frac{u}{\sqrt{z^2+(1-z)^2}}\right)\,\mathrm{d}z\,,\qquad u\in\mathbb R$
[$\Phi$ denoting the CDF of standard normal distribution].
But is that all I can say? What should be my final answer to the distribution of $U$?