What is $\Pr(X + Y < 0)$ where $X \sim U(0,1)$ and $Y \sim N(0, 1)$? $X$ and $Y$ are independent This is what I have attempted so far:
\begin{align}
    f_X = 1 \\
    f_Y = \frac{1}{\sqrt{2\pi}}\exp(-0.5y^2) 
\end{align}
Then let $Z = X + Y$ and we have
\begin{align}
    f_Z(z) = \int_0^1 f_X(x) f_Y(z - x) \, dx \\
    f_Z(z) = \int_0^1 1 \cdot f_Y(z - x) \, dx \\
    = \int_0^1 \frac{1}{\sqrt{2\pi}}\exp(-0.5(x - z)^2) \, dx
\end{align}
So
\begin{align}
Pr(Z \leq 0) = \int_{-\infty}^0 \int_0^1 \frac{1}{\sqrt{2\pi}}\exp(-0.5(x-z)^2) \, dx \, dz \\
= \int_{-\infty}^0 \int_0^1 \frac{1}{\sqrt{2\pi}}\exp(-0.5x^2) \exp(- 0.5z^2) \exp(0.5xz)\,dx\,dz \\
\end{align}
This looks like it's going to be a tedious integral to evaluate. I am not sure if I am taking the right approach. Is there an easier method for this?
 A: Assuming $X,\,Y$ are independent:
We want to $Y$-average $Pr(X<-Y)$, which at fixed $Y$ is $0$ if $Y\ge0$, $1$ if $Y<-1$ and $-Y$ otherwise. The average is$$\int_{-\infty}^{-1}f_Y(y)dy-\int_{-1}^0yf(y)dy=\Phi(-1)+\tfrac{1-e^{-1/2}}{\sqrt{2\pi}}\approx0.315.$$
A: It is a very big error avoiding to state that X and Y are independent. As it is written, the exercise cannot be solved.
So, assuming independence, first observe that if $Y<-1$ it is always true that $X+Y<0$ and this happens with probability $\Phi(-1)\approx 15.87\%$
For the rest, when $Y>-1$ the integral to be solved is
$$\int_{-1}^{0}\phi(y)dy\int_{0}^{-y}dx=-\frac{1}{\sqrt{2\pi}}\int_{-1}^{0}ye^{-\frac{y^2}{2}}dy=\frac{1}{\sqrt{2\pi}}[e^{-\frac{y^2}{2}}]_{-1}^{0}=\frac{1-e^{-0.5}}{\sqrt{2\pi}}$$
It is the integral in the purple area below

A: I think it's better to derive the full distribution of $Z=X+Y$ using convolution formula for CDFs. When I used convolution for pdfs, I got
$$
f_Z(z) = \Phi(z)-\Phi(z-1), -\infty<z<\infty
$$
which is very hard to integrate, so I used convolution for CDFs instead. Not that if $Y \sim R(0,1)$, then $F_Y(y) = P(Y<y) = P(Y<z-x)$, hence:
$$
F_Y(z-x)= \left\{
\begin{array}{lr}
0 & x>z\\
z-x & 0<z-x<1\\
1 & x<z-1
\end{array} 
\right.
$$
So we can ignore the pdf of $X$ if $ X>z$. For the second case, we have the following bounds: $z-1<x<z$, and the CDF of $Y$ is $z-x$, for the third case, the CDF of $Y$ is $1$, so we just take the pdf of $X$ for $-\infty<x<z-1$. Since $-\infty <z<\infty$, we just put these three cases together:
\begin{align}
P(Z<z) &= F_Z(z) = \int_{z-1}^{z}(z-x)\varphi_X(x)dx + \Phi(z-1) \\
&= z(\Phi(z)-\Phi(z-1)) - \int_{z-1}^{z}x\varphi(x)dx + \Phi(z-1), -\infty <z< \infty
\end{align}
where $\varphi, \Phi$ are density and cdf of standard normal distribution. By plugging $z=0$ you get the result. Note that this CDF makes sense, because
$$
\lim_{z \to \pm \infty} z(\Phi(z)-\Phi(z-1)) = 0 \ \ (1)\\
\lim_{z\to \infty} \int_{z-1}^{z}x\varphi(x)dx  = 0 \ \ (2)\\
\lim_{z \to \infty} F_Z(z) = 1\\
\lim_{z \to -\infty} F_Z(z) = 0
$$
Here both (1) and (2) can be proven by taking upper and lower bounds on $z$ and $x$ for the corresponding intervals, and then taking the limit. Also note $z-x$ is always positive, so the whole expression is always positive. Now take the derivative wrt $z$ (careful with the signs) to get
$$
f_Z(z) = \Phi(z) - \Phi(z-1), \ -\infty <z< \infty
$$
Also check the limits.
