If X and Y are two independent standard normal r.v.s, calculate $P(X+Y \in [0,1] \mid X \in [0,1])$ I'm not sure how to solve this. This is my attempt so far.
So if X,Y two independent standard normal r.v.s, we have:
$$\mathbb{P}(X+Y\in [0,1] \mid X \in [0,1])=\frac{\mathbb{P}(\{X+Y\in [0,1]\} \cap \{X \in [0,1]\})}{\mathbb{P}(X \in [0,1])}.$$
Moreover, we have:
\begin{split} \mathbb{P}(\{X+Y\in [0,1]\} \cap \{X \in [0,1]\}) = {} &  \int_0^{1}dx\frac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}{2}}\int_{-x}^{1-x} dy\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}.
\end{split}
To calculate the integral, it looks like it might be better to switch to polar coordinates (?). Then we have:
\begin{split} \int_0^{1}dx\frac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}2}\int_{-x}^{1-x} dy \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}  = & \frac{1}{2\pi}\int_{-\frac\pi4}^{0} \, d \varphi \int_0^{\frac{1} {  \cos \varphi}}r e^{-\frac{r^2}2} \, dr \\
& +  \frac{1}{2\pi}\int_{0}^{\frac\pi2} \, d \varphi \int_0^{\frac1 { \sin \varphi + \cos \varphi}}r e^{-\frac{r^2}2} \, dr \\
= &- \frac{1}{2\pi}\int_{-\frac\pi4}^{0} \, d \varphi \int_0^{-\frac1 { 2\cos^2 \varphi}} e^t \, dt \\
&  - \frac{1}{2\pi}\int_{0}^{\frac{\pi}{2}}d \varphi \int_{0}^{-\frac{1}{2 ( \sin \varphi + \cos \varphi) ^2}} e^t\,dt, \\
\end{split}
and I don't know what to do from here. Although, I'm not even sure if my procedure doesn't have any mistakes. Thanks for any insights.
 A: Write
$$ \{ X +Y \in [0, 1], X \in [0, 1]\} = \{ (X, Y) \in D_1 \} \cup \{ (X, Y) \in D_2 \}, $$
where
\begin{align*}
D_1 &= \{(x, y) : 0 \leq x \leq 1 \text{ and } 0 \leq y \leq 1 \text{ and } 0 \leq x+y \leq 1 \} \\
D_2 &= \{(x, y) : 0 \leq x \leq 1 \text{ and } -1 \leq y \leq 0 \text{ and } 0 \leq x+y \leq 1 \}.
\end{align*}
$\hspace{14.5em}$
Then by symmetry,
\begin{align*}
\Bbb{P}(X +Y \in [0, 1], X \in [0, 1])
&= \Bbb{P}((X, Y) \in D_1) + \Bbb{P}((X, Y) \in D_2) \\
&= \tfrac{1}{4}\Bbb{P}((X, Y) \in \tilde{D}_1) + \tfrac{1}{8}\Bbb{P}((X, Y) \in [-1,1]^2)
\end{align*}
where
$$ \tilde{D}_1 = \{(x, y) : -1 \leq x + y \leq 1 \text{ and } -1 \leq x - y \leq 1 \} $$
$\hspace{11em}$
is the square with corners $(\pm 1, 0)$ and $(0, \pm1)$. Finally, using the fact that the law of $(X, Y)$ is rotation invariant, we can replace $\tilde{D}_1$ by its $45^{\circ}$ rotation without affecting the probability:
\begin{align*}
\Bbb{P}(X +Y \in [0, 1], X \in [0, 1])
&= \tfrac{1}{4}\Bbb{P}((X, Y) \in [-\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}]^2) + \tfrac{1}{8}\Bbb{P}((X, Y) \in [-1,1]^2) \\
&= \Bbb{P}(X \in [0, \tfrac{1}{\sqrt{2}}])^2 + \frac{1}{2}\Bbb{P}(X \in [0, 1])^2 \\
&= \left( \Phi\left(\tfrac{1}{\sqrt{2}}\right) - \tfrac{1}{2}\right)^2 + \tfrac{1}{2}\left( \Phi(1)- \tfrac{1}{2}\right)^2.
\end{align*}
A: The desired probability is $$\Pr[0 \le X+Y \le 1 \mid 0 \le X \le 1] = \frac{\int_{x=0}^1 \int_{y=-x}^{1-x} f_{X,Y}(x,y) \, dy \, dx}{\int_{x=0}^1 f_X (x) \, dx}.$$  It is not necessary to separate the cases $X \le 1/2$ versus $X > 1/2$.  We find the denominator easily enough:  $$\int_{x=0}^1 f_X(x) \, dx = \Phi(1) - \tfrac{1}{2}.$$  The numerator is harder:
$$\begin{align*} \int_{x=0}^1 f_X(x) \int_{y=-x}^{1-x} f_Y(y) \, dy \, dx 
&= \int_{x=0}^1 f_X(x) \left(\Phi(1-x) - \Phi(-x)\right) \, dx \\ 
&= \int_{x=0}^1 f_X(x) (\Phi(1-x)+\Phi(x) - 1) \, dx \\ 
&=\left[\frac{\Phi(x)^2}{2} \right]_{x=0}^1 - \left(\Phi(1) - \frac{1}{2}\right) + \int_{x=0}^1 f_X(1-x) \Phi(x) \, dx \\
&= \frac{\Phi(1)^2}{2} + \frac{3}{8} - \Phi(1) + \int_{x=0}^1 f_X(1-x) \Phi(x) \, dx.
\end{align*}$$  Here, the last integral does not seem to have an elementary closed form.  The approximate value of the numerator through numeric integration is $0.125988$; the resulting probability is approximately $0.369094$.
A: Here is a simulation of your problem in  R statistical software, which
illustrates and confirms the results of @heropup and (just now) @SangChul Lee. (both +1)
I generate a million realizations of $X, Y,$ and $S = X + Y,$ and
focus on the values of $S|X \in (0,1),$ and then find the proportion
of these conditional $S$'s in $(0,1),$ which is  0.3695 correct to about
three places.
m = 10^6;  x = rnorm(m);  y = rnorm(m);  s = x+y
cond = (abs(x-.5) < .5)
mean(abs(s[cond]-.5) < .5)
## 0.3695232

In order to visualize this graphically, I reduce the number of simulated
values to 100,000 (to keep the scatterplot from being too crowded). 
 
The scatterplot at left suggests the bivariate distribution of $S$ and $X.$
The vertical blue band shows the points representing the conditional distribution
of $S.$ The denominator of the desired probability is the proportion of points
in the vertical blue band, and its numerator is the proportion of points in
the square bounded by red and blue lines. 
The (blue) conditional points of $S$ are shown in the histogram at the
right, and the desired probability is the area under the histogram between
the vertical red lines.
