$Y_1, Y_2$ is a random sample of size $2$ from the uniform distribution on the interval $(0,1).$ Find the p.d.f. of $U = Y_1+Y_2$
I know the answer to the question, but don't understand how to get it. Particularly, I don't understand why the picture for the region $1 < u \le 2$ has the top right corner as $y_2<u-y_1$ and that the sides of the triangle is (2-u).
To answer your first conceptual question intuitively, the picture of the (triangular) PDF becomes more clear when we recall that for a proper probability distribution, we need the area under the PDF curve to be exactly $1$. Certainly for a uniform distribution on $[0,1]$, the area is simply the area of the unit square. But for a triangular distribution on $[0,2]$, the area is a triangle with base length 2 and height of 1.
To answer your last question specifically, consider the PDF of the triangular distribution is split as:
$$f(u) = \begin{cases} u & \text{for $0 < z < 1$} \\ 2-u & \text{for $1 \le z < 2$} \\ 0 & \text{otherwise}. \end{cases}$$
The reason why $f(u) = 2 - u$ for $1 < u \le 2$ is because the probability density linearly decreases from the midpoint of $1$.
Here's some qualitative intuition for the shape of $Y_1 + Y_2$ here. We know the endpoints of the PDF, $0$ and $2$, since those are the minimum and maximum values that the sum could take (the minimum value being attained at $Y_1 + Y_2 = 0$ from $Y_1 = Y_2 = 0$ and the maximum at $Y_1 + Y_2 = 2$ from $Y_1 = Y_2 = 1.$ We know that the distribution should be symmetric because it is the sum of two i.i.d. symmetric random variables. Intuitively, we know that most of the mass of the PDF should be concentrated on the midpoint $1$ since there are many "combinations" of $Y_1$ and $Y_2$ that can sum to middle values whereas there are much fewer "combinations" of $Y_1$ and $Y_2$ that can sum to extreme values.
As for a mathematical proof, this post addresses the convolution in detail: density of sum of two uniform random variables $[0,1]$
