PDF of sum of random variables (with uniform distribution) How can I solve this:
Random variables $X,Y$ ~ Unif$(0, 1)$ are independent. Calculate the probability density function of sum $X + 3Y$.
I couldn't find a sum for uniformally  distributed random variables. I assume I have to go straight to the PDF and solve it that way.

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I am going to solve the above problem and hence you could follow the same for any similar problem such as this with not too much confusion.
$Z = X+ 3Y$ where X and Y are U(0,1).
I am going to define a new variable W where W is distributed according to U(0,3)
Thus $Z = X + 3Y = X+ W$ where X is U(0,1) and W is U(0,3).
Now I am going define the bounds
$t_{X_0} = 0$
$t_{X_1} = 1$
$t_{W_0} = 0$
$t_{W_1} = 3$
Thus $$f_Z(z) = 0, z \le t_{X_0}+t_{W_0} ,$$
$$f_Z(z) = \int_{max(t_{W_0}, t-t_{X_1})}^{min(t_{W_1}, t-t_{X_0})} f_W(w)f_X(z-w)dw,    \text{     }     t_{X_0}+t_{W_0} \le z \le t_{X_1}+t_{W_1},$$
$$f_Z(z) = 0, z \ge t_{X_1}+t_{W_1} ,$$
These translate to the following:
$$f_Z(z) = 0, z \le 0 ,$$
$$f_Z(z) = \int_{max(0, z-1)}^{min(3, z)} f_W(w)f_X(z-w)dw,    \text{     }     0\le z \le 4,$$
$$f_Z(z) = 0, z \ge 4 ,$$
$f_W(w) = \frac{1}{3}$ as $W$ is $U(0,3)$.
$f_X(x) = 1 $ as $X$ is $U(0,1)$,
The middle one needs to be split into three intervals, and they are a) $0\le z\le 1$, b) $1\le z\le 3$, and  c) $3\le z\le 4$.
Thus 
$f_Z(z) = \int_{0}^{z}\frac{1}{3}dw = \frac{z}{3}$, $0\le z\le 1$
$f_Z(z) = \int_{z-1}^{z}\frac{1}{3}dw = \frac{1}{3}$, $1\le z\le 3$
$f_Z(z) = \int_{z-1}^{3}\frac{1}{3}dw = \frac{4-z}{3}$, $3\le z\le 4$
Sanity check is to find if $\int_{0}^{4} f_Z(z) = 1$ which it is in this case and hence the solution.
Goodluck
