Sum of discrete and continuous random variables with uniform distribution Could you tell me how to find the distribution of $Z = X+Y$ if $X$ is a random variable with uniform distribution on $[0,1]$ and $Y$ has uniform distribution on $\{-1,0,1\}$?
$X$ and $Y$ are independent.
I know how to find distributions of sums of random variables if both are discrete or both are continuous. But here we have a mix.
I guess I should consider $P(Z<-1) = P(\{X<-1, Y<0\} \cup \{X<0, Y<-1\})=0$
$P(Z \in [-1,0)) = P(X \in [-1,0), Y=0 \ \text{or} \ X \in [0,1), Y=-1 \ \text{or} \ X \in [-2, -1), Y=1) = \frac{1}{3}$
$P(Z \in [0,1)) = P(X \in [0,1), Y=0 \ \text{or} \ X \in [1,2), Y=-1 \ \text{or} \ X \in [-1, 0), Y=1) = \frac{2}{3}$
$P(Z \in [1,2)) = P(X \in [1,2), Y=0 \ \text{or} \ X \in [2,3), Y=-1 \ \text{or} \ X \in [0, 1), Y=1) = 1$
And so $P(Z \ge 1) = 1$
Is that correct?
 A: You can extend the convolution method for summing continuous independent variables if you identify the "density" of a discrete variable as a sum of Dirac deltas. Here you find that the density of $X$ is $1_{[0,1]}$ while the "density" of $Y$ is $\frac{1}{3} \sum_{i=-1}^1 \delta_i$. So you now need to compute the convolution:
$$g(x) = \frac{1}{3} \sum_{i=-1}^1 \int_{-\infty}^\infty \delta_i(y) 1_{[0,1]}(x-y) dy \\
= \frac{1}{3} \sum_{i=-1}^1 \int_{x-1}^x \delta_i(y) dy$$
By the definition of the Dirac delta, the $i$th summand is $1$ if $i \in (x-1,x)$ and zero otherwise. So exactly one summand will be $1$ if and only if $x \in (-1,2)$, so the density here is $\frac{1}{3} 1_{(-1,2)}$.
In this case this is pretty much the same as the other answer; this method is nicer in more complicated examples.
A: Notice, that events $Z \in [-1,0)$, $Z \in [0,1)$ and $Z \in [1,2)$ have empty intersection. But the according to your solution: 
$$P(Z \in [-1,0)) + P(Z \in [0,1)) + P(Z \in [1,2)) > 1$$
which clearly could not be correct. 
So lets count, the desired probabilities once again, using the fact that X,Y are independent: 
\begin{align}
P(Z \in [-1,0)) &= P(X \in [0,1))P(Y=-1) = 1\times\frac{1}{3}=\frac{1}{3}, \\
P(Z \in [0,1)) &= P(X \in [0,1))P(Y=0) + P(X = 1)P(Y=-1) = 1\times\frac{1}{3} + 0\times\frac{1}{3} = \frac{1}{3}, \\
P(Z \in [1,2)) &= P(X \in [0,1))P(Y=1) + P(X = 1)P(Y=0) = 1\times\frac{1}{3} + 0.\times\frac{1}{3} = \frac{1}{3}.
\end{align}
Of course $P(Z < - 1)= 0$, but we have already counted 
$$P(Z \geq 1) = P(Z \in [1,2)) + P(Z \geq 2) = \frac{1}{3} + 0.$$
