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?
Best Answer
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.