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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?

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    $\begingroup$ Are you assuming (or have you been told) that $X$ and $Y$ are independent but you have opted to not share this information with us? $\endgroup$ Commented Feb 28, 2015 at 14:08
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    $\begingroup$ I'm so sorry. Yes, they are independed. I haven't noticed that I forgot to write this. $\endgroup$
    – Hagrid
    Commented Feb 28, 2015 at 14:36

2 Answers 2

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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.

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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 $$P(Z \geq 1) = P(Z \in [1,2)) + P(Z \geq 2) = \frac{1}{3} + 0.$$

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    $\begingroup$ I see where I made a mistake, in the second summands we only take one value of $X$, because the probability that $X$ has values outside $[0,1]$ is zero. Thank you! $\endgroup$
    – Hagrid
    Commented Feb 28, 2015 at 15:10
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    $\begingroup$ I have one more question, though. Is this distribution continuous? $\endgroup$
    – Hagrid
    Commented Feb 28, 2015 at 15:52
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    $\begingroup$ In you count the distribution function : $$P(Z \leq z) = \frac{1}{3}P(X \leq z+1) + \frac{1}{3}P(X \leq z) + \frac{1}{3}P(X \leq z-1)$$ for all $z \in [0,1]$, you should see that Z has uniform distribution on $[-1,2]$. $\endgroup$
    – iiivooo
    Commented Feb 28, 2015 at 16:03

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