Distribution of sum of discrete and uniform random variables

Question

Let $X$ be the number on a roll of a fair 6-sided die. Let Y $\sim$ Uniform $(0,1)$, independent of $X$. Let $Z = 10X + 10Y$. What is the distribution of Z?

My attempt:

The PDF of $X$ is $$f_X(x) = \sum_{x_i \in \mathcal{X}} p_X (x_i) \, \delta (x - x_i), \mathcal{X}=\{x_i=i:i=1,2,3,4,5,6\}$$

and the PDF of Y is obviously

$$f_Y(y)=\begin{cases}1, & x\in(0,1)\\0,&x\notin(0,1)\end{cases}$$

We know $X$ and $Y$ are independent, and therefore the joint density is

$$f_{(X,Y)}(x,y)=f_X(x)f_Y(y)$$

Let $Z^{'} = X+Y$ Then

$$f_{Z^{'}}(z^{'})=\int_{-\infty}^\infty{f_{(X,Y)}(x,z^{'}-x)dx}=\int_{-\infty}^\infty{f_{X}(x)f_Y(z^{'}-x)dx}$$

Now, if $Z =10Z^{'}$, then by a linear change of variables for densities we find that

$$f_Z(z)=\frac{1}{10}f_{Z^{'}}(\frac{z}{10})$$

I am unclear how to utilize this method because the bounds of the integral cause me some problems as well as not covering the convolution in class. The solution doesn't even use convolution, only a mere "check" to convince the reader the distribution is uniform. Any way to do this without convolution?

I have found a couple helpful posts (1,2), but can't come up with a final solution. My key tells me $Z \sim$ Uniform $(10,70)$.

Answer 1

From your stats.stackexchange link, you want to use the "formula": $$f_Z(z) = \sum_{x_i\in \mathcal{X}}f_Y(z-x_i)P(X = x_i)$$ We have that the dice is fair, so $P(X = x_i) = \frac{1}{6}$. So, we have that: $$f_Z(z) = \frac{1}{6}\sum_{x_i\in\mathcal{X}} f_Y(z-x_i)$$ As $x_i$ takes integer values, for at most one $x_i$ we will have that $z-x_i\in (0,1)$. So, we get that: $$f_Z(z) = \frac{1}{6}f_Y(y-x_i') = \frac{1}{6}$$ Note that this holds if $z-x_i\in(0,1)$, so we want $z\in(x_i,1+x_i)$. This is $(1,2)\cup(2,3)\cup\dots\cup(6,7)$. This agrees with $Z\sim\text{Unif}(1,7)$ except for on a set of probability $0$. So, we have that $Z\sim\text{Unif}(1,7)$, so $10Z\sim\text{Unif}(10,70)$.

Note you could get to the "formula" I used from your work. We have that: \begin{align*} \int_{-\infty}^\infty f_X(x)f_Y(z'-x)\mathrm{d}x & = \int_{-\infty}^\infty\sum_{i = 1}^6p_X(i)\delta(x-i)f_Y(z'-x)\mathrm{d}x \\ & = \sum_{i = 1}^6\int_{-\infty}^\infty p_X(i)f_Y(z'-x)\delta(x-x')\mathrm{d}x \\ & = \sum_{i= 1}^6 p_X(i)f_Y(z'-i) \end{align*} Here, we used that $\int_A f(x)\delta(x-a) = f(a)$ if $a\in A$.

Answer 2

The distribution of $Z$ is given by $P(Z \le z)$ so let's calc that:

\begin{align*} P(Z \le z) &= P(X+Y \le z) \\ &= \sum_{k=1}^6 P(X+k \le z,Y=k) \\ &= \sum_{k=1}^6 P(X+k \le z)P(Y=k) \\ &= \frac{1}{6} \sum_{k=1}^6 P(X \le z-k)\end{align*}