Distribution of function of two random variables Let $X$ be the number on a die roll, between 1 and 6. Let $Y$ be a random number which is uniformly distributed on $[0,1]$, independent of $X$. Let $Z = 10X + 10Y$.
What is the distribution of $Z$?
 A: Suppose $Y$ is another discrete random variables on $[0,1]$ with $\Pr(Y=0) = \Pr(Y=1) = \frac{1}{2}$. It helps to build the table of possible values of $Z = 10 X + 10 Y$:
$$
    \begin{array}{c|cccccc} Z(X,Y)  & X=1 & X=2 & X=3 & X=4 & X=5 & X+6 \cr \hline 
                            Y=0 & Z=10 & Z=20 & Z=30 & Z=40 & Z=50 & Z=60 \cr
                      Y=1 & Z = 20 & Z=30 & Z=40 & Z=50 & Z=60 & Z=70
    \end{array}
$$
Since there are 7 possible outcomes we compute their probabilities manually, e.g.:
$$ \begin{eqnarray}
  \Pr(Z=10) &=& \Pr(X=1,Y=0) = \frac{1}{6} \cdot \frac{1}{2} \\
  \Pr(Z=20) &=& \Pr(X=2,Y=0) + \Pr(X=1,Y=1) = \frac{1}{6} \cdot \frac{1}{2} + \frac{1}{6} \cdot \frac{1}{2} 
\end{eqnarray}
$$
and so on...
A: Hint:
$$
F_Z(x) = P(Z < x) = P(X + Y < x/10)
$$
Work it out by cases from here based on the potential values of $x$. For instance, if $x < 10$ then $x/10 < 1$, so $F_Z(x) = 0$ as $X \geq 1$. Another sample case: if $10 \leq x < 20$, then the value of $X$ in the right hand expression must be $1$ (in order for $X + Y < x/10$ to hold), and if $20 \leq x < 30$ then the value of $X$ must be either $1$ or $2$, each of which it takes with equal probability.
