Adding two discrete distributions I am taking a probability course and I am having trouble adding two discrete distributions.  The two distributions given are:
$X$ has a discrete uniform distribution on the integers $0,1, ... ,9$.
$Y$ is independent, and has the probability distribution $\Pr(Y = k) = a_k$ for $ k=0,1, ...$
I am asked to find $Z = X + Y \mod 10$.
Attempt so far:
Reference: http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf
$\Pr(Z = z) = \sum_{k=-\infty}^{\infty} \Pr(X = k)\cdot \Pr(Y = z - k) = \sum_{k=-\infty}^{\infty} 1/9 \cdot a_{z-k}$
I'm slightly confused on whether or not I've finished the question.  It seems right per my reference, and that does define a probability distribution, but why is there a mod 10?  It seems like the textbook knows something I don't...
 A: Hint: the probability that $X+Y\equiv n\pmod{10}$ is
$$
\sum_{k=0}^9\Pr[X=k]\Pr[Y\equiv n-k\pmod{10}]
$$
Now consider the sum
$$
\sum_{k=0}^9\Pr[Y\equiv n-k\pmod{10}]
$$
A: Your attempt almost solves the question, but indeed there is a reason for the mod $10$. Your solution does not take into account that $z$ can only be a number between $0$ and $9$. Here is my solution:
Let $z \in \lbrace 0,..., 9 \rbrace$. Then we have due to independence:
$$ Pr(Z= z ) = Pr(X+Y = z) = \sum_{i=0}^{9}{Pr(X = i) Pr(Y = z- i \textrm{ mod } 10)} = \otimes$$
Now we have to take into account that $Y$ has to be equal to a number mod $10$. Hence we continue:
$$ \otimes = \sum_{i=0}^{9}{\frac{1}{10} \sum_{j=0}^{\infty}{Pr(Y = j \cdotp 10 + z-i)}} = \sum_{j=0}^{\infty}{\sum_{i=0}^{9}{\frac{1}{10} Pr(Y = j \cdotp 10 + z - i)}} $$ Thus we can conclude:
$$ P(Z = z) = \sum_{j=0}^{\infty}{\sum_{i=0}^{9}{\frac{1}{10} a_{j \cdotp 10 + z - i}}}$$
One can then check (for one's own amusement) that if we take the sum over all $z \in \lbrace 0,...,9 \rbrace$ of these expressions we find that it's equal to 1 (what is needed for a probability distribution).
A: Let me just comment on your approach, and note the difference between what you do and the $\mod 10$ problem. Remember what the ranges of your random variables $X$ and $Y$ are: The $X$ can attain the values in the set $\{0,1,\ldots,9\}$, and $Y$ can attain the values $0,1,\ldots$
In your sum, $\text{Pr}(X=k)$ only has a non-zero value for $k\geq 0$, and $\text{Pr}(Y=z-k)$ only has a non-zero value for $z-k\geq 0$ (meaning $k\leq z$). Thus, you can remove a lot of the terms in your sum.
Also, there are $10$ numbers in the set of values that $X$ can attain (because $0$ is there), so you want to replace your $\frac{1}{9}$ with $\frac{1}{10}$.
Thus the sum reduces to
$$P(Z=z) = \sum_{k=0}^{z} \frac{1}{10}a_{z-k}=\frac{1}{10}\sum_{k=0}^z a_{z-k}=\frac{1}{10}\sum_{k=0}^z a_k.$$
This is the distribution of $Z=X+Y$, but it looks like your problem wants the distribution of $Z= X + Y\mod 10$. The variable $Z=X+Y\mod 10$ can only attain the values $0,1,2,\ldots,10$, so for this version of the problem, you only need to figure out $P(Z=z)$, when $0\leq z\leq 10$. This amounts to thinking about the probability $P(X+Y\mod 10 = z)$.
