[Math] Adding two discrete distributions

Question

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…

Answer 1

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 $$ \sum_{k=0}^9\Pr[Y\equiv n-k\pmod{10}] $$