# Parallel Composition of ($\epsilon, \delta$) differential privacy

I know that if there are $n$ functions $M_1, M_2, \cdots, M_n$ computed on disjoint subsets of the private database whose privacy guarantees are $\epsilon_1, \cdots ,\epsilon_n$ differential privacy, respectively, then any function $g$ of them: $g(M_1, \dots,M_n)$ is $$(\max_i \epsilon_i)$$differentially private.

This is known as the Parallel Composition Theorem.

My question is: does this Parellel Composition Theorem extend to $(\epsilon, \delta)$ differential privacy? In other words, is the following statement true?:

If there are $n$ functions $M_1, M_2, \cdots, M_n$ computed on disjoint subsets of the private database whose privacy guarantees are $(\epsilon_1, \delta_1), \cdots ,(\epsilon_n, \delta_n)$ differential privacy, respectively, then any function $g$ of them: $g(M_1, \dots,M_n)$ is $$(\max_i \epsilon_i, \max_i \delta_i)$$differentially private.

If this statement is NOT true, then what bounds are there are on the privacy of the composition function $g$.

Your conjecture seems correct. If the sets are disjoint then the mapping $\bar{M} = (M_1,\dots,M_n)$ is $(\max(\epsilon_i),\max(\delta_i))$-DP. Note that for every database $D$ and any $x$, $\bar{M}(D)$ differs from $\bar{M}(D\setminus {x})$ in only one coordinate by the disjoint assumption, w.l.o.g., they differ in the $j$'th coordinate. Therefore, the difference between the density functions of $\bar{M}(D)$ and $\bar{M}(D\setminus {x})$ is the same as the difference between $M_j(D)$ and $M_j(D\setminus {x})$.