In the Wikipedia page on differential privacy, the section on $\varepsilon$-differential privacy for the Laplace mechanism says that the Laplace mechanism is specified for some output as $\mathcal{T}_{\mathcal{A}}(x)=f(x)+ Y$, where $Y = (Y_1, \ldots, Y_n)$ and the $Y_i \sim \mathrm{Lap}(\lambda)$. Here's my question: Why can't we have $Y = (Y_1, \ldots, Y_1)$? It certainly seems simpler, and is still randomized. Furthermore, (I think) this definition still preserves $\varepsilon$-differential privacy (or at least I can't prove it doesn't). Where am I getting this wrong?



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