Differential privacy guarantees of Gaussian noise, when each coordinate has different sensitivity Suppose you have a function $f$ that takes a dataset $D$ as input and returns an output in $\mathbb{R}^d$.
If this function has $L^2$-sensitivity $\Delta$, then the analytical Gaussian mechanism (Theorem 8 in this paper) says that if you add Gaussian noise of variance $\sigma^2$ to each coordinate of result, with:
$$ \Phi\left(\frac{\Delta}{2\sigma}-\frac{\epsilon\sigma}{\Delta}\right)-e^\varepsilon\Phi\left(-\frac{\Delta}{2\sigma}-\frac{\epsilon\sigma}{\Delta}\right) \le \delta$$
where $\Phi$ is the Gaussian CDF, then you obtain an $(\varepsilon,\delta)$-differentially private mechanism.
Now, suppose that there is a finer way of describing the sensitivity of $f$. Rather than only knowing a bound on the maximum $L^2$ norm of $f(D_1)-f(D_2)$ for neighboring $D_1$ and $D_2$, we have a per-coordinate sensitivity bound: we know that the first coordinate of $f(D_1)-f(D_2)$ is always below $\Delta_1$ (in absolute value), the second below $\Delta_2$, etc., and $\Delta_d$ bounds the sensitivity along the $d$-th coordinate.
In this case, intuitively, adding the same magnitude of noise along each coordinate doesn't seem like the best solution. For example, if $\Delta_1$ is much smaller than the other per-coordinate sensitivities, then we will likely add too much noise to the first coordinate for it to be useful. Thus my question: is there an equivalent analytical result where we can add Gaussian noise proportional to each coordinate sensitivity?
I know I could be using Laplace noise instead, but then the per-coordinate noise magnitude grows in $O(d)$ instead of $O(\sqrt{d})$ (unless $d$ is sufficiently big to use the Advanced Composition Theorem, but that only makes a big difference for large values of $d$), so I'm interested in a Gaussian noise formula hoping that it will work well for not-too-high values of $d$ (say, $5&lt;d&lt;50$).
I haven't read your full question, but the answer to:

Is there an equivalent analytical result where we can add Gaussian noise proportional to each coordinate sensitivity?

and (implicitly)

Can the noise scale better than $O(d)$ for $d$-dimensional output?

Then the answer is yes.
The following should be able to be easily adapated for essentially any mechanism $\mathcal{M}$, but I'll focus on the case that the mechanism is adding independent mean-zero Gaussian noise, as this is your case.
Moreover, this argument should work provided that the sensitivity is &quot;homogenous&quot; in a certain sense --- in particular, any sensitivity defined with respect to a norm should work, so any $\ell_p$ sensitivity (for $p\geq 1$ I think).
Throughout this, I will write $\vec{\Delta} = (\Delta_1,\dots, \Delta_d)$ to be a vector of the per-coordinate sensitivities of $f(x)$.
I will write the $i$th coordinate of $f(x)$ as $f(x)_i$.
I will also write $\mathsf{diag}(\vec{\Delta})$ to denote the $d\times d$ diagonal matrix with $(i, i)$ entry $\Delta_i$.
The idea here is to transform $f(x)$ as follows:
$$g(x) = \mathsf{diag}(\vec{\Delta})^{-1}f(x)$$
If you compute the $\ell_2$sensitivity of $g(x)$, you get that:
\begin{align*}
\Delta_2(g) &amp;= \max_{x, y} \lVert g(x) - g(y)\rVert\\
&amp;= \max_{x, y}\lVert \mathsf{diag}(\vec{\Delta})^{-1} (f(x)- f(y))\rVert\\
&amp;= \max_{x, y}\sqrt{\sum_{i\in[d]} \left(\frac{|f(x)_i-f(y)_i|}{\Delta_i}\right)^2}\\
&amp;\leq \sqrt{\sum_{i\in[d]} \left(\max_{x, y}\frac{|f(x)_i-f(y)_i|}{\Delta_i}\right)^2}\\
&amp;\leq \sqrt{\sum_{i\in[d]} \left(\frac{\Delta_i}{\Delta_i}\right)^2}\\
&amp; = \sqrt{d}
\end{align*}
So, we have that $\Delta_2(g)\leq \sqrt{d}$.
I'm pretty sure this should extend in a straightforward way to give you that $\Delta_p(g) \leq \sqrt[p]{d}$, which would explain why you can only get the upper bound of $O(d)$ for the $\ell_1$ sensitivity.
Anyway, now we can just apply (essentially any) differentially private mechanism to $g(x)$.
In particular, the Gaussian mechanism states that:
$$h(x) = g(x) + \mathcal{N}^d(0, 2\ln(1.25/\delta)n/\epsilon^2)$$
Is $(\epsilon, \delta)$ differentially private.
For what follows, recall that differential privacy is closed under post-processing, so &quot;we're done&quot; on the topic of privacy.
$h(x)$ has the undesirable property that $\mathbb{E}[h(x)] = \mathbb{E}[g(x)] + 0 = \mathbb{E}[\mathsf{diag}(\vec{\Delta})^{-1}f(x)] \neq f(x)$, so we have (potentially significantly) biased the output of your function.
Fortunately, we can fix this by having our output be $\mathsf{diag}(\vec{\Delta})h(x)$ (which you can quickly check leads to an unbiased output).
In total, the mechanism which outputs:
$$f(x) + \mathcal{N}\left(0, \frac{2\ln(1.25/\delta)n}{\epsilon^2}\mathsf{diag}(\vec{\Delta})^2\right)$$
Is therefore $(\epsilon, \delta)$ differentially private.
This is what you would expect --- if you have sensitivities which are different in each coordinate, then instead of &quot;spherical&quot; Gaussian noise, one should use ellipsoidal Gaussian noise &quot;matched&quot; to each coordinate, which is precisely what the mechanism does when you write it all out.
Note that the gaussian noise is no longer the product of $d$ i.i.d. gaussians (but still has quite simple covariance structure).