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How can one formalize the fact that the law of $X+Z$ where $X \in \mathbb{R}^d$ is any vector-valued random variable and $Z\sim \mathcal{N}(0, \sigma^2 \mathbf{I}_d)$ closely resembles the law of $Z$ if $\sigma^2$ is sufficiently large ? $X$ and $Z$ are supposed independent. Ideally, I would like to prove that some distance/divergence between the law of $X+Z$ and $Z$ approaches zero as $\sigma^2\to \infty$.

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If 𝑋 were Gaussian, for example, with a non-zero mean, $𝜇_{𝑋+𝑍}=𝜇_𝑋+𝜇_𝑍≠𝜇_𝑍.$

$\displaystyle \lim_{\sigma\rightarrow \infty} (\mu_{X+Z} - \mu_{Z})=\mu_X \ne 0.$

So the difference between the law of $X+Z$ and the law of $Z$ does not approach zero.

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  • $\begingroup$ This was meant as a comment ... Not an "answer" for now, but let's see how the discussion goes. Thanks. $\endgroup$ – mjw Mar 26 at 18:47
  • $\begingroup$ $X$ is any fixed vector-valued random variable, i.e. the law of $X$ cannot change or depend on the law of $Z$. $\endgroup$ – Nocturne Mar 26 at 19:18
  • $\begingroup$ $X$ is constant? or a random variable? $\endgroup$ – mjw Mar 26 at 19:39
  • $\begingroup$ Okay, $X\sim \mathcal{N}(0, \sigma_0^2 \mathbf{I}_d)$, for example. A fixed distribution, whereas you want to show what happens as the distribution of $Z$ spreads out, yes? $\endgroup$ – mjw Mar 26 at 19:43
  • $\begingroup$ Well, if $X$ were Gaussian with a non-zero mean, $\mu_{X+Z} = \mu_X + \mu_Z \ne \mu_Z$. $\endgroup$ – mjw Mar 26 at 19:45

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