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Suppose $\bar X$ (respectively $ \bar Y$) is an $ R^m$ (resp. $R^n$) valued random vector with density f(.) and g(.). Let $\phi: R^n \rightarrow R^m$ be $C_1$. If X,Y are independent, then show that $(\bar X + \phi(\bar Y ) | \bar Y = \bar a) \rightarrow^d \bar X + \phi(\bar a)$.

I thought of using convolutions, but how would that even work in n (m?) dimensions? Would the density of $\phi(Y)$ be $g(\phi ^{-1} (X)). J(\phi^{-1})$? I can't seem to wrap my head around it.

AvHz
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  • Don't bother with the convolution and densities; go with the characteristic function. – kimchi lover Nov 03 '19 at 18:14
  • Could I possibly use MGFs instead? Characteristic functions haven't been covered, but the approach seems (and I use the term very loosely) similar. – AvHz Nov 03 '19 at 18:18

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