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I have the following question:

If two random vectors $X$ and $Y$ have the same distribution and $X$ has independent components, does $Y$ have also independent components?

If it is not true, why is it true when $X$ is a $d$-Gaussian vector?

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    $\begingroup$ The measure induced by $Y$ is a product measure so the components of $Y$ are necessarily independent. Normality is not required. $\endgroup$ Commented May 28, 2021 at 9:57
  • $\begingroup$ The measure induced by Y is a product measure because it is the same as the one induced by $X$, and this last one is a product measure as the components of $X$ are independent. Isn't it? $\endgroup$
    – Eparoh
    Commented May 28, 2021 at 10:13
  • $\begingroup$ Exactly. You seem to have a neat proof already. $\endgroup$ Commented May 28, 2021 at 10:14

1 Answer 1

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$\quad \mathbb P(Y_1 \le y_1, Y_2 \le y_2, \ldots , Y_n \le y_n)$ the CDF of $\mathbf Y$

$=\mathbb P(X_1 \le y_1, X_2 \le y_2, \ldots , X_n \le y_n)$ since $\mathbf{X}$ and $\mathbf Y$ have the same distribution

$=\mathbb P(X_1 \le y_1) \mathbb P(X_2 \le y_2) \cdots \mathbb P(X_n \le y_n)$ by independence of the components of $\mathbf X$

$=\mathbb P(Y_1 \le y_1) \mathbb P(Y_2 \le y_2) \cdots \mathbb P(Y_n \le y_n)$ since $\mathbf{X}$ and $\mathbf Y$ have the same distribution

So the components of $\mathbf Y$ are also independent and this does not depend on the shapes of the individual marginal distributions of the components

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    $\begingroup$ Why is the last equality true? If the distribtion on two random vectors are the same, is then the distribution of evety component the same? $\endgroup$
    – Eparoh
    Commented May 28, 2021 at 10:16
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    $\begingroup$ @Eparoh The marginal distribution of each component is determined by the overall distribution so yes. For example $\mathbb P(Y_1 \le y_1) = \mathbb P(Y_1 \le y_1, Y_2 \lt +\infty,,\ldots, Y_n \lt +\infty )$ $= \mathbb P(X_1 \le y_1, X_2 \lt +\infty,,\ldots, X_n \lt +\infty ) = \mathbb P(X_1 \le y_1)$ $\endgroup$
    – Henry
    Commented May 28, 2021 at 12:47

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