# Two random vectors with the same distribution and one with independent components

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?

• The measure induced by $Y$ is a product measure so the components of $Y$ are necessarily independent. Normality is not required. Commented May 28, 2021 at 9:57
• 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? Commented May 28, 2021 at 10:13
• Exactly. You seem to have a neat proof already. Commented May 28, 2021 at 10:14

$$\quad \mathbb P(Y_1 \le y_1, Y_2 \le y_2, \ldots , Y_n \le y_n)$$ 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
• @Eparoh The marginal distribution of each component is determined by the overall distribution so yes. For example $\mathbb P(Y_1 \le y_1) = \mathbb P(Y_1 \le y_1, Y_2 \lt +\infty,,\ldots, Y_n \lt +\infty )$ $= \mathbb P(X_1 \le y_1, X_2 \lt +\infty,,\ldots, X_n \lt +\infty ) = \mathbb P(X_1 \le y_1)$ Commented May 28, 2021 at 12:47