If $\mathbf{X}_1$ and $\mathbf{X}_2$ are i.i.d. random vectors, are $\mathbf{a}\cdot \mathbf{X}_1$ and $\mathbf{a}\cdot \mathbf{X}_2$ i.i.d.?

Question

If $\mathbf{X}_1 = (X_{11}, \ldots, X_{1k})$ and $\mathbf{X}_2 = (X_{21}, \ldots, X_{2k})$ are i.i.d. random vectors, how do we know that $\mathbf{a}\cdot \mathbf{X}_1$ and $\mathbf{a}\cdot \mathbf{X}_2$ are i.i.d. (for $\mathbf a \in \mathbb R^k$)? The below proof seems to use this fact, and I don't know how to prove it, because I don't think the $X_{11}, \ldots, X_{1k}$ are necessarily independent, just that the two vectors must be.

This source: https://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf says that if $X$ and $Y$ are independent, their sum's distribution function is only dependent on $X$ and $Y$'s distributions (i.e. in this case, if $X =_d X'$ and $Y =_ d Y'$ and $X$ and $Y$ are independent and $X'$ and $Y'$ are independent, we have that $X+Y =_d X'+Y'$). But I don't know what to do in the case that they might not be independent.

Answer 1

We have that two variables $X,Y$ are independent if the cumalative distribution functions

$$F_{X \times Y}(X_1\leq x_1, ... , X_n\leq x_n, Y_1\leq y_1, ... , Y_n\leq y_n)\\ = F_{X}(X_1\leq x_1, ... , X_n\leq x_n) F_{Y}(Y_1\leq y_1, ... , Y_n\leq y_n );$$

therefore if $X,Y$ are independent then we have that

$$F_{aX \times aY}(aX_1\leq x_1, ... , aX_n\leq x_n, aY_1\leq y_1, ..., aY_n\leq y_n )\\ = F_{X\times Y}(X_1\leq a^{-1}x_1, ... , X_n\leq a^{-1}x_n, Y_1\leq a^{-1}y_1, ... , Y_n\leq a^{-1}y_n )\\ =F_{X}(X_1\leq a^{-1}x_1, ..., X_n\leq a^{-1}x_n) F_{Y}(Y_1\leq a^{-1}y_1, ... , Y_n\leq a^{-1}y_n )\\ = F_{aX}(aX_1\leq x_1, ... , aX_n\leq x_n) F_{aY}(aY_1\leq y_1, ..., aY_n\leq y_n)$$

Similarly for the identical part if

$$ F_{X}(X_1\leq c_1, ... , X_n\leq c_n) = F_{Y}(Y_1\leq c_1, ... , Y_n\leq c_n )$$

then we must have that

$$ F_{aX}(aX_1\leq c_1, ... , aX_n\leq c_n) = F_{aY}(aY_1\leq c_1, ... , aY_n\leq c_n )$$

by a similar arguement.