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Let $\mathbf{X} = (X_1,\ldots,X_n)$ and $\mathbf{Y} = (Y_1,\ldots,Y_n)$ be random vectors, and let $f_{\mathbf{X}}(x_1,\ldots,x_n)$ and $f_{\mathbf{Y}}(y_1,\ldots,y_n)$ be their respective pdfs or pmfs. If $X_i$ and $Y_i$ are independent for each $i$, i.e. $$f_{X_i,Y_i}(x,y) = f_{X_i}(x) f_{Y_i}(y) \;\;\;\text{for} \; i=1,2,\ldots,n$$ (where $f_{X_i}(x),f_{Y_i}(x)$ are the marginal pdfs/pmfs of $X_i$ and $Y_i$, respectively, and $f_{X_i,Y_i}(x,y)$ is the joint pdf/pmf of $X_i$ and $Y_i$), then is it necessarily true that

$$f_{\mathbf{X},\mathbf{Y}}(x_1,\ldots,x_n,y_1,\ldots,y_n) = f_{\mathbf{X}}(x_1,\ldots,x_n) f_{\mathbf{Y}}(y_1,\ldots,y_n) \; ?$$ ($f_{\mathbf{X}}$ and $f_\mathbf{Y}$ being the pdfs/pmfs of $\mathbf{X}$ and $\mathbf{Y}$, respectively, and $f_{\mathbf{X},\mathbf{Y}}$ being the joint pdf/pmf of $\mathbf{X}$ and $\mathbf{Y}$.) This seems like it should be true, and I've tried to prove it by induction on $n$, but I was not successful.

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No, this is not true. Here's a counterexample: You flip two fair coins (independently). $H_1$ is 1 if the first coin is heads and 0 if it's tails, while $T_1$ is 1 if the first coin is tails and 0 otherwise. We use analogous notation for the result of the second flip, $H_2$ and $T_2$.

Now consider $X = (H_1, H_2)$ and $Y = (T_2, T_1)$. Each component is independent (i.e. $H_1$ is independent of $T_2$, and $H_2$ is independent of $T_1$). However $f_{X,Y}(H_1=1, H_2=1, T_2=1, T_1=1) = 0$, rather than $f_{X}(H_1=1, H_2=1)f_Y(T_2=1, T_1=1) = (1/4)(1/4) = 1/16$.

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  • $\begingroup$ Thanks for this nice simple example! $\endgroup$ – Leonidas Jul 18 at 17:46
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As a sort of elaboration of @mlphd31's nice example (+1), suppose $X_1, X_2, \dots, X_{50}$ and $Y_1, Y_2, \dots, Y_{50}$ are all independently distributed as $\mathsf{Unif}(0,50),$ and that $X_{51}, X_{52}, \dots, X_{100}$ and $Y_{51}, Y_{52}, \dots, Y_{100}$ are all independently distributed as $\mathsf{Unif}(50,100).$ This scenario is simulated in R:

set.seed(716)
a=rep(c(0,50),50)
x = runif(100, a, a+50)
y = runif(100, a, a+50)

Then $\mathbf{X} = (X_1, \dots, X_{100})$ is not independent of $\mathbf{Y} = (Y_1, \dots, Y_{100}),$ as shown by non-zero correlation and a 2-dimensional plot.

cor(x, y)
[1] 0.7454673
plot(x, y, pch=20)

enter image description here

Perhaps closer to a practical example, for each $i = 1, 2, \dots, 100,$ let $X_i, Y_i$ be independently distributed as $\mathsf{Norm}(i+30, 5).$ [We may be weighing 100 standard specimens of known weights $i+30$ on two different low-quality scales in order to assess whether the two scales are of equal accuracy or precision.]

set.seed(2020)
mu = 31:130
x = rnorm(100, mu, 5)
y = rnorm(100, mu, 5)
cor(x, y)
[1]0.962425
plot(x, y, pch=20)

enter image description here

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  • $\begingroup$ Thanks for these thoughtful examples + plots! $\endgroup$ – Leonidas Jul 18 at 17:48

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