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From "The probability Tutoring Book" by Carol Ash,

If a joint density function $f_{x,y}(x,y)$ is separable and the universe is a rectangle, then $X$ and $Y$ are independent.

My question is why is this true?

I know that a function is separable if it can be written as the product of two functions, each of which is only dependent on one variable.

The second requirement simply means $a \leq X \leq b$ and $c \leq Y \leq d$, right?

But I'm not understanding why these two conditions make X and Y independent...

Also, if I have three variables, then does this "theorem" still hold true if the density is separable and $a \leq X \leq b\text{ and }c \leq Y \leq d\text{ and } e \leq Z\leq f$?

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  • $\begingroup$ asking for clarification on your question ... as an alternative, for example, to the rectangle part of the question, if X and Y were limited to the upper right half triangle of the rectangle bound by (a,c) and (b,d) then the two random variables would not be independent even if $f_{x,y}(x,y)=f_x(x)f_y(y)$? $\endgroup$ Jul 22, 2020 at 19:58
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    $\begingroup$ @phdmba7of12 The theorem requires the joint density $f_{X,Y}(x,y)$ to exist and be a product of two functions $g(x) h(y)$ for all real numbers $x,y$. $\endgroup$ Jul 22, 2020 at 20:53
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    $\begingroup$ You can define those functions "piecewise", so you might have $g(x) = 0$ outside some interval $[a,b]$ and $h(y) = 0$ outside some interval $[c,d]$. $\endgroup$ Jul 22, 2020 at 21:18

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If the joint density $f_{X,Y}(x,y) = g(x) h(y)$ for some functions $g$ and $h$, we have in particular $$ \eqalign{1 &= \int_{-\infty}^\infty \int_{-\infty}^\infty\; dx \; dy\; g(x) h(y) \cr &= \int_{-\infty}^\infty dx\; g(x)\; \int_{-\infty}^\infty dy\; h(y) }$$ If $\int_{-\infty}^\infty dx\; g(x) = a $, $ \int_{-\infty}^\infty dy\; h(y) = 1/a$. We can replace $g(x)$ by $g(x)/a$ and $h(y)$ by $a h(y)$, and we get $\int_{-\infty}^\infty g(x) = 1$ and $\int_{-\infty}^\infty h(y) = 1$. Now for any $x_0$, $$P(X \le x_0) = P(X \le x_0, Y < \infty) = \int_{-\infty}^{x_0} dx\; g(x)\; \int_{-\infty}^\infty dy\; h(y) = \int_{-\infty}^{x_0} dx\; g(x)$$ $$P(Y \le y_0) = P(X < \infty, Y \le y_0) = \int_{-\infty}^{\infty} dx\; g(x)\; \int_{-\infty}^{y_0} dy\; h(y) = \int_{-\infty}^{y_0} dx\; h(x)$$ and $$P(X \le x_0, Y \le y_0) = \int_{-\infty}^{x_0} dx\; g(x)\; \int_{-\infty}^{y_0} dy\; h(y) = P(X \le x_0) P(Y \le y_0) $$ From this you can obtain $P(X \in A, Y \in B) = P(X \in A) P(Y \in B)$ for all Borel sets $A$ and $B$, and therefore $X$ and $Y$ are independent.

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