0

If two random variables $X$ and $Y$ are not independent, is it possible to find a nontrivial function $f$ so that $f(X)$ and $f(Y)$ are independent? Thanks.

(I am trying to understand if the simple uniform assumption for a hash function requires the keys to be independent.)

Tim
  • 47,382
  • 2
    Nontrivial must mean non-constant here, e.g. https://math.stackexchange.com/questions/1585379/independence-between-a-constant-random-variable-and-another-random-variable – Matthew Towers Nov 13 '22 at 20:08

1 Answers1

1

In some cases yes. Not a one-to-one function though, because if a (Borel measurable) left inverse function $g$ exists (i.e. $g(f(t)) = t$ for all possible values $t$ of $X$ and $Y$), then if $f(X)$ and $f(Y)$ are independent, $X = g(f(X))$ and $Y = g(f(Y))$ are independent.

Robert Israel
  • 448,999
  • Thanks. Could you also try to see if you can explain the assumption of simple uniform hashing in https://math.stackexchange.com/questions/4575891/how-shall-i-understand-the-uniform-hashing-assumption-of-a-hash-function – Tim Nov 13 '22 at 21:36