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Feb 5, 2023 at 7:29 comment added Glen_b Uncorrelated is not the same as independent. One way to think about independence of two variables is "knowing the value of one variable tells you nothing about the distribution of the other that you didn't already know before specifying that value". When $Y=X^2$ that's not that case. For example, if you don't know $X$, then $Y$ is somewhere between $0$ and $1$ (but more often closer to $0$ than $1$). Now if you know $X=0$, $Y$ must be $0$, but if $X=1$, then $Y=1$ ... that's very strong dependence since knowing $X$ tells you everything about $Y$.
Feb 5, 2023 at 2:08 comment added kjetil b halvorsen @whuber: Yes, very uninteresting, but neither very difficult to find.
Feb 5, 2023 at 0:00 history tweeted twitter.com/StackStats/status/1622022212080340999
Feb 4, 2023 at 21:33 comment added Sextus Empiricus If you think of stochastic functions, like the functions that map the distribution of $X_k$ to the distribution of $X_{k+1}$ describing stochastic processes such as auto-regressive processes or markov chains, then depending on the type of variable $X_k$ the distribution of $X_{k+1}$ can be independent of $X_{k}$, also for more interesting functions. (a situation described here has causal dependency, but not statistical dependency)
Feb 4, 2023 at 20:55 comment added Sextus Empiricus For specific distributions more interesting functions can be possible. For instance, $Y = \sin(2\pi X)$ is independent from $X$ if the domain of $X$ is integer values only. But yes, it will still be the not so interesting map where all values in the domain of $X$ need to be mapped to a single point. It is because $f(X)$ given $X=x$ is a singular distribution in the point $f(x)$. Independence requires that the distribution of $f(X)$ is the same distribution for every value of $X$, $f(x)$ must be a single value for every $x$.
Feb 4, 2023 at 19:39 comment added Aksakal it's strange question. if you observed the realization of X then you certainly know what is X^2. why would anyone even doubt they are dependent? they even have the same X in the definition of the variables.
Feb 4, 2023 at 19:20 answer added Sextus Empiricus timeline score: 2
Feb 4, 2023 at 18:00 comment added whuber @Kjetil Certainly. Not very interesting, are they? ;-)
Feb 4, 2023 at 17:25 comment added kjetil b halvorsen @whuber: one such function $h$ is $h(x)=0$. Another one is $h(x)=5$.
Feb 4, 2023 at 15:55 comment added whuber I find it challenging to find any function $h$ for which $X$ and $h(X)$ are independent.
Feb 4, 2023 at 14:54 history edited Richard Hardy CC BY-SA 4.0
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Feb 4, 2023 at 13:15 answer added Massimo Ortolano timeline score: 7
Feb 4, 2023 at 13:12 answer added Henry timeline score: 3
Feb 4, 2023 at 12:01 history became hot network question
Feb 4, 2023 at 9:05 comment added Dave @RaySiplao I would focus on the picture here.
Feb 4, 2023 at 8:46 vote accept Ray Siplao
Feb 4, 2023 at 5:26 history edited User1865345
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Feb 4, 2023 at 4:32 answer added User1865345 timeline score: 11
Feb 4, 2023 at 4:29 answer added Zhanxiong timeline score: 11
Feb 4, 2023 at 4:15 comment added Galen Plotting simulated values of $X$ and $Y$ will be enlightening, but you can also work out the density of $Y$ via a change of variables from $X$ in order to compare the probabilities in terms of definition of statistical independence.
Feb 4, 2023 at 4:08 comment added Ray Siplao @Dave RVs are independent if $P(X\le a, Y\le b)=P(X\le a) P(Y\le b)$ for any real number $a,b$. I am asking this question because I don't understand some answers in this post.
Feb 4, 2023 at 4:02 comment added Dave Welcome to Cross Validated! What’s the definition of independence?
S Feb 4, 2023 at 3:55 review First questions
Feb 4, 2023 at 4:12
S Feb 4, 2023 at 3:55 history asked Ray Siplao CC BY-SA 4.0