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If $X$ and $Y$ has the Spearman correlation $r_s=1$, is it possible that the Pearson correlation between $X$ and $Y$ is 0?

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Yes if by $1$ and $0$ you mean $\approx 1$ and $\approx 0$. Here is one example in R with a stupid outlier

x=c(1:1000)
y=c(1:999,-100000)

> cor(x,y,method="spearman")
[1] 0.994006
> cor(x,y,method="pearson")
[1] 0.03571434
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  • $\begingroup$ I suspect you could find an example with $r_s\approx 1$ and $r_p=0$ but not $r_s= 1$ and $r_p\approx 0$ $\endgroup$
    – Henry
    Commented Dec 7, 2022 at 10:41
  • $\begingroup$ Can we construct a monotone function $Y=f(X)$ such that $f$ is monontone, But $Cov(X,Y)=0$? $\endgroup$ Commented Dec 7, 2022 at 10:41
  • $\begingroup$ @mathbeginner What do you think? $\endgroup$
    – Henry
    Commented Dec 7, 2022 at 10:43
  • $\begingroup$ I mean that if $x$ and $y$ has the strictly monotone relationship, for example, y=x^3, can we have cov(x,y)=0 $\endgroup$ Commented Dec 7, 2022 at 10:48
  • $\begingroup$ A Spearman correlation of 1 implies that all data lie on a rising monotonic curve (in fact on many such curves). I can't see that is compatible with Pearson correlation zero. $\endgroup$
    – Nick Cox
    Commented Dec 7, 2022 at 11:41

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