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There are some papers and some R packages providing exact calculations for the CDF and the inverse-CDF of the (sample) Spearman rank correlation coefficient.

My question is: How difficult would it be to calculate the exact distribution of the Spearman rank coefficient of correlation for a given setting like this (just an example):

  • $X$ and $Y$ continuous with known marginal distributions (not necessarily normal).
  • $X$ and $Y$ are correlated with a known (population) Pearson correlation coefficient different from zero.
  • Of course, the value of the sample size $n$ is known, and, since both $X$ and $Y$ are considered to be continuous, the non-ties case can be assumed.

If not, I was thinking of estimating the distribution by using Monte-Carlo simulation. What do you think? Any other idea?

By the way, I know that the Spearman coefficient is a distribution-free method, meaning that its probability distribution does not depend on the concrete distribution of $X$ and $Y$ under the (null) hypothesis of independence. But I think it's clear that its distribution does depend on the concrete distribution of $X$ and $Y$ under the (alternative) hypothesis of non-zero correlation (as it happens with many distribution-free methods, anyway).

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    $\begingroup$ Basically, impossible, because knowing the marginal distributions and a correlation coefficient is not sufficient to know the joint distribution, which would be necessary to do this. Even knowing it, however, would probably not help you in practice, since Spearman is a rank correlation, so you would have to convert the joint distribution of $X$ and $Y$ to a joint distribution of the ranks of a sample of size $n$, which seems to me to not be a practical thing to do for any significant $n$ and almost any distributions, esp. continuous ones. $\endgroup$
    – jbowman
    May 14 at 18:29
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    $\begingroup$ @jbowman: Agreed, but there is actually a paper: jstor.org/stable/2285508?seq=1 or amstat.tandfonline.com/doi/abs/10.1080/… $\endgroup$ May 14 at 20:27
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    $\begingroup$ This setting is kind of self-contradictory: to obtain a result you have to make parametric assumptions about the joint distribution, whereas one uses the Spearman correlation specifically to avoid making such assumptions. $\endgroup$
    – whuber
    May 14 at 21:23
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    $\begingroup$ @whuber It is common that non-null distributions of distribution-free method statistics actually depend on the underlying distribution. The paper they cited above makes it clear. $\endgroup$
    – Vicent
    May 14 at 21:26
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    $\begingroup$ I cannot make any sense of that remark because it sounds circular. It's also too vague: the issue here concerns specifying the alternative hypothesis. Your question cannot be answered without a more specific and narrower definition than "non-null." $\endgroup$
    – whuber
    May 14 at 21:29

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It's not possible to do this exactly, as knowing the marginal distributions and a correlation coefficient is not sufficient to determine the joint distribution, which would be necessary to do this. Even knowing it, however, would probably not help you in practice, since Spearman is a rank correlation, so you would have to convert the joint distribution of $X$ and $Y$ to a joint distribution of the ranks of a sample of size $n$, which seems to me to not be a practical thing to do for any significant $n$ and almost any distributions, especially continuous ones.

However, given the joint distribution, simulation becomes (in many practical cases) a feasible alternative, although requiring that you abandon the goal of knowing the exact distribution under the alternative. Of course, if you have a point null and a point alternative hypothesis, the Spearman correlation coefficient is not likely to be the best statistic to use to discriminate between them, especially given the Neyman-Pearson lemma.

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    $\begingroup$ +1 One possibility is to investigate properties across a range of potential/more-or-less plausible joint alternatives (e.g. choose some specific copulas indexed by a parameter indicating stronger monotonic association and investigate the power as a function of the parameter at a number of sample sizes), noting that it's not necessary to worry about marginals with rank-correlations, only the copula matters. This is of course a non-trivial exercise, but has the added advantage that it can be helpful with interpreting the statistic more generally. $\endgroup$
    – Glen_b
    May 15 at 1:16

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