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Suppose that I have M draws from a unit normal distribution N(0,1). By construction, the sum of the M draws equals zero, and the sum of the squares equals M. Given these constraints, what is the conditional fourth moment of each random draw?

I can easily picture the constraints for M = 3. The constraint on the sum is a plane, and the constraint on the squares is a sphere. Accordingly, the distribution of each random draw is a circle in three-dimensional space. I can deduce the conditional distribution on each random draw x as f(x) = 1 / (pi * sqrt(2 - x^2)). Integration yields the associated fourth moment to be 3/2.

My intuition suggests that the formula as a function of M may be 3(M-2)/(M-1). But I have not been able to derive a formula for the general case. I've tried forming the joint distribution of the M draws (easy), doing change of variables to get the joint of the first M-2 draws, the sum of the draws, and the sum of squares (messy), and then form the joint distribution of the first M-2 draws conditioned on the sum of the draws and the sum of the squares (messier still!). There must be an easier way.

I've also tried forming the fourth moment by taking the expectation of the sum of the draws, raised to the fourth power. This expectation -- which equals zero, given the constraint on the sum of the draws -- is a function of various products, each of which I can then relate to the fourth moment of a single draw. Unfortunately, all the terms cancel out, and all I can prove is that 0 = 0.

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    $\begingroup$ For $M=2$, the conditional distribution would be two point masses at $(-1,1)$ and $(1,-1)$ such each forth moment $E X_i^4$ would be 1 which is different from $3(M-2)/(M-1)=0$ so your formula doesn't hold in that case. I suspect the marginal conditionals become shifted beta distributions in the general case as in stats.stackexchange.com/a/520811/77222 $\endgroup$
    – Jarle Tufto
    Commented Aug 21, 2021 at 8:26
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    $\begingroup$ Could you clarify what "by construction" might mean? How exactly do you assure these two constraints hold? Are you perhaps asking for distributions conditional on these requirements? If so, your problem reduces to the uniform distribution on the sphere (of radius $\sqrt M$) in one less dimension. $\endgroup$
    – whuber ♦
    Commented Aug 21, 2021 at 14:46
  • $\begingroup$ Jarle Tufto, thanks for your M = 2 insight and the link. Yes, whuber, I am asking for the distribution conditional on these two constraints. $\endgroup$
    – Tom Parkinson
    Commented Aug 22, 2021 at 4:46
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    $\begingroup$ @whuber... I think I see how to get there now. The constraint on the mean of the random variables, coupled with the constraint on the sum of the squares, forces any M-1 of the random variables to lie on an ellipsoid. Via an appropriate coordinate transformation, I can define a set of random variables that lie on a sphere. Then I should be able to use the logic you laid out in <stats.stackexchange.com/questions/520498/…>. That should enable me to get to an answer. Thanks! $\endgroup$
    – Tom Parkinson
    Commented Aug 22, 2021 at 7:08
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    $\begingroup$ @JarleTufto, the marginal distribution is indeed a shifted beta distribution, but with parameters (M/2-1, M/2-1). The additional constraint on the mean reduces the degrees of freedom by one. The fourth moment turns out to be 3(M-1)/(M+1). Thanks again! $\endgroup$
    – Tom Parkinson
    Commented Aug 22, 2021 at 9:28

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