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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.

Tom Parkinson
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    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 https://stats.stackexchange.com/a/520811/77222 – Jarle Tufto Aug 21 '21 at 08:26
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    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. – whuber Aug 21 '21 at 14:46
  • Jarle Tufto, thanks for your M = 2 insight and the link. Yes, whuber, I am asking for the distribution conditional on these two constraints. – Tom Parkinson Aug 22 '21 at 04:46
  • @whuber... After further thought, I don't think the joint distribution is uniform across a sphere. I believe that the constraint on the mean of the samples induces a correlation between any two random variables that affects the shape. For example, for M = 3, the feasible region for any two random variables is an ellipse whose axes are at 45 degrees to the axes of the coordinate system. – Tom Parkinson Aug 22 '21 at 05:14
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    @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 . That should enable me to get to an answer. Thanks! – Tom Parkinson Aug 22 '21 at 07:08
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    @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! – Tom Parkinson Aug 22 '21 at 09:28
  • I don't see any (proper) ellipsoids here: the cross section determined by the sum-to-zero constraint is a sphere. But that is because I am interpreting your question as concerning a conditional distribution. It still isn't clear what you really mean by a "constraint." – whuber Aug 22 '21 at 14:16
  • @whuber... Yes, I did mean the conditional distribution, conditioned on: (1) sum of random variables equals zero; and (2) sum of squares equals M. I apologize. I am new here, and I haven't yet learned how to post equations. Let me try to explain where I think the ellipsoid arises. Suppose M = 3, and let the variables be x, y, and z. From (1), z = -x -y. From (2), x^2 + y^2 + (-x -y)^2 = 2x^2 + 2xy + 2y^2 = M. Similarly, for M > 3, I think the same logic defines an ellipsoid in M-1 dimensions. By the way, I forgot to say that your derivation in the other thread is brilliant! – Tom Parkinson Aug 23 '21 at 02:36
  • @whuber... Ok, I think I see the issue, but my intuition suffers severely when M > 3. For the M = 3 case, the sum of squares constraint forces the points to lie on a sphere. The sum constraint forces them to lie in a plane. The intersection of the plane and the sphere is a circle -- or, in M > 3 dimensions, a sphere. But, given the orientation of the plane, the projection of the circle onto any two of the three dimensions is an ellipse. Similarly, I think the projection of the sphere onto any lower number of dimensions would be an ellipsoid. – Tom Parkinson Aug 23 '21 at 07:53
  • There's only one flaw in your argument, Tom: there is no projection going on. The *intersection* of a hyperplane with a hypersphere is a codimension-2 sphere (or a point or empty). – whuber Aug 23 '21 at 12:56
  • @whuber... I have no doubt that what you say is correct. But I think we are talking about two separate problems (or, at least, two very different statements of the same problem). I am obviously unable to explain my point of view clearly. Thank you for your time! – Tom Parkinson Aug 24 '21 at 08:01

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