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For a random vector $\mathbf{X} \in \mathbb{R}^n$ uniformly distributed on the surface of a sphere of radius $r$, the PDF is the inverse of the surface $$f_\mathbf{X}(\mathbf{x}) = 2\pi^{-n/2}\Gamma(n/2)r^{1-n} \tag{1}$$

To derive the marginal distribution, I use the law of total probability. For $\mathbf{X_i} = [X_1 ... X_{n-i}]$, $$f_\mathbf{X_1}(\mathbf{x_1}) = \int_{-r}^{r} f_\mathbf{X}(\mathbf{x}) dx_n = 2\pi^{-n/2}\Gamma(n/2)r^{1-n} \times 2r$$ $$f_\mathbf{X_2}(\mathbf{x_2}) = \int_{-r}^{r} f_\mathbf{X_1}(\mathbf{x_1}) dx_{n-1} = 2\pi^{-n/2}\Gamma(n/2)r^{1-n} \times (2r)^2$$ and go on $$f_{X_1}(x_1) = 2\pi^{-n/2}\Gamma(n/2)r^{1-n} \times (2r)^{n-1} = 2^{n-2}\pi^{-n/2}\Gamma(n/2) \tag{2}$$

Equation (2) shows that the marginal PDF is uniform and does not depend on $r$. This is incorrect.

Could anyone show me where I am wrong?

Cath Maillon
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    The distribution is found at https://stats.stackexchange.com/questions/85916/distribution-of-scalar-products-of-two-random-unit-vectors-in-d-dimensions/85977#85977, *inter alia.* The connection is that the marginal distribution is the scalar product of $X$ with a unit coordinate vector. Your calculations are incorrect because you are not using the correct volume element for a sphere: you are integrating over a cube of side $2r$ and the factor of $r^{1-n}$ is wrong, too. – whuber May 21 '18 at 16:30
  • @whuber thanks for answer. Could you tell me which factor $r^{1-n}$ is wrong? The one in (1) or the one in the result of integration because I did not use the correct volume element for a sphere? I mean I don't expect that (1) is wrong because the density should be the inverse of the surface of the sphere, shouldn't it? – Cath Maillon May 21 '18 at 23:04
  • I would suggest ignoring $r$ altogether, because it merely establishes the linear unit of measurement. You may therefore take it to equal $1.$ That will help you focus on the essential ideas. – whuber May 22 '18 at 13:18
  • @whuber Let me try once more time. I am interested in $f_{X_1,...,X_{n-1}}(x_1,...,x_{n-1})$ and I write $f_\mathbf{X}(\mathbf{x}) = 2\pi^{-n/2}\Gamma(n/2) \delta(\left\lVert \mathbf{x}\right\rVert = 1)$ where $\delta()$ is the delta function. Then $$f_{X_1,...,X_{n-1}}(x_1,...,x_{n-1}) = \int_{-1}^1 f_\mathbf{X}(\mathbf{x}) dx_n \\= 2\pi^{-n/2}\Gamma(n/2) \int_{-1}^1 \delta(x_n^2 = 1 - x_1^2-...-x_{n-1}^2) dx_n = 2\pi^{-n/2}\Gamma(n/2) \times 2$$ because there is 2 points $x_n^2 = \pm\sqrt{1 - x_1^2-...-x_{n-1}^2}$. – Cath Maillon May 24 '18 at 13:54
  • This PDF does not depend on $x_i$ either and if I continue like that, the marginal distribution $f_{X_1}(x_1)$ will be something does not depend on $x_1$ which is different from the correct distribution in your link. – Cath Maillon May 24 '18 at 13:55
  • You're not integrating over the sphere correctly. It has an "area element" that cannot be ignored. Either taking the geometrical approach or using spherical coordinates are much easier and more reliable ways to obtain the result. – whuber May 24 '18 at 15:14

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