I need to complete the following problem:
Let $~(X_1, X_2, X_3)~$ be a continuous random vector,
with uniform density on the unit sphere $~\{(x_1,~ x_2,~ x_3) ~∈~ \mathbb R^3 ~:~ x_1^2 + x_2^2 + x_3^2 = 1\}~$. What would be the marginal pdf of $~X_1~$?
I have to follow the next steps:
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Express the density in spherical coordinates.
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Think about the symmetry of the problem. Think about spherical coordinates. Then think how can you integrate the effects of two variables while keeping the other fixed in spherical coordinates.
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After 2, you will have the marginal in one angle. Then you need to do a transformation of random variables ("a change of variables in the integral of the density so that the limits become (-1, 1)") to transform the angle to the Cartesian coordinate you are interested in obtaining the marginal.
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The marginal pdf of X1 should be one that is intuitive for this problem (yet that intuition will fail if the sphere was a circle or a high-dimensional sphere!).
I have completed the first step by:
$$f(\theta, \varphi) = {\sin(\varphi)\over4\pi}$$
where $\theta \in [0,2\pi]$ and $\varphi \in [0,\pi]$
I'm not sure if the second step is correct by:
$$f\varphi(\varphi) = \int_0^{\pi} {\sin(\varphi)\over4\pi}\,d\varphi= {1\over 2\pi}$$
Or the second option:
$$f\theta(\theta) = \int_0^{2\pi} {\sin(\varphi)\over4\pi}\,d\theta= {\frac12\sin(\varphi)}$$
Best Answer
My take on the problem is that you are supposed to let $x_1=\cos\phi$, $x_2=\sin\phi\cos\theta$, and $x_3=\sin\phi\sin\theta$. Then looking at the integral $$\int_0^{\pi}f(\phi)d\phi=\int_0^{\pi}\frac12\sin\phi\,d\phi=\int_1^{-1}\frac12(-dx_1)=\int_{-1}^1\frac12dx_1=\int_{-1}^1f(x_1)dx_1$$ So that $f(x_1)=\frac12$, i.e. uniformly distributed with respect to $x_1$. In $n$ dimensions the "circumference" of the "circle" at a given $x_1$ is $K_{n-1}(1-x_1^2)^{\frac{n-2}2}$ and the arc length along the surface from $x_1$ to $x_1+dx_1$ is $\frac{dx_1}{\sqrt{1-x_1^2}}$ so $$f(x_1)dx_1=C_n(1-x_1^2)^{\frac{n-3}2}dx_1=\frac{\operatorname{\Gamma}\left(\frac n2\right)(1-x_1^2)^{\frac{n-3}2}}{\sqrt{\pi}\operatorname{\Gamma}\left(\frac{n-1}2\right)}dx_1$$ So you can see that only in $3$ dimensions is it uniform.