b*****e
发帖数: 474 | 1 假设所有变量都是[0, 1] 之间均匀分布的随机变量,
则（x1^2+x2^2+x3^2+x4^2）^0.5 小于 r 的几率是
半径为 r 的四维超球被以 (v1, v2, v3, v4) 为顶
点的超立方所截的体积, 其中v1, v2, v3, v4 为 0
或 1. 这是一个直观的理解. (仅限非负象限).
具体计算时可用四维球极坐标:
dx1 = d(r cos(a))
dx2 = d(r sin(a)cos(b))
dx3 = d(r sin(a)sin(b)cos(c))
dx4 = d(r sin(a)sin(b)sin(c)) | P***a
发帖数: 9 | 2
I think this question can only be solved in terms of its CDF
or PDF, but not in terms of assigning it with an specific
well-known named distribution.
I am working on it. It can be translated to a integration
problem. The CDF(X) is the integration of 1 respect to x1,
x2, x3, x4, with the constraints that
x1^2+x2^2+x3^2+x4^2<=x^2 and 0<=x1,x2,x3,x4<=1.
When x belongs to [0,1], the CDF is Pi^2 * x^4 / 32, or PDF
is Pi^2 * x^3 / 8. When x is less than 0, the CDF is 0 and
the PDF is 0.
When x is lar | b*****e
发帖数: 474 | 3 仍然是半径为 r 的四维超球被 16 个超平面切割后求体积的问题.
假定 x_i \in [u_i, v_i], u_i >= 0, 同样的四维球极坐标:
x1 = r cos(a)
x2 = r sin(a)cos(b)
x3 = r sin(a)sin(b)cos(c)
x4 = r sin(a)sin(b)sin(c)
积分时, r 从 \sum(u_i) 到 \sum(v_i),
a 从 arc cos(u_1/r) 到 arc cos(v_1/r),
b, r sin(a) >= v_2 时,
从 arc cos(u_2/(r sin(a))) 到 arc cos(v_2 /(r sin(a)))
v_2 > r sin(a) >= u_2 时,
从 arc cos(u_2/(r sin(a))) 到 arc cos(1)
c 的情况更复杂些, 但一样分段解决.
u_i, v_i 不同正负时可以分象限积分.
【在 P***a 的大作中提到】
:
: I think this question can only be solved in terms of its CDF
: or PDF, but not in terms of assigning it with an specific
: well-known named distribution.
: I am working on it. It can be translated to a integration
: problem. The CDF(X) is the integration of 1 respect to x1,
: x2, x3, x4, with the constraints that
: x1^2+x2^2+x3^2+x4^2<=x^2 and 0<=x1,x2,x3,x4<=1.
: When x belongs to [0,1], the CDF is Pi^2 * x^4 / 32, or PDF
: is Pi^2 * x^3 / 8. When x is less than 0, the CDF is 0 and
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