设二维随机变量（x，y）服从x^2+y^2<=R^2上的均匀分布，求点（x，y）到圆心的距离的数学期望 设二维随机变量(X,Y)服从单位圆上的均匀分布, 则P(X^...
\u4e3a\u4ec0\u4e48fX(x)\u7684\u5b9a\u4e49\u57df\u662f-R<x<R\uff1a\u8bbe\u4e8c\u7ef4\u968f\u673a\u53d8\u91cf(X,Y)\u670d\u4ece\u5706\u57dfG:x^2+y^2\u2264R^2\u4e0a\u7684\u5747\u5300\u5206\u5e03\uff0c\u6c42\u5173\u4e8eX\u4ee5\u53ca\u5173\u8fd9\u662f\u4e2a\u9762\u79ef\u4e3a\u03c0R^2\u7684\u5706\u5f62\uff0c\u5747\u5e03\u5728\u5706\u5185(dx dy)\u7684\u6982\u7387\u503c\u4e3a1/\u03c0R^2\u3002 \u5982\u679c\u6c42\u8fb9\u7f18\u5206\u5e03\u7684\u8bdd\uff0c\u4e5f\u5c31\u662f\u6c42f(x)\u548cf(y)\uff0c\u7531\u5bf9\u79f0\u6027\u53ef\u770b\u51fa\u5b83\u4fe9\u5f62\u5f0f\u4e00\u6837 f(x) \u7684\u503c\u57df\u662f-1\u52301, \u800c\u5bf9\u5e94\u4e00\u4e2a\u786e\u5b9ax\u7684y\u7684\u503c\u57df\u662f(-sqrt(1-x^2),sqrt(1-x^2)) \u6240\u4ee5f(x) = 2sqrt(1-x^2)
P=S\u5c0f\u5706/S\u5355\u4f4d\u5706=\uff081/2\uff09\u5e73\u65b9\u5140/1\u5e73\u65b9\u5140=1/4
^\u7531\u9898\u610f\u77e5\uff1aX^2+Y^2=1\uff0c\u6240\u4ee5\u53ef\u8bbe\uff1aX=cos\u03b8\uff0cY=sin\u03b8\uff0c\u03b8\u4e3a[-\u03c0\uff0c\u03c0]\u4e0a\u5747\u5300\u5206\u5e03\u7684\u968f\u673a\u53d8\u91cf\u3002
E(X)=(1/2\u03c0)\u222b(-\u03c0\u2192\u03c0)cos\u03b8d\u03b8=0\uff1b
E(Y)=(1/2\u03c0)\u222b(-\u03c0\u2192\u03c0)sin\u03b8d\u03b8=0\uff1b
E(X^2)=(1/2\u03c0)\u222b(-\u03c0\u2192\u03c0)(cos\u03b8)^2d\u03b8=1/2\uff1b
E(Y^2)=(1/2\u03c0)\u222b(-\u03c0\u2192\u03c0)(sin\u03b8)^2d\u03b8=1/2\uff1b
D(X)=E(X^2)-[E(X)]^2=1/2\uff1b
D(Y)=E(Y^2)-[E(Y)]^2=1/2\uff1b
E(XY)=(1/2\u03c0)\u222b(-\u03c0\u2192\u03c0)cos\u03b8sin\u03b8d\u03b8=0\uff1b
\u534f\u65b9\u5deeCov(X\uff0cY)=E(XY)-E(X)E(Y)=0\uff1b \u6240\u4ee5\uff0c\u76f8\u5173\u7cfb\u6570\u03c1XY=Cov(X\uff0cY)/[(\u221aD(X))(\u221aD(Y))]=0\uff1b\u6240\u4ee5X\u3001Y\u4e0d\u76f8\u5173\uff1b \u53e6\u5916\uff0c\u663e\u7136\u6709P{0\uff1cX\uff1c1/2}\u22600\uff0c P{0\uff1cY\uff1c1/2}\u22600\uff1b
\u6240\u4ee5\uff1aP{0\uff1cX\uff1c1/2}P{0\uff1cY\uff1c1/2}\u22600\uff0c \u4f46\u662f\uff0c0\uff1cX\uff1c1/2\u548c0\uff1cY\uff1c1/2\u540c\u65f6\u53d1\u751f\u7684\u6982\u7387\u4e3a\u96f6\uff0c\u5373\uff1aP{0\uff1cX\uff1c1/2\uff0c0\uff1cY\uff1c1/2}=0\uff0c \u6240\u4ee5P{0\uff1cX\uff1c1/2\uff0c0\uff1cY\uff1c1/2}\u2260P{0\uff1cX\uff1c1/2}P{0\uff1cY\uff1c1/2}\uff0c\u6240\u4ee5X\u3001Y\u4e0d\u72ec\u7acb\u3002
\u6269\u5c55\u8d44\u6599\uff1a
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可以这样想：
一条线段，端点为(0,0)和(R,0)，这条线段上所有点到原点的距离的期望，是R/2.(每一个点都能在线段上找到一个关于(R/2,0)对称的点，即这一对点到原点的平均距离为R/2,所有点都是这样的)
紧接着，把这条线段复制无限次，旋转排布，变成圆。
结论依然同上，期望是R/2
(x,y)与圆心距离为：d=√(x²+y²)
E(d)=1/(πR²)∫∫ √(x²+y²) dxdy
极坐标
=1/(πR²)∫∫ r² drdθ
=1/(πR²)∫[0→2π]dθ∫[0→R] r² dr
=(2/R²)(1/3)r³ |[0→R]
=(2/3R³)/R²
=2R/3
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解法如下：
根据题意：概率密度f(x,y)=1/πR^2 -R< x<R, -R< y<R,
f(x,)=0 其他
点（x，y）到圆心的距离：
L=√x^2+y^2
E(L)=∫(-∞,+∞)√x^2+y^2f(x,y)dxdy
=∫∫(-∞,+∞)√x^2+y^2f(x,y)dxdy
=∫∫(-∞,+∞)√x^2+y^2f(x,y)dxdy 转化为极坐标积分:r==√x^2+y^2
=∫(0,2π)dθ∫(0,R)(r^2/πR^2)dr
=2R/3
(x,y)与圆心距离为：d=√(x²+y²)
E(d)=1/(πR²)∫∫ √(x²+y²) dxdy
极坐标
=1/(πR²)∫∫ r² drdθ
=1/(πR²)∫[0→2π]dθ∫[0→R] r² dr
=(2/R²)(1/3)r³ |[0→R]
=(2/3R³)/R²
=2R/3
距离为R，圆心为坐标原点，则服从X^2+Y^2=R^2
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