问一下为什么连续型随机变量超过积分上限F(x)都为1 设连续型随机变量X的概率密度函数为为f(x)=1/2*e^(...

\u5df2\u77e5\u8fde\u7eed\u578b\u968f\u673a\u53d8\u91cfX\u7684\u5bc6\u5ea6\u51fd\u6570\u4e3a f(x)=x, 0<=x<1 2-x, 1<=x<a 0, \u5176\u5b83

\u5df2\u77e5\u8fde\u7eed\u578b\u968f\u673a\u53d8\u91cfX\u7684\u5bc6\u5ea6\u51fd\u6570\uff0c\u90a3\u4e48\u5bf9\u5176\u5728\u8d1f\u65e0\u7a77\u5230\u6b63\u65e0\u7a77\u4e0a\u8fdb\u884c\u79ef\u5206\u7684\u503c\u4e3a1
\u6240\u4ee5
\u222b(\u4e0a\u96501\uff0c\u4e0b\u96500) x dx + \u222b (\u4e0a\u9650a\uff0c\u4e0b\u96501) 2-x dx
= [0.5x² (\u4ee3\u5165\u4e0a\u96501\uff0c\u4e0b\u96500)] + [2x-0.5x² (\u4ee3\u5165\u4e0a\u9650a\uff0c\u4e0b\u96501)]
=0.5 + 2a -0.5a² -1.5
= 2a -0.5a² -1 =1\uff0c

\u5373a² -4a+4=0\uff0c\u89e3\u5f97a=2

\u800c
P{X>1} = 1 - P{x\u22641} =1 - 0.5 *1² = 0.5

\u4e00\u3001\u5bf9\u6982\u7387\u5bc6\u5ea6\u51fd\u6570\u79ef\u5206\u5c31\u53ef\u4ee5\u5f97\u5230\u5206\u5e03\u51fd\u6570\uff0c
\u5f53x<0\u65f6\uff0c
f(x)=1/2*e^x
\u6545\u5206\u5e03\u51fd\u6570
F(x)
=\u222b(\u4e0a\u9650\u5ea6x\uff0c\u4e0b\u9650-\u221e) 1/2 *e^x dx
=1/2 *e^x [\u4ee3\u5165\u4e0a\u9650x\uff0c\u4e0b\u9650-\u221e]
=1/2 *e^x
\u5f53x>=0\u65f6\uff0c
f(x)=1/2*e^(-x)
\u6545\u5206\u5e03\u51fd\u6570
F(x)
=F(0)+ \u222b(\u4e0a\u9650x\uff0c\u4e0b\u96500) 1/2 *e^(-x) dx
=F(0) - 1/2 *e^(-x) [\u4ee3\u5165\u4e0a\u9650x\uff0c\u4e0b\u96500]
=F(0) - 1/2 *e^(-x) +1/2
\u800cF(0)=1/2
\u6545F(x)=1 -1/2 *e^(-x)
\u6240\u4ee5
F(x)= 1 -1/2 *e^(-x) x>=0
1/2 *e^x x<0
\u4e8c\u3001\u4f8b\u5982\uff1a
(1) f(x)\u662f\u5076\u51fd\u6570, \u5219, xf(x)\u662f\u5947\u51fd\u6570. \u6240\u4ee5 E{X} = \u222bzhidao[-\u221e,\u221e] xf(x)dx = 0
x(|\u4e13x|)f(x)\u4e5f\u662f\u5947\u51fd\u6570.
X\u4e0e|X|\u7684\u534f\u65b9\u5dee = E{X(|X|)}-E{X}E(|X|) = E{X(|X|)}-(0)E{|X|}
=\u222b[-\u221e,\u221e] x(|x|)f(x)dx = 0
X\u4e0e|x|\u4e0d\u76f8\u5173
(2) \u4f46X\u4e0e|X|\u4e0d\u72ec\u7acb.\u4e00\u4e2a\u4f8b\u5b50\u5c31\u591f. \u5f53 X=1\u662f, |X|\u4e00\u5c5e\u5b9a\u4e5f\u7b49\u4e8e1\u3002

\u6269\u5c55\u8d44\u6599\uff1a


\u7531\u4e8e\u968f\u673a\u53d8\u91cfX\u7684\u53d6\u503c \u53ea\u53d6\u51b3\u4e8e\u6982\u7387\u5bc6\u5ea6\u51fd\u6570\u7684\u79ef\u5206\uff0c\u6240\u4ee5\u6982\u7387\u5bc6\u5ea6\u51fd\u6570\u5728\u4e2a\u522b\u70b9\u4e0a\u7684\u53d6\u503c\u5e76\u4e0d\u4f1a\u5f71\u54cd\u968f\u673a\u53d8\u91cf\u7684\u8868\u73b0\u3002\u66f4\u51c6\u786e\u6765\u8bf4\uff0c\u5982\u679c\u4e00\u4e2a\u51fd\u6570\u548cX\u7684\u6982\u7387\u5bc6\u5ea6\u51fd\u6570\u53d6\u503c\u4e0d\u540c\u7684\u70b9\u53ea\u6709\u6709\u9650\u4e2a\u3001\u53ef\u6570\u65e0\u9650\u4e2a\u6216\u8005\u76f8\u5bf9\u4e8e\u6574\u4e2a\u5b9e\u6570\u8f74\u6765\u8bf4\u6d4b\u5ea6\u4e3a0\uff08\u662f\u4e00\u4e2a\u96f6\u6d4b\u96c6\uff09\uff0c\u90a3\u4e48\u8fd9\u4e2a\u51fd\u6570\u4e5f\u53ef\u4ee5\u662fX\u7684\u6982\u7387\u5bc6\u5ea6\u51fd\u6570\u3002
\u8fde\u7eed\u578b\u7684\u968f\u673a\u53d8\u91cf\u53d6\u503c\u5728\u4efb\u610f\u4e00\u70b9\u7684\u6982\u7387\u90fd\u662f0\u3002\u4f5c\u4e3a\u63a8\u8bba\uff0c\u8fde\u7eed\u578b\u968f\u673a\u53d8\u91cf\u5728\u533a\u95f4\u4e0a\u53d6\u503c\u7684\u6982\u7387\u4e0e\u8fd9\u4e2a\u533a\u95f4\u662f\u5f00\u533a\u95f4\u8fd8\u662f\u95ed\u533a\u95f4\u65e0\u5173\u3002\u8981\u6ce8\u610f\u7684\u662f\uff0c\u6982\u7387P{x=a}=0\uff0c\u4f46{X=a}\u5e76\u4e0d\u662f\u4e0d\u53ef\u80fd\u4e8b\u4ef6\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u6982\u7387\u5bc6\u5ea6\u51fd\u6570

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