设二维离散型随机变量xy的联合分布律 如下 大一工程数学：设二维离散型随机变量（X、Y）的联合分布律为
\u8bbe\u4e8c\u7ef4\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf\uff08X,Y\uff09\u7684\u8054\u5408\u5206\u5e03\u5f8b\u4e3a\u5982\u4e0b \u8bd5\u5206\u522b\u6839\u636e\u4e0b\u5217\u6761\u4ef6\u6c42a\u548cb\u7684\u503ca+0.2=0.3,\u6545a=0.1;0.3+0.4+0.1+b=1\uff0c\u6545b=0.2.
P(X=-1)=0.3,P(X=0)=0.4,P(X=2)=0.3;
P(Y=1)=0.5,P(Y=3)=0.5
\u4ee4\uff1ba+1/6+1/12+
+1/6+1/6+1/6+
+1/12+1/6+b=1,\u5f97\uff1a
a+b+1=1\uff0c\u5373\uff1aa+b=0\u3002
\u56e0\u4e3aa>=0, b>=0,\u6545\u77e5\u9053\u5fc5\u6709\uff1a
a=0\uff0cb=0\u3002
\u6240\u6c42\u6982\u7387P=0+1/6+1/12+
+1/6+1/6+1/6=3/4\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u5f53\u968f\u673a\u53d8\u91cf\u7684\u53ef\u53d6\u503c\u5168\u4f53\u4e3a\u4e00\u79bb\u6563\u96c6\u65f6\u79f0\u5176\u4e3a\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf\uff0c\u5426\u5219\u79f0\u5176\u4e3a\u975e\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf\uff0c\u8fd9\u662f\u5f88\u5927\u7684\u4e00\u4e2a\u7c7b\uff0c\u5176\u4e2d\u6709\u4e00\u7c7b\u662f\u6781\u5176\u5e38\u89c1\u7684\uff0c\u968f\u673a\u53d8\u91cf\u7684\u53d6\u503c\u4e3a\u4e00(n)\u7ef4\u8fde\u7eed\u7a7a\u95f4\uff0c\u79f0\u5176\u4e3a\u8fde\u7eed\u578b\u968f\u673a\u53d8\u91cf\u3002
\u80fd\u6309\u4e00\u5b9a\u6b21\u5e8f\u4e00\u4e00\u5217\u51fa\uff0c\u5176\u503c\u57df\u4e3a\u4e00\u4e2a\u6216\u82e5\u5e72\u4e2a\u6709\u9650\u6216\u65e0\u9650\u533a\u95f4\uff0c\u8fd9\u6837\u7684\u968f\u673a\u53d8\u91cf\u79f0\u4e3a\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf\u3002\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf\u4e0e\u8fde\u7eed\u578b\u968f\u673a\u53d8\u91cf\u4e5f\u662f\u7531\u968f\u673a\u53d8\u91cf\u53d6\u503c\u8303\u56f4\uff08\u6216\u8bf4\u6210\u53d6\u503c\u7684\u5f62\u5f0f\uff09\u786e\u5b9a\uff0c\u53d8\u91cf\u53d6\u503c\u53ea\u80fd\u53d6\u79bb\u6563\u578b\u7684\u81ea\u7136\u6570\uff0c\u5c31\u662f\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf
E(X)=0\u00d7\uff080.20+0.05+0.10\uff09+1\u00d7\uff080.05+0.10+0.25\uff09+2\u00d7\uff080+0.15+0.10\uff09=0.9
E\uff08Y\uff09=-2\uff080.20+0.05+0\uff09+0\uff080.05+0.1+0.15\uff09+1\uff080.1+0.25+0.10\uff09=-0.05
XY=0\uff1a0\u00d7\uff08-2\uff09\uff0c0\u00d70,0\u00d71\uff1b1\u00d70,2\u00d70\uff0c\u6982\u7387=0.2+0.05+0.1+0.1+0.15=0.6
-2:1\u00d7\uff08-2\uff09\uff0c\u6982\u7387=0.05\uff0c
-4:2\u00d7\uff08-2\uff09\uff0c\u6982\u7387=0\uff1b
1:1\u00d71\uff0c\u6982\u7387=0.15\uff1b
2\uff1a2\u00d71\uff0c\u6982\u7387=0.1
E(XY\uff09=-4\u00d70-2\u00d70.05+0\u00d70.6+1\u00d70.15+2\u00d70.1=0.25
X²+Y²\uff1a0:0²+0²\uff0c\u6982\u7387=0.05\uff1b
1\uff1a0²+1²\uff0c1²+0²\uff0c\u6982\u7387-0.1+0.1=0.2
2:1²+1²\uff0c\u6982\u7387\uff1a0.25
4\uff1a0+\uff08-2\uff09²\uff0c2²+0²\uff0c\u6982\u7387=0.20+0.15=0.35
5:1+\uff08-2\uff09²\uff0c2²+1²\uff0c\u6982\u7387=0.05+0.10=0.15
8:2²+\uff08-2\uff09²\uff0c\u6982\u7387=0
E\uff08X²+Y²\uff09=0\u00d70.05+1\u00d70.2+2\u00d70.25+4\u00d70.35+5\u00d70.15+8\u00d70=2.85
\u6982\u5ff5\u8fa8\u6790\uff1a
\u80fd\u6309\u4e00\u5b9a\u6b21\u5e8f\u4e00\u4e00\u5217\u51fa\uff0c\u5176\u503c\u57df\u4e3a\u4e00\u4e2a\u6216\u82e5\u5e72\u4e2a\u6709\u9650\u6216\u65e0\u9650\u533a\u95f4\uff0c\u8fd9\u6837\u7684\u968f\u673a\u53d8\u91cf\u79f0\u4e3a\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf\u3002\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf\u4e0e\u8fde\u7eed\u578b\u968f\u673a\u53d8\u91cf\u4e5f\u662f\u7531\u968f\u673a\u53d8\u91cf\u53d6\u503c\u8303\u56f4\uff08\u6216\u8bf4\u6210\u53d6\u503c\u7684\u5f62\u5f0f\uff09\u786e\u5b9a\uff0c\u53d8\u91cf\u53d6\u503c\u53ea\u80fd\u53d6\u79bb\u6563\u578b\u7684\u81ea\u7136\u6570\uff0c\u5c31\u662f\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf\u3002
\u4ee5\u4e0a\u5185\u5bb9\u53c2\u8003\uff1a\u767e\u5ea6\u767e\u79d1-\u79bb\u6563\u578b\u968f\u673a\u53d8\u91cf
+1/6+1/6+1/6+
+1/12+1/6+b=1,得：
a+b+1=1，即：a+b=0。
因为a>=0, b>=0,故知道必有：
a=0，b=0。
所求概率P=0+1/6+1/12+
+1/6+1/6+1/6=3/4。
绛旓細瑙ｉ杩囩▼濡備笅鍥撅細
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