概率论二项分布可加性证明
概率论二项分布可加性证明
数学人气：184 ℃时间：2020-03-23 14:17:03
优质解答
用随机变量的特征函数证明最简单,若直接证为设X服从B(p,m),Y服从B(p,n)(下面∑(l;0,k)为0到k对l求和)P(X+Y=k)=∑(l;0,k)P(X=l,y=k-l)=∑(l;0,k)[P(x=l)*P(Y=k-l)]=∑(l;0,k)[C(m,l)p^l*q^(m-l)*C(n,k-l)p^(k-l)*q^(n...
我来回答
类似推荐
猜你喜欢
- 1小学词语填空-什么样的灵魂
- 2Did you konw ( )? A.how long she has been to Shanghai B.if she will come next year or not
- 3一件商品按50%的利润定价,然后又打九折出售,结果每件获利20元,这件商品的成本价是多少?
- 4Simon often＿＿＿＿＿ ＿＿＿＿＿ ＿＿＿＿＿＿＿ Tom to swim togother after school
- 5英语翻译
- 6是I don't think there will be more pollution.对,还是I not think there will be more pollution.
- 7一名著名的演员用英语怎么说?
- 8main(){ char *p,*q; p=(char*)malloc(sizeof(char)*20);q=p; scanf("%s%s",p,q);printf("%s%s",p,q); }
- 9"晚饭不吃了"用英语怎么说
- 10猫既饱且安,率不捕鼠