概率论与数理统计！设随机变量X的分布律为：X分别是-1,0,1,2；P分别对应0.2，0.3，0.4，0.1。求数学期望 概率论与数理统计问题求解!!!!!
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设Y=1/(1+X^2)，则Y分别是（1/2, 1, 1/5）；P分别对应（0.2+0.4=0.6，0.2 ，0.1 ）所以EY=1/2*0.6+1*0.2+1/5*0.1=0.52。
绛旓細(1/4)*鈭慪i锝濶锛0,1锛,鎵浠(x-2)²鏈嶄粠鑷敱搴︿负1鐨剎²鍒嗗竷,锛1/16)*鈭慪i²鏈嶄粠鑷敱搴︿负涓鐨剎²鍒嗗竷,鎵浠鏈嶄粠F(1,1)鍒嗗竷,11,姒傜巼璁轰笌鏁扮悊缁熻鐨勯鐩,璁鹃殢鏈哄彉閲廥锝濶锛2,1锛,Y₁,Y₂,Y₃,Y₄鍧囨湇浠嶯锛0,4锛変笖X,Yi(i=1,2,3...
绛旓細X^2~X²(1) 鍗℃柟鍒嗗竷 D(X^2)=2
绛旓細D锛x锛=E锛X^2锛-[E锛圶锛塢^2=1 E锛圶^2锛=1
绛旓細P{X鈮3}=F(3)=(3-0)/(5-0)=3/5
绛旓細棰樼洰鏄細璁鹃殢鏈哄彉閲廥 鏈嶄粠鍙傛暟涓哄叆鐨勬硦鏉惧垎甯冿紝涓攑(x=2)=p(x=4) 姹傚叆 鍚с傝В锛(鍏2/2!)*e^(-鍏)=(鍏4/4!)*e^(-鍏)(鍏2/12)=1 鍏=2鈭3
绛旓細D(X^2)=E(X^4)-(E(X^2))^2 E(X^2)=D(X)+(E(X))^2 E(X^4)灏卞緱绉垎浜嗭紝鐢ㄤ袱娆″垎閮ㄧН鍒嗗氨鍙互浜嗭紝鍐欒捣鏉ユ瘮杈冮夯鐑﹀氨涓嶅啓浜
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绛旓細4.璁鹃殢鏈哄彉閲廥~N锛0,1锛夛紝鍒欐牴鎹垏姣旈洩澶笉绛夊紡P{涓╔涓ㄢ墺3}鈮わ紵1/9 5.璁綪锛圓锛=0.1锛孭锛圔锛=0.3锛屼笖A涓嶣浜掍笉鐩稿锛屽垯P锛圓骞禕锛=锛0.4 6.璁鹃殢鏈哄彉閲廥~P(1),鍒橢(X²)=锛烢(X²)=DX-(EX)^2=1-1^2=0 7.璁句簨浠禕鍖呭惈浜嶢锛孭(A)=0.4,P(B)=0.3,鍒橮...
绛旓細鐢卞垎甯冨嚱鏁版眰姒傜巼瀵嗗害鍑芥暟涓篻(z)=3/2(1-z^2)浜屻佸疄闄呬笂鍦ㄨ繖閲岀敾鍑哄浘鍗冲彲锛屽垎甯冨尯鍩熶负D:X+Y>1锛x灞炰簬(0,1)锛寉灞炰簬(0,1)闈㈢НS=1/2锛岃岀敾鍑篨+Y>1鐨勭洿绾匡紝涓庡垎甯冨尯鍩熺浉浜ゅ緱鍒 鍗(1/2 ,1/2),(1,0)鍜(0,1)涓夌偣缁勬垚鐨勪笁瑙掑舰锛岄偅涔堟樉鐒堕潰绉负1/4锛屾墍浠(X+Y>1)= (1/4) ...
绛旓細濡傚浘
