设二维随机变量(X,Y)在区域D={(x,y)丨x>=0,y>=0,x+y
f(x,y) = 1/2, x>0, y>0, x+y绛旓細Z=X+Y 鍏紡: f(z) = (璐熸棤绌峰埌姝ｆ棤绌风Н鍒) f(x,z-x)dx f(z)=(0 鍒 z 绉垎)(1/2)dx = (1/2)z, 0<z<1; =0, 鍏跺畠 绂绘暎鍨闅忔満鍙橀噺鐨勫垎甯冨緥鍜屽畠鐨勫垎甯冨嚱鏁版槸鐩镐簰鍞竴鍐冲畾鐨勩傚彲浠ョ敤鏉ユ弿杩扮鏁ｅ瀷闅忔満鍙橀噺鐨勭粺璁¤寰嬫э紝浣嗗垎甯冨緥姣斿垎甯冨嚱鏁版洿鐩磋绠鏄庯紝鍥犳锛屼竴鑸槸鐢ㄥ垎甯冨緥鑰屼笉鏄...
绛旓細鍥犱负浜岀淮闅忔満鍙橀噺锛圶锛孻锛夊湪鍖哄煙D涓婃湇浠庡潎鍖鍒嗗竷锛屾墍浠ュ綋锛坸锛寉锛夆垐D鏃讹紝姒傜巼瀵嗗害f锛坸锛寉锛変负鍖哄煙D鐨勯潰绉殑鍊掓暟锛屽綋锛坸锛寉锛変笉鍦―鍐呮椂锛宖锛坸锛寉锛変负0銆侱锛0<=x<=2锛0<=y<=2鏄竟闀夸负2鐨勬鏂瑰舰鍖哄煙锛屾墍浠鐨勯潰绉负4锛屾晠姒傜巼瀵嗗害涓篺锛坸锛寉锛=1/4锛岋紙x锛寉锛夆垐D0锛屽叾瀹冨張鍥犱负鐐癸紙...
绛旓細瑙ｏ細鈭礑={(x,y)涓0<x<1锛-x<y<x}锛屸埓D鏄痽=x銆亂=-x銆亁=1鍥存垚鐨勪笁瑙鍖哄煙锛屽叾闈㈢НSD=1*2/2=1銆傗埓鎸夌収鍧囧寑鍒嗗竷鐨剒hi瀹氫箟锛锛坸,y锛鐨勫瘑搴﹀嚱鏁颁负daof(x,y)=1/SD=1锛(x,y)鈭圖銆乫(x,y)=0锛(x,y)∉D銆(1)fX(x)=鈭(-鈭,鈭)f(x,y)dy=鈭(-x,x)dy=2x锛屽叾...
绛旓細瑙ｉ鎬濊矾锛氱敱宸茬煡鍑哄彂寰楀埌鎯宠鐨勪俊鎭啀杩涗竴姝ヨВ绛斻傞渶瑕佹敞鎰忕殑鏄細鍗曠函鐨勮姒傜巼瀵嗗害娌℃湁瀹為檯鐨勬剰涔夛紝瀹冨繀椤绘湁纭畾鐨勬湁鐣屽尯闂翠负鍓嶆彁銆傚彲浠ユ妸姒傜巼瀵嗗害鐪嬫垚鏄旱鍧愭爣锛屽尯闂寸湅鎴愭槸妯潗鏍囷紝姒傜巼瀵嗗害瀵瑰尯闂寸殑绉垎灏辨槸闈㈢Н锛岃岃繖涓潰绉氨鏄簨浠跺湪杩欎釜鍖洪棿鍙戠敓鐨勬鐜囷紝鎵鏈夐潰绉殑鍜屼负1銆
绛旓細鍖哄煙D瀹為檯涓婃槸浠ワ紙-1锛0锛夛紝锛1锛0锛夛紝锛0锛1锛夛紝锛0锛-1锛変负椤剁偣鐨勬鏂瑰舰鍖哄煙锛孌鐨勯潰绉负2锛浜岀淮闅忔満鍙橀噺锛圶锛孻锛鐨勮仈鍚堟鐜囧瘑搴(x锛寉)锛12锛(x锛寉)鈭圖0锛屽叾瀹冿紙鈪狅級鈶犳牴鎹竟缂樻鐜囧瘑搴︾殑瀹氫箟fX(x)锛濃埆+鈭?鈭瀎(x锛寉)dy鈭村綋-1鈮鈮0鏃讹紝fX(x)锛濃埆1+x?1?x12dy锛1+x锛涘綋...
绛旓細鑱斿悎姒傜巼瀵嗗害:f(x,y)=1/(b-a)(d-c)鍦ㄧ煩褰㈠煙a<x<b,c<y<d涓,0锛 鍏朵粬 杈圭紭姒傜巼:fX(x)=1/(b-a)鍦ㄥ尯闂碼<x<b,涓,0锛 鍏朵粬 fY(y)=1/(d-c)鍦ㄥ尯闂碿<y<d涓,0锛 鍏朵粬
绛旓細绉垎姹傚嚭鍖哄煙G鐨勯潰绉傛墍浠ュ綋0<x^2<y<x<1鏃讹紝鍗冲尯鍩熷湪G鍐咃紝锛圶锛孻锛鐨勮仈鍚堟鐜囧瘑搴(x,y)灏辩瓑浜庡尯鍩烥鐨勯潰绉垎涔嬩竴锛屽叾浠栨儏鍐典笅锛岃仈鍚堟鐜囧瘑搴(x,y)灏辩瓑浜0.銆傝В寰楀尯鍩烥鐨勯潰绉槸1/6.鎵浠ワ紙X锛孻锛夌殑鑱斿悎姒傜巼瀵嗗害f(x,y)=6锛堝湪G鍖哄煙鍐咃級锛宖(x,y)=0锛屼笉鍦℅鍖哄煙鍐呫
绛旓細f(y)=0 1鈮<+鈭 E(Z)=E(X)+2E(Y)+3=鈭(0,1)2x^2dx+0+3=2/3+3=11/3 D(Z)=D(X)+4D(Y)=[E(X^2)-(E(X))^2]+[E(Y^2)-(E(Y))^2]=鈭(0,1)2x^3dx-(2/3)^2+鈭(-1,0)y^2(1+y)dy+鈭(0,1)y^2(1-y)dy=(1/2)-(4/9)+1/6=2/9 ...
绛旓細銆愮瓟妗堛戯細褰搙鈮1锛0鈮锛1鏃讹紝F(x锛寉)=y(2-y)锛涘綋0鈮锛1锛寉鈮鏃讹紝F(x锛寉)=x2锛涘綋0鈮锛渪锛1鏃讹紝F(x锛寉)=y(2x-y)锛
绛旓細璇烽噰绾
