设二维随机变量(X，Y)的概率密度为f (x，y)= 则P{X+Y≤1}= 设二维随机变量(X，Y)的概率密度为f (x，y)=6x,0...
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P(X+Y<=1)=(1*1/2)/2=1/4
0.25
(X,Y)服从二维均匀分布，把联合概率密度非零的区域画出来，
其中X+Y≤1的区域占总区域的比例就是所求
绛旓細EX=鈭埆[0<=y<=x<=1] xf(x,y)dxdy=鈭玔0->1]鈭玔0->x] 12xy²dydx=4/5 EY=鈭埆[0<=y<=x<=1] yf(x,y)dxdy=鈭玔0->1]鈭玔0->x] 12y³dydx=3/5 E(X²+Y²)=鈭埆[0<=y<=x<=1] (x²+y²)f(x,y)dxdy=鈭玔0->1]鈭玔0...
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绛旓細璇︾粏姝ラ鏄細鈶犲厛姹傚嚭X銆乊鐨勮竟缂樺垎甯冦傛寜鐓у畾涔夛紝fX(x)=鈭(-鈭,鈭)f(x,y)dy銆傗埓fX(x)=鈭(0,x)15xy²dy=5x^4锛屽叾涓0<x<1銆傚悓鐞嗭紝fY(y)=鈭(y,1)15xy²dx=15(y²-y^4)/2锛屽叾涓0<y<1銆傗憽姹傚嚭E(X)銆丒(Y)銆丒(X²)銆丒(Y²)銆侲(X)=鈭(...
绛旓細1=鈭(0~2)鈭(-x~x) kx(x-y)dydx 1=鈭(0~2) kx(xy-y^2/2)|(-x~x) dx 1=鈭(0~2)2kx^3dx 1=2kx^4/4(0~2)1=8k k=1/8 鐢诲浘鍙煡鑼冨洿 fy(y)=鈭(|y|~2) kx(x-y)dx =kx^3/3-kyx^2/2 | |y|~2 =1/8(8/3-2-(y^2|y|/3-y^3/2))=1/12-y^2|y...
绛旓細a=3,瀵筙鐨勮竟缂姒傜巼瀵嗗害涓3/2,瀵Y鐨杈圭紭涓3y^2;X鐨杈圭紭*Y鐨勮竟缂=9y^2/2涓嶇瓑浜庤仈鍚堝瘑搴,鎵浠ヤ笉鐙珛.
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绛旓細璁句簩缁撮殢鏈哄彉閲(X,Y)鐨勬鐜瀵嗗害涓篺(x,y)=4.8y(2-x),0 鎴戞潵绛 1涓洖绛 #鐑# 浣滀负濂虫,浣犵敓娲讳腑鏈夋劅鍙楀埌鈥滀笉瀹夊叏鎰熲濈殑鏃跺埢鍚?宕斿够澶 2022-05-09 路 TA鑾峰緱瓒呰繃107涓禐 鐭ラ亾绛斾富 鍥炵瓟閲:115 閲囩撼鐜:75% 甯姪鐨勪汉:81.8涓 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 ...
绛旓細浜岀淮闅忔満鍙橀噺(X锛孻)鏄竴涓潎鍖鍒嗗竷锛屽潎鍖鍒嗗竷姹姒傜巼锛屽彧闇闈㈢Н鐩搁櫎灏辫浜 P(X+Y<=1)=(1*1/2)/2=1/4
绛旓細璁句簩缁撮殢鏈哄彉閲(X,Y)鐨勬鐜瀵嗗害涓: 璁句簩缁撮殢鏈哄彉閲(X,Y)鐨勬鐜囧瘑搴︿负:f(x,y)=[cx^2yx^2<y<10鍏朵粬姹(1)璇曠‘瀹氬父鏁癈(2)姹傝竟缂樻鐜囧瘑... 璁句簩缁撮殢鏈哄彉閲(X,Y)鐨勬鐜囧瘑搴︿负:f(x,y)=[cx^2y x^2<y<1 0 鍏朵粬 姹(1)璇曠‘瀹氬父鏁癈(2)姹傝竟缂樻鐜囧瘑 灞曞紑 ...
绛旓細鏍规嵁姒傜巼瀵嗗害鍑芥暟鐨勭Н鍒=1锛屽彲浠ョ畻鍑篈鐨勫笺傚嵆锛氣埆 鈭 f(x,y) dx dy = 1 锛堚埆 鍧囦粠-鈭炵Н鍒嗗埌+鈭烇級銆傚垯浠庨涓凡鐭ユ潯浠跺彲寰楋紝+鈭 蟺/2 鈭 鈭 f(x,y) dx dy = 鈭 鈭 A sin(x+y) dx dy (x,y 鍧囦粠0鍒跋/2绉垎)-鈭 0 蟺/2 = A 鈭 -[cos(蟺/2+y) - co...
