设随机变量X~N(0.1),求Y=4-x^2的概率密度

e^[(y-4)/2]/[根号(pi（8-2y)) sigma^2] 好人做到底吧

这就是简便方法

一般是求大F然后求导

等上电脑再给你行吗

是求大F换底求导 FY(y)=P(Y<=y)= P(4-X^2<=y)=P(X^2>=(4-y))=P(X>=根号(4-y))+P(X<=-根号(4-y)) =∫(根号(4-y)~无穷) f(x) dx+∫(负无穷~-根号(4-y)) f(x) dx x=根号(4-y) 右半边，恰相反 dx=-1/(2根号(4-y)) dy dx=1/(2根号(4-y)) dy 范围转换是用4减去原先的数的平方 (根号(4-y)~无穷)->(4-(4-y)~4-无穷) (负无穷~-根号(4-y))->(4-无穷~4-(4-y)) FY(y)=∫(y~负无穷)f(x)(-1/(2根号(4-y)) dy+∫(负无穷~y)f(x)(1/(2根号(4-y)) dy =∫(负无穷~y)f(x)/(2根号(4-y)) dy+∫(负无穷~y)f(x)(1/(2根号(4-y)) dy =∫(负无穷~y){f(x)/(2根号(4-y)) +f(x)/(2根号(4-y))} dy =∫(负无穷~y){f(x)/根号(4-y)} dy 根据y求导，自然得到 fY(y)=dFY(y)/dy=f(x)/根号(4-y) (y<=4)