Topic: 有学统计的没有,请出手相助: Determine constants A and B such that the function f(x) given below is a valid PDF(probability distribution fuction)whose mean (u) is equal to 0.9 times its mode(Xm). f(x)=Ax-Bx^3, 0<=x<=2 @华人论坛:枫下论坛 The Rolia Forum of

工作学习 / IT杂谈 / 有学统计的没有,请出手相助: Determine constants A and B such that the function f(x) given below is a valid PDF(probability distribution fuction)whose mean (u) is equal to 0.9 times its mode(Xm). f(x)=Ax-Bx^3, 0<=x<=2 -soochow(©乡屋宁™); 2003-9-23 (#1381689@0)

算你运气好，统计我最拿手了。mode(Xm)指在X的定义域内，某个出现凭率最高的值。比如你的例子，0<=X<=2，X若在此范围内取指0, 0, 1, 1.5, 1.5, 1.5, 1.7, 2，那么mode(Xm)=1.5
明白？ -jqian(Q_Q); 2003-9-23 {143} (#1381949@0)

2a-4b=1,8a/3-32b/5=sqrt(a/b)/10, it seems it's not easy to solve it. -raymondo(raymondooo); 2003-9-23 (#1382402@0)
solution:f(x) to be a valid PDF if and only if:
integration of f(x) = 1, x from 0 to 2 -- equation 1 of A and B

mean of x = integration of x*f(x), x from 0 to 2
( which is a function of A and B)

mode of x:
the x which makes f(x) to be the maximum between [0,2] -- Xm
df(x)/dx=A-3Bx^2=0 -> Xm=(A/3B)^(1/2)

u=0.9*Xm --equation 2 of A and B

two equations and two variables, can solve for A & B -eagles(海阔天空); 2003-9-23 {506} (#1382457@0)

@