SOLUTION: please helps in this let the pdf f(x) be positive at x=-1,0,1 and zero else where 1. if f(0) = 0.25 , find E(x^2) 2. if f(0) = 0.25 and if E(X) =0.25 , determine f(-1) and f(1

Question 59211: please helps in this
let the pdf f(x) be positive at x=-1,0,1 and zero else where
1. if f(0) = 0.25 , find E(x^2)
2. if f(0) = 0.25 and if E(X) =0.25 , determine f(-1) and f(1)

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
let the pdf f(x) be positive at x=-1,0,1 and zero else where
1. if f(0) = 0.25 , find E(x^2)
Let f(-1)=x
The f(1)=0.75-x
E(X^2)=(0^2)*0.25+(-1)^2*x+(1)^2*(0.75-x)
E(X^2)=0+x+(0.75-x)
=0.75
------------
2. if f(0) = 0.25 and if E(X) =0.25 , determine f(-1) and f(1)
Let f(-1)=x
Then f(1)=(0.75-x)
E(X)=0*0.25+(-1)x+(1)(0.75-x)=0.25
-x+0.75-x=0.25
-2x=-0.5
x=0.25
So, f(-1)=x=0.25
f(1)=(0.75-x)=0.75-0.25=0.50
Cheers,
Stan H.
RELATED QUESTIONS
Let X be a random variable with pdf f (x) = kx^2 where 0 ≤ x ≤ 1. (a) Find k. (b)... (answered by CPhill)

Let X be a random variable with pdf f (x) = kx^2 where 0 ≤ x ≤ 1. (a) Find k. (b)... (answered by CPhill)

If f(x)=3x^2-0-1,find f(0),f(-2),and... (answered by JBarnum)

Please help me with this problem. I'm not really good at probability and statistics. A... (answered by stanbon)

Let X be a random variable with pdf f(x) = kx^2 where 0 is less than or equal to x which... (answered by Fombitz)

A discrete random variable has pdf f(x). (a) If f(x) = k(1/2)^x for x= 1, 2, 3, and zero (answered by math_tutor2020)

Let f(x) be a polynomial. Find the remainder when f(x) is divided by x(x - 1)(x - 2), if (answered by CPhill)

The random variables (X,Y) have joint pdf f(x,y) = 1/x for the region 0 < x < 1, 0 (answered by CPhill)
The random variables (X,Y) have joint pdf f(x,y) = 1/x for the region 0 < x < 1,... (answered by CPhill)