Answer:
a) c=3/4
b) for -1<x<1
F(x) = (3/4) ( x + (2-x^3)/3 )
c)
E(x)=0
Var(x) = 1/5
Step-by-step explanation:
Hi!
a)
In order to f(x) be a probability density function it must be normalized, which means that its integral must be equal to 1:
[tex]1 = \int\limits^{\infty}_{-\infty} {f(x)} \, dx[/tex]
Since f(x) is zero for x>1 and x<-1
[tex]1 = \int\limits^1_{-1} {f(x)} \, dx = \int \limits^1_{-1} {c(1-x^2)}\, dx \\\\\\1 = c ((1-\frac{1}{3}) - (-1 - \frac{-1}{3})) = c(\frac{2}{3} - \frac{-2}{3} ) = c\frac{4}{3}[/tex]
Therefore:
c=3/4
b)
The cumulative distribution fucntion F(x) can be obtained integrating f(x) from -∞ to x:
Since f(x) = 0 for x<-1
F(x) = 0 for x<-1
for -1<x<1:
[tex]F(x) = \int \limits^x_{-1} f(x')\, dx' = c(x'-\frac{x'^3}{3})^x_{-1} = c (x-\frac{x^3}{3} + \frac{2}{3})\\\\F(x) = \frac{3}{4}(x + \frac{2-x^3}{3})[/tex]
for x>1
F(x)=1
c)
The mean E(x) can be found integrating xf(x)
[tex]E(x) = \int \limits_{-1}^1 {x f(x)}\, dx[/tex]
We can easily infer that the mean must be zero, since f(x) is an even function and thus xf(x) is an odd function integrated in a simetric interval:
E(x) = 0
The variance of x, Var(x), can be evaluated integrating (x-E(x))^2f(x), since E(x)=0:
[tex]Var(x) = \int \limits^{1}_{-1} {x^2f(x)}\, dx\\Var(x) = \int \limits^{1}_{-1} {c (x^2 - x^4 )dx}\\Var(x) = c \frac{4}{15}[/tex]
Since c=3/4
Var(x) = 1/5
