Question

Let X be a random variable with probability density function fX(x) = ( c(1 − x 2 ) if − 1 < x < 1 0 otherwise. a) What is the value of c? b) What is the cumulative distribution function of X? c) Compute E(X) and Var(X).

Answer 1

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:

$1 = \int\limits^{\infty}_{-\infty} {f(x)} \, dx$

Since f(x) is zero for x>1 and x<-1

$1 = \int\limits^1_{-1} {f(x)} \, dx = \int \limits^1_{-1} {c(1-x^2)}\, dx \\\\\\1 = c ((1-\frac{1}{3}) - (-1 - \frac{-1}{3})) = c(\frac{2}{3} - \frac{-2}{3} ) = c\frac{4}{3}$

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:

$F(x) = \int \limits^x_{-1} f(x')\, dx' = c(x'-\frac{x'^3}{3})^x_{-1} = c (x-\frac{x^3}{3} + \frac{2}{3})\\\\F(x) = \frac{3}{4}(x + \frac{2-x^3}{3})$

for x>1

  F(x)=1

c)

The mean E(x) can be found integrating  xf(x)

$E(x) = \int \limits_{-1}^1 {x f(x)}\, dx$

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:

$Var(x) = \int \limits^{1}_{-1} {x^2f(x)}\, dx\\Var(x) = \int \limits^{1}_{-1} {c (x^2 - x^4 )dx}\\Var(x) = c \frac{4}{15}$

Since c=3/4

Var(x) = 1/5