Question

Let X be a random variable with CDF given byFX(t) =0 for t < 1,1 /2 for ?1 t < 11/ 2 t for 1 t < 21 for t 2Calculate E[X]

Answer 1

`\mathbf{E(X) = (3)/(4)}`

Explanation:

From the given data, the cumulative distribution function of a random variable can be represented as:

`F_X(t) =\left\{ \begin{array}{c}0........... t <-1 \\ (1)/(2) ... -1 \leq t < 1\\ (1)/(2) ....... 1 \leq t < 2 \\ 1 .............. t \geq 2\\\end{array}\right.`

The objective is to estimate E(X), to do that, let's first evaluate the probability density function by differentiating the cumulative distribution function from above.

`f_X(x) =\left \{ {{(1)/(2) .......1 \leq x \leq 2 } \atop {0..... otherwise }} \right.`

∴

`f_X(t) =\left\{ \begin{array}{c} (d)/(dx)(0)=0........... <-1 \\ (d)/(dx)((1)/(2) ) =0... -1 \leq t < 1\\ (d)/(dx)((1)/(2)x) = (1)/(2)....... 1 \leq x < 2 \\ (d)/(dx)(1) = 0 .............. x \geq 2\\\end{array}\right.`

The expected value of x i

.e E(X) can now be estimated by taking the integral:

`E(X) = \int ^(\infty)_(\infty) x f(x) \ dx`

`E(X) = \int ^(1)_(- \infty) x 0 dx + \int^2_1 \ x (1)/(2)\ dx + \int ^(\infty)_2 \ x0dx`

`E(X) = \int ^(2)_(1) x (1)/(2) dx`

`E(X) = (1)/(2)[(x^2)/(2)]^2_1`

`E(X) = (1)/(2)[(4)/(2)-(1)/(2)]`

`E(X) = (1)/(2) * [(3)/(2)]`

`\mathbf{E(X) = (3)/(4)}`