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

Answer:

$\mathbf{E(X) = \dfrac{3}{4}}$

Step-by-step explanation:

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

$F_X(t) =\left\{ \begin{array}{c}0........... t$

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.

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

∴

$f_X(t) =\left\{ \begin{array}{c} \dfrac{d}{dx}(0)=0...........$

The expected value of x i

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

$E(X) = \int ^{\infty}_{\infty} x f(x) \ dx$

$E(X) = \int ^{1}_{- \infty} x 0 dx + \int^2_1 \ x \dfrac{1}{2}\ dx + \int ^{\infty}_2 \ x0dx$

$E(X) = \int ^{2}_{1} x \dfrac{1}{2} dx$

$E(X) = \dfrac{1}{2}[\dfrac{x^2}{2}]^2_1$

$E(X) = \dfrac{1}{2}[\dfrac{4}{2}-\dfrac{1}{2}]$

$E(X) = \dfrac{1}{2} \times [\dfrac{3}{2}]$

$\mathbf{E(X) = \dfrac{3}{4}}$