Question

A discrete random variable X has possible values xi=i^2 i=1,2,3,4,5, which occour with probabilities 0.4 0.25 0.15 0.1 and 0.1 respectively
a) write the probability density fx(x) and probability distrubution Fx(x) functions for X random variable
b)find mean value E[X]

Answer 1

Task. A discrete random variable $X$ has possible values $x_{i} = i^{2}$ , $i = 1,2,3,4,5$ , which occur with probabilities 0.4, 0.25, 0.15, 0.1, and 0.1 respectively.

a) write the probability density $f_{X}(t)$ and probability distribution $F_{X}(t)$ functions for $X$ random variable

b) find mean value $E[X]$

Solution. The distribution of probabilities of values of $X$ is given in the following table:

Since $X$ is discrete, its probability density function is the sum of delta-functions:

$$f _ {X} (t) = 0. 4 \delta (t - 1) + 0. 2 5 \delta (t - 4) + 0. 1 5 \delta (t - 9) + 0. 1 \delta (t - 1 6) + 0. 1 \delta (t - 2 5),$$

and its probability distribution $F_{x}(t)$ is given by the formula:

$$F _ {X} (t) = \left\{ \begin{array}{l l} 0, & t < 1 \\ 0. 4, & 1 \leq t < 4 \\ 0. 4 + 0. 2 5 = 0. 6 5, & 4 \leq t < 9 \\ 0. 6 5 + 0. 1 5 = 0. 8, & 9 \leq t < 1 6 \\ 0. 8 + 0. 1 = 0. 9, & 1 6 \leq t < 2 5 \\ 1, & t \leq 2 5 \end{array} \right.$$

Thus

$$F _ {X} (t) = \left\{ \begin{array}{l l} 0, & t < 1 \\ 0. 4, & 1 \leq t < 4 \\ 0. 6 5, & 4 \leq t < 9 \\ 0. 8, & 9 \leq t < 1 6 \\ 0. 9, & 1 6 \leq t < 2 5 \\ 1, & t \leq 2 5 \end{array} \right.$$

The mean value $E[X]$ is defined by the formula:

$$\begin{array}{l} E [ X ] = \sum_ {i = 1} ^ {5} x _ {i} p (x _ {i}) = 1 * 0. 4 + 4 * 0. 2 5 + 9 * 0. 1 5 + 1 6 * 0. 1 + 2 5 * 0. 1 \\ = 0. 4 + 1 + 1. 3 5 + 1. 6 + 2. 5 = 6. 8 5. \\ \end{array}$$