Question

If random variable (x) takes vales 1, 2, 3, 4 such

that 2 P(x=1) = 3 P (x=2) = P(x=3) = 5 P (x=4)

Find the probability distribution function (PDF) and cumulative distribution function

(CDF), also state the properties of PDF and CDF.

Answer 1

2 P(x=1) = 3 P (x=2) = P(x=3) = 5 P (x=4) =t

Then P(x=1)=t/2, P (x=2)=t/3, P(x=3)=t, P (x=4) =t/5

1=P(x=1) + P(x=2) + P(x=3)+ P(x=4) = t/2+t/3+t+t/5 = 61t/30

Therefore, t=30/61

P(x=1) = 15/61

P(x=2) = 10/61

P(x=3) = 30/61

P(x=4) = 6/61

Probability distribution function (PDF) is

$P(x=t)=\begin{Bmatrix} 15/61 & if\ t=1 \\ 10/61 & if\ t=2 \\ 30/61 & if\ t=3 \\ 6/61 & if\ t=4 \\ 0 & else \end{Bmatrix}$

Its properties:

1) This PDF takes non-zero values in a finite set of points (discreteness).

2) All values of PDF are non-negative real numbers, not greater than 1.

3) The sum of all non-zero values of PDF equals 1.

Cumulative distribution function (CDF) is $P(x\leq t)$.

If t<1 then CDF(t)=0, because of x does not take any values t<1.

If $1\leq t<2$ then $P(x\leq t)=P(x=1)=15/61$

If $2\leq t<3$ then $P(x\leq t)=P(x=1 \vee x=2)=P(x=1)+P(x=2)=15/61+10/61=25/61$

If $3\leq t<4$ then $P(x\leq t)=P(x=1)+P(x=2)+P(x=3)=15/61+10/61+30/61=55/61$

If $t>4$ then $x\leq t$ in any case and$P(x\leq t)=1$.

Finally,

$P(x\leq t)=\begin{Bmatrix} 0 & if\ t<1 \\ 15/61 & if\ 1\leq t<2 \\ 25/61 & if\ 2\leq t<3 \\ 55/61 & if\ 3\leq t<4 \\ 1 & if\ t\geq 4 \end{Bmatrix}$

Its properties:

1) CDF takes its values in the segment [0,1].

2) CDF is monotoneously increases everywhere

3) CDF has limits: $\lim\limits_{t\to-\infty}P(x\leq t)=0$ and $\lim\limits_{t\to+\infty}P(x\leq t)=1$.

4) CDF is continuous at every point t, such that PDF(t)=0.