Question

Explain why each of the following is or is not a valid probability distribution for a discrete random variable x:

a.

x	0	1	2	3
P(x)	.1	.3	.3	.2

b.

x	-2	-1	0
P(x)	.25	.50	.25

c.

x	4	9	20
P(x)	-3	.4	.3

d.

x	2	3	5	6
P(x)	.15	.15	.45	.35

Short answer

a. It is not valid.

b. It is valid.

c. It is not valid

d. It is valid.

Step-by-step solution

Identifying the criteria to test the validity

a. The main criterion to find out whether the probability distribution is valid is to check the total probability associated with each value. The total probability will have to be equal to 1 and not any other number in case of .valid probability distribution.

Identifying the total probability

The summation of the associated probabilities, 0.1, 0.3, 0.3, and 0.2, is shown below:

$\begin{array}{rcl}\mathit{T}\mathit{o}\mathit{t}\mathit{a}\mathit{l}\mathbf{\text{probability}}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{1}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{2}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{9}\\ & & \end{array}$

The summation is 0.9, which is less than 1. So, it is not a valid probability distribution.

Identifying the total probability

b.

The summation of the associated probabilities, 0.25, 0.50, and 0.25 is shown below:

$\begin{array}{rcl}\mathit{T}\mathit{o}\mathit{t}\mathit{a}\mathit{l}\mathbf{\text{probability}}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{25}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{25}\\ & \mathbf{=}& \mathbf{1}\\ & & \end{array}$

The summation is 1 and so it is a valid probability distribution.

Identifying the total probability

c.

The associated probability of 4 is -3 which is not possible as the value of a probability cannot be negative. Therefore, there is no chance that the summation will be 1 and so the probability distribution is invalid.

Identifying the total probability

d.

The summation of the associated probabilities, 0.15, 0.15, 0.45, and 0.35 is shown below:

$\begin{array}{rcl}\mathit{T}\mathit{o}\mathit{t}\mathit{a}\mathit{l}\mathbf{\text{probability}}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{15}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{45}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{35}\\ & \mathbf{=}& \mathbf{1}\\ & & \end{array}$

The summation is 1 and so the distribution is completely valid.