Question

determine if the following are probability distribution (if no, state why)

X=1,2,3,4 and 5

P(x)= 3/10, 1/10, 1/10, 2/10 and 3/10

Answer 1

By using the conditions of this problem, we may construct a probability space by the following:

1) Take $\Omega=\{1,2,3,4,5\}$ as a sample space;

2) Take $\mathbb{F}_{\Omega}=2^{\Omega}$, which is the sigma-algebra of all subsets of $\Omega$, as an event space;

3) Take $Pr:\mathbb{F}_{\Omega}\to[0,+\infty)$, $A\mapsto \sum\limits_{x\in A}P(x)$ as a probability function.

This function is additive: for all disjoint subsets $A,B\subset\Omega$, we have

$Pr(A\cup B)=\sum\limits_{x\in A\cup B}P(x)=\sum\limits_{x\in A}P(x)+\sum\limits_{x\in B}P(x)$

$=Pr(A)+Pr(B)$.

Since $\Omega$ is a finite set, the finite additiveness of $Pr$ is equivalent to its sigma-additiveness.

Since for all $x\in\Omega$, $P(x)>0$, the function $Pr$ is non-negative.

Since $Pr(\Omega)=\sum\limits_{x\in \Omega}P(x)=P(1)+P(2)+P(3)+P(4)+P(5)$

$=\frac{3}{10}+\frac{1}{10}+\frac{1}{10}+\frac{2}{10}+\frac{3}{10}=1$, the total probability is equal to 1.

Therefore, such defined a function $Pr$ is indeed a probability function, that is, the function $P(x)$ correctly defines a probability distribution on the set {1, 2, 3, 4, 5}.