Q. 3.22. - A First Course in Probability

Question

3.22. Prove or give counterexamples to the following statements:

(a) If E is independent of F and E is independent of G, then E is independent of $F\cup G$

(b) If E is independent of F, and E is independent of G, and$FG=\varnothing$, then E is independent of$F\cup G$.

(c) If E is independent of F, and F is independent of G, and E is independent of F G, then G is independent of EF.

Step-by-Step Solution

Answer

a)$E$ is independent of $F$ and $E$ is independent of $G$, then $E$ is independent of$F\cup G$ is  False

b) $E$ is independent of $F$, and $E$ is independent of $G$, and $FG=\varnothing$, then E is independent of $F\cup G$. is Correct, use characterization of independence with conditional probability

c)$E$ is independent of $F$, and $F$ is independent of $G$, and $E$ is independent of $F G$, then G is independent of $EF$ is Correct, use characterization of independence with probability of intersection

Step1: prove E is independent of F and E is independent of G , then E is independent of F ∪ G (part a)

$E$ and $F$ independent

$E$ and $G$ independent

$\Rightarrow E$ independent of$F\cup G$

This is false

Counterexample: Two fair dice are rolled.

$E$ - the sum of the results is even.

$F$- result on the second die is $3 .$

$G$- result on the first die is$4$

Since $P(E\mid F)=\frac{1}{2}$, and $P(E\mid G)=\frac{1}{2}$and $P\left(E\right)=\frac{1}{2}$ (this can be obtained by conditioning on the number on the first die).

 E and F and E and G are independent.

$P(E\mid F\cup G)=\frac{6}{11}$- method of counting.

Thus:

$P(E\mid F\cup G)\ne P\left(E\right)$

.So $E$ and $F\cup G$are not independent.

Step2: Prove E is independent of F , and E is independent of G , and F G = ϕ , then E is independent of(part b)

$E$ and $F$ independent

$E$ and $G$ independent

$F\cap G=\varnothing$

$\Rightarrow E$ independent of $F\cup G$

A and B are independent $\iff P(A\mid B)=P\left(A\right)/P(B\mid A)=P\left(B\right)$

$P(E\mid F\cup G)=\frac{P(E\cap (F\cup G\left)\right)}{P(F\cup G)}$

$=\frac{P(EF\cup EG))}{P(F\cup G)}$

$F\cap G=\varnothing ,\Rightarrow EF\cap EG=\varnothing$

$=\frac{P\left(EF\right)+P\left(EG\right)}{P\left(F\right)+P\left(G\right)}$

Independence

$=\frac{P\left(E\right)P\left(F\right)+P\left(E\right)P\left(G\right)}{P\left(F\right)+P\left(G\right)}$

$=\frac{P\left(E\right)\left[P\right(F)+P(G\left)\right]}{P\left(F\right)+P\left(G\right)}$

$=P\left(E\right)$

And$P(E\mid F\cup G)=P\left(E\right)$ proves independence

Step3: prove E is independent of F, and F is independent of G, and E is independent of F G, then G is independent of E (part c)

E and F independent

F and G independent

E and FG independent

$F\cap G=\varnothing$

This is correct

Use

A and B are independent $\iff P\left(AB\right)=P\left(A\right)P\left(B\right)$

$P\left(EFG\right)=P\left(E\right)P\left(FG\right)$

E,FG independent

$=P\left(E\right)P\left(F\right)P\left(G\right)$

F,G independent

$=P\left(EF\right)P\left(G\right)$

E,F independent

And $P(E F G)=P\left(G\right) P(E F)$ proves independence of G and E F