E and F being events, F is independent of E but E is not independent of F. Any example?

Question

I was doing independent events and there I found that any two events(say $E$ and $F$) are independent when: $$P(F|E)=P(F)$$ provided $P(E)\ne0$

and $$P(E|F)=P(E)$$ provided $P(F)\ne0$

I'm having problem in understanding that why we have to work out the above two equations in order to see that $E$ and $F$ are independent.

I think that when $F$ is independent of $E$ then $E$ is also independent of $F$.

How can this happen that $E$ is independent of $F$ but $F$ is dependent of $E$.

Kindly provide me an example in which first event is independent of second but second is dependent on first.

Answer 1

It is not possible. If $E$ is independent of $F$, then the converse is true as well ($E$ and $F$ are independent). To see why, recall that $$ \Pr[E\mid F] = \frac{\Pr[E\cap F]}{\Pr F} = \frac{\Pr[E\cap F]}{\Pr E}\frac{\Pr E}{\Pr F} = \Pr[F\mid E]\frac{\Pr E}{\Pr F} $$ so that $\Pr[E\mid F] = \Pr[E]$ if and only if $\frac{\Pr[F\mid E]}{\Pr F} = 1$, i.e. $\Pr[F\mid E] = \Pr F$. Note that the "usual" (equivalent) definition of independence is $\Pr[E\cap F] = \Pr[E]\cdot\Pr[F]$.

Answer 2

Formally $P\left(F\mid E\right)$ is defined by the condition: $$P\left(F\mid E\right)P\left(E\right)=P\left(E\cap F\right)$$ If $P\left(E\right)\neq0$ this comes to the same as $P\left(F\mid E\right)=\frac{P\left(E\cap F\right)}{P\left(E\right)}$.

However if $P\left(E\right)=0$, then, since $(E\cap F) \subset E$, it follows that $P(E\cap F) \leq P(E)$ and so $P(E\cap F)$ also equals $0$. Thus, the definition of $P\left(F\mid E\right)$ as a fraction becomes of the form $\frac{0}{0}$ and cannot be used to decide the value of $P\left(F\mid E\right)$. Nor does $P(F \mid E)\cdot P(E) = P(E\cap F)$ help because $P(F \mid E)\cdot 0 = 0$ is true for all choices of value for $P(F \mid E)$.

Two events $E$ and $F$ are by definition independent if: $$P\left(E\cap F\right)=P\left(E\right)P\left(F\right)$$ Note that this is an expression symmetric in $E$ and $F$. Note also that if either $P(E)$ or $P(F)$ equals $0$, then $P(E\cap F)$ also necessarily equals $0$ and so $P(E\cap F)=P(E)P(F)$ automatically holds: an event of probability $0$ is independent of all other events.

Statement "$E$ is (in)dependent of $F$" must be read as: "$E$ and $F$ are (in)dependent".

If $P\left(E\right)\neq0$ then it comes to the same as $P\left(F\mid E\right)=P\left(F\right)$ and if $P\left(F\right)\neq0$ then it comes to the same as $P\left(E\mid F\right)=P\left(E\right)$.

I advice you not to use these conditional expressions as definition of independence. The special cases $P\left(E\right)=0$ and/or $P\left(F\right)=0$ are somehow spoiling.