Proof that $\operatorname{Pr}(E|EUF)\ge\operatorname{Pr}(E|F)$

Question

I tried rewriting the equation to $\frac{\operatorname{Pr}(E \cap (EUF))}{\operatorname{Pr}(EUF)}\geq\frac{\operatorname{Pr}(E\cap F)}{\operatorname{Pr}(F)}$,

but cannot get any further. Any hints/ideas?

Answer 1

Use the definition of conditional probability:

We want to prove that $P(E \text{ and } (E \text{ or } F)) / P(E \text{ or } F) \geq P(E \text{ and } F) / P(F)$

Now, $E \text{ and }(E \text{ or } F) = E$ by the absorption property.

Therefore, we have to prove $P(E) / P(E \text{ or } F) \geq P(E \text{ and } F) / P(F)$

Clear denominators. Now, it is sufficient to prove $P(E) P(F) \geq P(E \text{ and } F)P(E \text{ or } F)$.

To prove this, we split up the probability space into $A = E - F, B = E \cap F, C = F - E$ and $D = E^c \cap F^c$. Note that they are all disjoint.

Therefore, we have $P(E)$ $P(F) = (P(A) + P(B))(P(B) + P(C))$, and $P(E \text{ and } F)P(E \text{ or } F) = P(B) (P(A) + P(B) + P(C))$.

From here you can solve the rest.

Answer 2

First, it is trivial to prove that if $a,b,c\ge 0, b\ne 0$ and $a\le b$, then $$\frac{a+c}{b+c}\ge \frac{a}{b}\tag{1}\label{1}$$

Then:

$$P(E|E\cup F)=\frac{P(E\cap(E\cup F))}{P(E\cup F)}=\frac{P(E)}{P(E\cup F)}=\frac{P(E\cap F)+P(E\setminus F)}{P(F)+P(E\setminus F)}\stackrel{(\ref{1})}{\ge} \frac{P(E\cap F)}{P(F)}=P(E|F)$$

Which was to be proved.