Show that for any events Eand F, P(E∣E∪F)≥P(E∣F)Hint: Compute P(E∣E∪F) by conditioning on whether F occurs.

The result is that P(E∣E∪F)is weighted average between P(E∣F) and 1

Q. 3.30 - A First Course in Probability

Show that for any events $E$ and $F$ ,

$P (E ∣ E \cup F) \geq P (E ∣ F)$

Hint: Compute $P (E ∣ E \cup F)$ by conditioning on whether F occurs.

Verified

Answer

The result is that $P (E ∣ E \cup F)$ is weighted average between $P (E ∣ F) a n d 1$

1Step 1: Prove

Demonstrate as all eventualities E,F

$P (E ∣ E \cup F) = P [E ∣ F \cap (E \cup F)] P (F ∣ E \cup F) + P [E ∣ F^{c} \cap (E \cup F)] P (F^{c} ∣ E \cup F)$

If it's not clear, evaluate that equations via enlarging the correct hand side by a probability density statement.

$F \cap (E \cup F) = F$

$F^{c} \cap (E \cup F) = E \cap F^{c}$

$\Rightarrow P (E ∣ F^{c} \cap (E \cup F)) = P (E ∣ E F^{c}) = 1$

2Step 2: Weighted Average

That's also sufficient to ascertain the disparity, as

$P (E ∣ E \cup F) = P (E ∣ F) P (F ∣ E \cup F) + 1 \cdot P (F^{c} ∣ E \cup F)$

$P (F ∣ E \cup F) + P (F^{c} ∣ E \cup F) = 1$

$P (E ∣ F) \leq 1$

thus $P (E ∣ E \cup F)$ is that the weighted combination of such constants $P (E ∣ F)$ and $1$ , so just between that two integers,

$P (E ∣ F) \leq P (E ∣ E \cup F) \leq 1$