Finding a minimum value of a probability

Question

Suppose we have r.v's $X_1,X_2,X_3$ so that $P(X_1 > X_2) = 0.7$ and $P(X_2 > X_3) = 0.6$. What is the minimum possible value of $P(X_1 > X_2 > X_3)$?

Try:

We can write

$$ P(X_1 > X_2 > X_3) = 1 - P(X_1 < X_2 < X_3) $$

And also $P(X_1 < X_2 < X_3) = P( \{ X_1 < X_2 \} \; \cap \; \{ X_2 < X_3 \} ) $. Thus,

$$ P(X_1 < X_2 < X_3) = P(X_1 < X_2) + P(X_2 < X_3) - P( \{ X_1 < X_2 \} \; \cup \; \{ X_2 < X_3 \} ) = 0.3+0.4 - - P( \{ X_1 < X_2 \} \; \cup \; \{ X_2 < X_3 \} ) $$

Thus,

$$ P(X_1 > X_2 > X_3) = 0.3 - P( \{ X_1 < X_2 \} \; \cup \; \{ X_2 < X_3 \} ) $$

This is where I get stuck. How can we evalue the probability in RHS?

Answer 1

I use $P(A \cup B) \le 1$.

Then $$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$

$$\to P(A \cap B) = 0.6 + 0.7 - P(A \cup B)$$

$$\to P(A \cup B) = 0.6 + 0.7 - P(A \cap B) \le 1$$

$$\to 0.6 + 0.7 - P(A \cap B) \le 1$$

$$\to 0.3 = 0.6 + 0.7 - 1 \le P(A \cap B)$$

Answer 2

That is similar to the probability that you will end up bounding.

As mentioned in the comments, your initial step is already wrong, since the two events $X_1 > X_2 > X_3$ and $X_3 > X_2 > X_1$ are not complements; there are loads of other cases like $X_2 > X_3 > X_1$, and $X_1 = X_2 > X_3$ and many more.

You already wrote the equation that you need to use the problem; you need to use the fact that $P(A \cap B) = P(A) + P(B) - P(A \cup B)$; note that $\{X_1 > X_2 > X_3\} = \{X_1 > X_2\} \cap \{X_2 > X_3\}$, and note that probabilities are always between 0 and 1. Can you finish from there?