# What can we say about P(X<Y and X<Z)?

Let $$X$$, $$Y$$ and $$Z$$ be independent random variables, and we are interested in estimating $$P(X.

If we say that $$X$$, $$Y$$, and $$Z$$ are identically distributed, then all permutations are equally possible, i.e.:

$$P(X $$=P(Z

Thus:

$$P(X

And:

$$P(X

The above is all very simple to calculate based on enumerating the permutations.

However, let's say we have a different situation. All three random variables are still independent, but only $$Y$$ and $$Z$$ are identically distributed (i.e. $$X$$ has a different distribution), and all we know about the relationship between the variables is:

$$P(X

Under this situation, what can we say about $$P(X?

We can decompose it as follows:

$$P(X

I don't believe it is possible to evaluate that conditional probability without making additional assumptions about the probability distributions. But it seems to me that we should at least be able to put bounds on what that probability can be, given our constraints (i.e. that $$Y$$ and $$Z$$ are i.i.d. and the marginal probabilities $$P(X and $$P(X are known). I don't have a very good idea of how to proceed from here, or even to what degree it is possible without additional assumptions.

Since $$Y$$ and $$Z$$ are exchangable, denote $$P(X
So the target can be rewritten as $$P(X.
According to the permutations and the additional condition, $$p_1,p_2,p_3$$ satisfy the following relations: $$\begin{cases} 2p_1+2p_2+2p_3=1\\ 2p_1+p_2=\frac{2}{3}\\ \end{cases}$$ Solve this linear system, will get $$2p_1=\frac{2}{3}-p_2\\ 2p_3=\frac{1}{3}-p_2$$ In order to make $$p_1\ge 0,p_2\ge 0,p_3\ge 0$$, $$p_2$$ must satisfy $$0\le p_2\le \frac{1}{3}$$, so that $$2p_1=\frac{2}{3}-p_2 \ge \frac{1}{3}$$ and $$2p_1 \le \frac{2}{3}$$. That is, $$\frac{1}{3} \le P(X