0
$\begingroup$

Let $X_1 .... X_K$ represent $K$ independent random variables. These var (of unknown distribution). I am trying to understand how to simplify the probability that $X_1 $ is the greatest of all. That is:

$$Pr(X_1 > X_2, X_1 > X_3 ... X_1 > X_K)$$

Notice that I am not excluding the chance that the variables are equal. They could be equal, so there is a non-zero probability of that.

My understanding is that, in general, the events of interest (the logical comparisons) are not independent. That being the case, I am trying to decompose the formula above into simple probabilities (not joint or conditional).

With the simple case of $K = 3$, and using the formula of conditional probabilities, I get at:

\begin{align}Pr(X_1 > X_2, X_1 > X_3) = Pr(X_1 > X_3 | X_1 > X_2) \cdot Pr(X_1 > X_2) \end{align}

Which I think can be simplified like:

\begin{align}Pr(X_1 > X_2, X_1 > X_3) =& Pr(X_1 > X_2)\cdot[Pr(X_3 \leq X_2)+Pr(X_3>X_2)Pr(X_1>X_3)]\\ & \cdot Pr(X_1 > X_2) \end{align}

What I am trying to understand is, first, if the above simplification makes sense, and second, how that could be extended to $K > 3$ in a generic way.

EDIT: I can impose the assumption that $X_1,...,X_K$ follow the same distribution family, but I cannot assume they follow exactly the same distributin, that is, with same distribution parameters.

$\endgroup$
  • $\begingroup$ @spaceisdarkgreen Sorry for missing that info, I will edit to add it: yes, the rvs could be equal. $\endgroup$ – Louis15 Jul 3 '19 at 21:57
1
$\begingroup$

One way to compute a generic answer is to condition on $X_1$. Given $X_1$, the comparisons do become independent, and we obtain $$ \mathbb P(X_1>X_2,\ldots,X_1>X_{K}\mid X_1)=\mathbb P(X_1>X_2\mid X_1)^{K-1}. $$ Even though you didn't say it explicitly, I am assuming that $X_1,\ldots,X_K$ all have the same distribution. Thus, $$ \mathbb P(X_1>X_2,\ldots,X_1>X_{K})=\mathbb E \left[\mathbb P(X_1>X_2\mid X_1)^{K-1}\right]. $$ Note, I am using the notation of conditional expectation as described here, and if you aren't familiar with this concept then the meaning of the expression I have written may be lost. It has another equivalent expression that may be clearer to you. Let $f(x)$ be (a variant of) the CDF of $X_2$, defined so that $f(x)=\mathbb P(X_2<x)$. Then the quantity $\mathbb P(X_1>X_2\mid X_1)$ is equal to $f(X_1)$. Note that it is random, obtained by substituting $X_1$ into the (deterministic) function $f(x)$. Using this notation, the formula can be equivalently written as follows: $$ \mathbb P(X_1>X_2,\ldots,X_1>X_{K})=\mathbb E \left[f(X_1)^{k-1}\right]. $$

$\endgroup$
  • $\begingroup$ Thanks for your answer. Unfortunately, while I can assume $X_1,...,X_K$ have a same distribution type, I cannot assume they have exactly the same distribution parameters. With that, I don't think I can even assume $ \mathbb P(X_1>X_2,\ldots,X_1>X_{K}\mid X_1)=\mathbb P(X_1>X_2\mid X_1)^{K-1}. $, right? $\endgroup$ – Louis15 Jul 7 '19 at 2:52
  • $\begingroup$ @Louis15 Of course the same technique works, just instead of $\mathbb P(X_1>X_2\mid X_1)^{K-1}$ you would have a product $\mathbb P(X_1>X_2\mid X_1)\cdot \mathbb P(X_1>X_3\mid X_1)\cdots \mathbb P(X_1>X_K\mid X_1)$. $\endgroup$ – pre-kidney Jul 7 '19 at 20:01
  • $\begingroup$ Oh, you are obviously right. And then the expectation of that product works the same too. In which case I guess if we let $f_i(x)$ be the modified CDF of variable $X_i$ (or more specifically the CDF without allowing $X_i = x$), then we could still end with $\mathbb{P}(X_1>X_2,\ldots,X_1>X_{K}) = \mathbb{E} [f_2(X_1) \cdot f_3(X_1) \cdot ... \cdot f_K(X_1)]$. If this makes sense to your line of thinking, the only thing left for me to understand and which I would appreciate if you could elaborate on, is why exactly in your answer $\mathbb P(X_1>X_2\mid X_1) = \mathbb P(X_2 < X_1)$? $\endgroup$ – Louis15 Jul 8 '19 at 5:43
  • $\begingroup$ @Louis15 in fact $\mathbb P(X_1>X_2\mid X_1)\not=\mathbb P(X_2<X_1)$ and they were never claimed to be equal in my answer. More to the point, $f(X_1)\not=\mathbb P(X_2<X_1)$: you cannot substitute $X_1$ in for $x$ inside a $\mathbb P$ symbol, since changing whether a quantity is random or deterministic will change which variables the probability is taken with respect to (i.e., which variables will be "summed over" or "integrated out" as the case may be). In my opinion it is an unfortunate aspect of the probabilistic notation but historical precedent can't be changed easily... $\endgroup$ – pre-kidney Jul 9 '19 at 5:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.