Suppose I have a sequence of iid random variables $(X_n)_{n\in \mathbb{N}}$ would I be correct in thinking that the following is true?

$$\mathbb{P}(X_k>X_1,\cdots, X_{k-1})=\mathbb{P}(X_k>X_1)^{n-1}$$

My thoughts for this is that $$\mathbb{P}(X_k>X_1,X_k>X_2,\cdots,X_k>X_{k-1})=\mathbb{P}(X_k>X_{k-1})\cdots\mathbb{P}(X_k>X_1)=\mathbb{P}(X_k>X_1)^{n-1}$$ where the first equality follows from independence and second equality follows from identically distributed.

However, there is also a part of me that goes against this argument, let me illustrate with an example:

Suppose we just consider $k=3$ then


where I obtained the square on $\mathbb{P}(X_3>X_1)$ is due to I splitted up $\mathbb{P}(X_3>X_2>X_1)=\mathbb{P}(X_3>X_2)\mathbb{P}(X_2>X_1).$ Then evidently two are not the same.

Where have I gone wrong in applying independence to this problem? Without further context on $X_n$ how can we reach a correct expression to the concerned probability? (which is $\mathbb{P}(X_k>X_1,\cdots, X_{k-1})$)

Many thanks!


1 Answer 1


If $P_1, P_2, P_3$ are i.i.d., then

$$\mathbb{P}(X_3>X_2>X_1) \neq \mathbb{P}(X_3>X_2)\mathbb{P}(X_2>X_1)$$

because the two events on the right are not independent.

The way we usually approach this is to observe that since the variables are i.i.d., no one variable is more likely to be greatest than any other:

$$\mathbb{P}(X_1>X_2,X_3) = \mathbb{P}(X_2>X_1,X_3) = \mathbb{P}(X_3>X_1,X_2). $$

If the variables have continuous distributions then the probability that any two are exactly equal is zero, and the probability is $1$ that there will be one variable that is greater than the other two:

$$\mathbb{P}(X_1>X_2,X_3) + \mathbb{P}(X_2>X_1,X_3) + \mathbb{P}(X_3>X_1,X_2) =1. $$

From these two equations you can find $\mathbb{P}(X_3>X_1,X_2).$

If the distribution is not continuous then the calculation is typically a bit more complicated because you have to account for the non-zero probability that no variable is greater than the other two.

  • $\begingroup$ Ah I see! That is a brilliant approach! Thank you so much! $\endgroup$ Commented Nov 12, 2020 at 16:39

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