If $X_1,\ldots,X_n$ are independent, does $\mathbb{P}\{X_n>\max\{X_1,\ldots,X_{n-1}\}\}=\mathbb{P}\{X_n>X_1)\cdots\mathbb{P}\{X_n>X_{n-1}\}$ hold?

Question

If $X_1, \ldots, X_n$ are independent but not necessarily identically distributed random variables, is it always true $\mathbb{P}\{X_n > \max\{X_1, \ldots, X_{n-1}\}\} = \mathbb{P}\{X_n > X_1, \ldots, X_n > X_{n-1}\} = \mathbb{P}\{X_n > X_1\}\cdots \mathbb{P}\{X_n > X_{n-1}\}$?

Answer 1

No you need an integral. The probability is equivalent to:

$$P\left(\bigcap_{i\neq n} X_n>X_i\right)=\int P(\cap_{i\neq n} X_n>X_i|X_n=x_n)dP_{X_n}(x_n)=\int\prod_{i\neq n}P(X_n>X_i|X_n=x_n)dP_{X_n}(x_n).$$

Answer 2

No.

Let $X_1,X_2,X_3$ be independent with $X_3$ uniform on $\{0,1\}$ and $X_1,X_2$ constant, equal to $0$ w.p. 1. Then, $\mathbb{P}\{X_3>X_1\}\cap \{X_3>X_2\} = \mathbb{P}\{X_3=1\} = \frac{1}{2}.$

But $\mathbb{P}\{X_3>X_1\}\cdot \mathbb{P} \{X_3>X_2\} = \mathbb{P}\{X_3=1\}^2 = \frac{1}{4}$.

The idea is that while the $X_i$'s are independent, the events $(\{X_n>X_i\})_{i\geq 2}$ may not be.

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NB: This can easily be adapted to have $X_1,X_2$ non-constant: take $X_3$ uniform on $\{0,2\}$ and $X_1,X_2$ uniform on $\{0,1\}$.