Q. Let $\{X_i\}_{i=1}^6$ be iid standard normals. What is the expected number of the $X_i$ that are greater than or equal to all $X_j$ such that $j < i$?

A (so far). I may be interpreting this wrong, but for every fixed $i \in \{1, \dots, 6\}$, we want to find the expected number of $X_j \leq X_i$. So, I believe we can denote this as $$Z := Z^1 + \dots + Z^6 \quad\text{s.t.}\quad Z^i := \sum_{j < i}\mathbb{1}\{X_j \leq X_i\},$$

where we want to find, $\mathbb{E}Z = \mathbb{E}Z^1 + \dots + \mathbb{E}Z^6$. Well, $\mathbb{E}Z^1 = 0$ by definition. For the rest they all follow this similar pattern:

$$\mathbb{E}Z^2 = \mathbb{E}\big[\mathbb{1}\{X_1 \leq X_2\}\big] = \mathbb{P}(X_1 \leq X_2) = \mathbb{P}(X_1 - X_2 \leq 0) = \mathbb{P}(\mathcal{N}(0, 2) \leq 0) = \frac{1}{2}.$$

That is, we are just adding an additional $\frac{1}{2}$ each time. So,

$$\mathbb{E}Z^3 = \frac{2}{2}, \quad \mathbb{E}Z^4 = \frac{3}{2}, \quad \mathbb{E}Z^5 = \frac{4}{2}, \quad \mathbb{E}Z^6 = \frac{5}{2},$$

and $\mathbb{E}Z = \frac{15}{2}$. Intuitively, the expected numbers seem ok. Is this right, or am I way off here?


1 Answer 1


My interpretation is the following: count the number of $X_i$ that are greater than all the previous numbers (your definition of $Z^i$ counts isn't very consistent with the description of your problem. Instead you should probably have $Z^i=\mathbb 1\{X_i \leq X_j, \forall j<i\}$).

For example $X_1$ is greater than all previous numbers (depending on your definition).

$X_2$ is greater than $X_1$ w.p. 1/2

$X_3$ is greater than both $X_1$,$X_2$ w.p. 1/3

so you have $1+1/2+1/3+1/4+1/5+1/6$


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