# IID standard normals $\{X_i\}_{i=1}^6$, expected number of $X_i \geq X_j$, $j < i$.

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?

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 $$Z^i=\mathbb 1\{X_i \leq X_j, \forall j).
For example $$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$$