# Conditional expectation for Haar random states

Let $$U$$ be an $$n$$ qubit Haar random circuit applied to $$|0^n \rangle$$. Thereafter, the state is measured in the standard basis. Let $$p_0$$ be the probability of getting $$0$$ in the first qubit. We know that, $$\begin{equation} \mathbb{E}_U[p_0] = \mathbb{E}_U\big[\text{Tr}(|0\rangle \langle 0| \otimes I_{n-1} ~U|0^n\rangle \langle0^n| U^{*}\big] = \frac{1}{2}, \end{equation}$$ where $$I_{n-1}$$ is the identity operator on the remaining $$n-1$$ qubits.

Let $$p(x_0 = 0 | x_1 = b_1, x_2 = b_2, \ldots, x_n = b_n)$$ be the conditional probability of getting $$0$$ on the first qubit, conditioned on the other qubits being $$b_1, b_2, \ldots, b_n$$ respectively, where each $$b_i \in \{0, 1\}$$.

Is $$p(x_0 = 0 | x_1 = b_1, x_2 = b_2, \ldots, x_n = b_n)$$ also $$\frac{1}{2}$$? How might we express and compute this quantity as a Haar integral?

Intuitively, that's the case: the vector being random, there is no reason to prefer $$|0\rangle$$ over $$|1\rangle$$ on the first qubit.
I don't think it requires to compute some integrals other than using the fact that being given a single Haar-random state, the density matrix is: $$\int U|0\rangle\langle0|U\,\mathrm{d}\mu(U)=\frac{1}{2^n}I_{2^n}$$ After having measured the $$n-1$$ last qubits, the state of the first qubit is $$\frac12I_2$$. Note that this state is unaware of the result of the previous measurements you've made. Thus, not only does $$\mathbb{P}\left[x_0=0\middle|x_1=b_1,\cdots,x_n=b_n\right]$$ equal $$\frac12$$, but in fact these two events are independant: $$\mathbb{P}\left[x_0=0\middle|x_1=b_1,\cdots,x_n=b_n\right]=\mathbb{P}\left[x_0=0\right]=\frac12$$ which is another way to see that this probability is equal to $$\frac12$$.