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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?

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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 $\frac12$.

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