# Dirichlet distribution: posteriors and priors of distribution

Let $$|\psi\rangle \in \mathbb{C}^{2n}$$ be a random quantum state such that $$|\langle z| \psi \rangle|^{2}$$ is distributed according to a $$\text{Dirichlet}(1, 1, \ldots, 1)$$ distribution, for $$z \in \{0, 1\}^{n}$$.

Let $$z_{1}, z_{2}, \ldots, z_{k}$$ be $$k$$ samples from this distribution (not all unique). Choose a $$z^{*}$$ that appears most frequently.

1. I am trying to prove:

$$\underset{|\psi\rangle}{\mathbb{E}}\big[|\langle z^{*}| \psi \rangle|^{2}\big] = \underset{|\psi\rangle}{\mathbb{E}}\bigg[\underset{m}{\mathbb{E}}\big[|\langle z^{*}| \psi \rangle|^{2} ~| ~m\big]\bigg] = \mathbb{E}\bigg[\frac{1+m}{2^{n}+k}\bigg],$$ where $$m$$ is a random variable that denotes the frequency of $$z^{*}$$.

1. I am also trying to prove that for the collection $$z_{1}, z_{2}, \ldots, z_{k}$$ $$\sum_{i \neq j}\mathrm{Pr}[z_{i} = z_{j}] = {n \choose k}\frac{2}{2^{n} + 1}.$$

Basically, I am trying to trace the steps of Lemma $$13$$ (page 10) of this quantum paper. I realize that my questions have to deal with posteriors and priors of the chosen distribution (though I do not understand how they have been explicitly derived or used here. An explicit derivation will be helpful). Is there any resource where I can find quick formulas for calculating these for other distributions, like the Binomial distribution?

Let $$p_i = |\langle i | \phi \rangle|^2 \sim Dir(a_1, .., a_{2^n}) = Dir(1, .., 1)$$ and $$m_i$$ the occurences of outcome $$|i\rangle$$ on samples $$z_1, .. z_k$$.

Since the Dirichlet distribution is the conjugate prior of the categorical (see here), meaning

$$\bf{p}$$ $$| Z, (1, .. 1),$$ $$\bf{m}$$ $$\sim Dir(2^n,$$ $$\bf{m} + 1$$)

and using the formula for the mean value of Dirichlet we get

$$\mathbb{E}[p_{z*} | m] = \frac{m+1}{\sum_{j=1}^{2^n} (m_j + 1)} = \frac{m+1}{2^n + k}$$

For the second claim, take $$i \neq j$$ and compute \begin{align*} \mathbb{P}[z_i = z_j] &= \int \mathbb{P}[z_i = z_j | (p_1, .. p_{2^n})] \cdot f(p_1, .. p_{2^n}) \\ &= \int \sum_{k=0}^{2^n} p_k^2 \cdot f(p_1, .. p_{2^n}) \\ &= \sum_{k=0}^{2^n} \mathbb{E}[p_k^2] \\ &= \sum_{k=0}^{2^n} \frac{2}{2^n(2^n + 1)} = \frac{2}{2^n + 1} \end{align*}

(the last equality holds since $$\bf{x}$$ $$\sim Dir(\bf{a}$$) $$\implies \mathbb{E}[x_i^2] = \frac{a_i(a_i + 1)}{a_0(a_0 + 1)}$$, $$a_0 = \sum_{i=1}^{N} a_i$$.

This means that $$\sum_{i \neq j} \mathbb{P}[z_i = z_j] = {k \choose 2} \frac{2}{2^n + 1}$$.

• By $Z$ i refer to the collections of all the samples $z_1, .. z_k$. Also $\sum_{j=1}^{2^n} (m_j + 1) = \sum_{j=1}^{2^n} m_j + \sum_{j=1}^{2^n}1 = k + 2^n$ since the first sum simply counts the number of samples. In general you need $2^n$ concentration paramaters $a_i$ for the Dirichlet distribution to generate points in the $2^n-1$ simplex. Finally, i only used different parameters for the prior and the posterior distributions Oct 6 '20 at 9:34
• I define $m_i$ to be the number of occurences of $|i\rangle$ and then i write $m_{z*} = m$. $\sum_{j=1}^{2^n} m_j = \sum_{j=1}^{2^n} \sum_{l=1}^{k} \delta_{z_l, j} = \sum_{l=1}^{k} \sum_{j=1}^{2^n} \delta_{z_l, j} = \sum_{l=1}^{k} 1 = k$ Oct 6 '20 at 10:12
• $\mathbb{P}[z_i = z_j | (p_1, .., p_{2^n})] = \sum_{k=0}^{2^n} \mathbb{P}[z_i = k | (p_1, .., p_{2^n}), z_j = k] \cdot \mathbb{P}[z_j = k | (p_1, .., p_{2^n})] = \sum_{k=0}^{2^n} p_k \cdot p_k = \sum_{k=0}^{2^n} p_k^2$ since $z_i, z_j$ independent. For the integral sign i used that $\int p_k^2 \cdot f(p_1, .. p_{2^n}) = \mathbb{E}[p_k^2]$ since $f$ is the p.d.f Oct 6 '20 at 10:21
• Well, each $z_k$ takes the value $|i \rangle$ with probability $p_i$. This means that $z_k |$ $\bf{p}$ is distributed according to the categorical. The "valid" distributions for $(p_1, .. p_{2^n})$ are multivariate distributions defined on a simplex, since it must hold that $\sum_{i} p_i = 1$. The binomial is not one of them. For other "valid" distributions, i don't believe that there are some nice formulas that you can write down but honestly, i don't know. Oct 6 '20 at 11:02
• We are taking the expectation condtional on $\bf{m}$. If we know Z, then we know $\bf{m}$. Oct 6 '20 at 11:39