# Expectation of the inverse of a linear function of the Dirichlet distribution

Let $\mathbf{X}=(X_1,X_2,\cdots,X_k)\in\mathbb{R}^k$ follow a Dirichlet distribution $X \sim \mathsf{Dir}(\alpha_1, \alpha_2, ..., \alpha_k)$, and let $\mathbf{c}\in\mathbb{R}_+^k$ be a vector of non-negative real numbers. Is there a way to compute/approximate/upper-bound the expectation $$\mathbb{E} \left[ \frac{1}{\mathbf{c}^\top \mathbf{X}} \right] = \mathbb{E} \left[ \frac{1}{c_1 X_1 + c_2 X_2 + \cdots + c_kX_k} \right]$$ taken with respect to $\mathbf{X}$?

$E[X_i^k]$ is easy to calculate. For example, if $k \geq 1$, we have $$E[X_i^k] = \frac{\alpha_i (\alpha_i + 1)...(\alpha_i + k - 1)}{(\sum \alpha_i) (\sum \alpha_i + 1)...(\sum \alpha_i + k - 1)}.$$ Same thing for $k < 0$: $$E[X_i^k] = \frac{(\sum \alpha_i - 1)...(\sum \alpha_i - k)}{(\alpha_i - 1)...(\alpha_i - k)}.$$ Calculating $E[X_1^{k_1}...X_n^{k_n}]$ is also easy: $$E[X_1^{k_1}...X_n^{k_n}] = \frac{\prod_i \prod_{j = 0}^{k_i - 1} (\alpha_i + j)}{\prod_{j = 0}^{\sum k_i - 1}(\sum \alpha_i + j)}.$$

Some preliminary inequalities to calculate the upper bound: \begin{align} & \frac{1}{\sum c_i X_i} \leq \frac{1}{\min c_i} \\ & \frac{1}{\sum c_i X_i} \leq \frac{1}{n} \left( \prod c_i X_i \right)^{-1/n} \leq \frac{\sum \frac{1}{c_i X_i}}{n^2} \\ & \frac{1}{\sum c_i X_i} \leq \frac{1}{(\sum c_i)^2} \sum c_i \frac{1}{X_i} \end{align}

The following potentially gives an infinite expansion. \begin{align} \frac{1}{\sum c_i X_i} & = \frac{\frac{1}{\max c_j}}{1 - \sum \frac{\max c_j - c_i}{\max c_j} X_i} = \frac{1}{\max c_j} \sum_{k = 0}^\infty \left( \sum \frac{\max c_j - c_i}{\max c_j} X_i \right)^k \end{align} There are several things that you can do with this expansion. For example, $$\sum \frac{\max c_j - c_i}{\max c_j} X_i \leq \frac{\max c_j - \min c_j}{\max c_j},$$ so the series vanishes exponentially and you can truncate the summation and get an upperbound to any precision.

Another possibility is to consider $$\left( \sum \frac{\max c_j - c_i}{\max c_j} X_i \right)^k \leq \left( \sum \frac{\max c_j - c_i}{\max c_j} \right)^{k - 1} \sum \frac{\max c_j - c_i}{\max c_j} X_i^k,$$ which can make the calculation easier.

Another possible upper bound: \begin{align} \frac{1}{\sum c_i X_i} & = \frac{\frac{1}{\min c_j}}{1 + \sum \frac{c_i - \min c_j}{\min c_j} X_i} \leq \frac{1}{\min c_j} \sum _{k = 0}^{2K} (-1)^k \left( \sum \frac{c_i - \min c_j}{\min c_j} X_i \right)^k. \end{align} However, the above does not converge as $K \to \infty$, unless $2\min c_i > \max c_i$, so this probably will not yield a good upper bound.

The two upperbounds above are special cases of the following. Notice that $\min c_i \leq \sum c_i X_i \leq \max c_i$, so we are indeed analyzing $f(x) = \frac{1}{x}$ for $x \in [\min c_i, \max c_i]$. Getting an upperbound of this $f$ is easy using Taylor expansion or other techniques.

• Thanks! The power series approach in the second bound is interesting, but unfortunately I cannot assume $2\min c_i > \max c_i$. As for the infinite expansion, I can only get a lower bound by truncating the series, but not an upper bound, right? Actually, the only upper-bound I have been able to derive is $$\mathbb{E} \left[\frac{1}{\mathbf{c}^\top \mathbf{X}} \right] \le \mathbb{E} \left[ \sum_i \frac{1}{c_i} X_i \right] = \sum_i \frac{1}{c_i} \mathbb{E} [X_i] = \sum_i \frac{1}{c_i} \frac{\alpha_i}{\alpha_1 + \alpha_2 + \cdots + \alpha_k},$$ which is probably quite loose. Nov 30, 2016 at 21:48