Let $g =(g_1, ..., g_n)$ denote a sequence of standard Gaussian variables. Let $p = (p_1, ..., p_n)$ denote a vector in the simplex $\mathcal{P}_n$, given by $$ \mathcal{P}_n = \{p \in \mathbb{R}^n : p_i \geq 0, \sum_{i=1}^n p_i = 1\}. $$ We can define the vector-valued function $\psi \colon \mathcal{P}_n \to \mathcal{P}_n$ with coordinates $\psi(p) = (\psi_1(p), .., \psi_n(p))$, such that $$ \psi_i(p) = \mathbb{E}_{g} \bigg[\frac{p_i g_i^2}{\sum_{j=1}^n p_j g_j^2}\bigg] $$ Question: What is a good explicit approximation of $\psi$ (e.g., in terms of elementary functions)? In particular, I am hoping for a function $\Psi \colon \mathcal{P}_n \to \mathbb{R}^n$ that provides a multiplicative approximation in the following sense: there exist $0 < c \leq C < \infty$ such that if $p \in \mathcal{P_n}$ then $$ c\, \Psi_i(p) \leq \psi_i(p) \leq C\, \Psi_i(p) \quad \mbox{for all}~i.$$ Ideally the constants $c, C$ could be made independent of $n$.

  • $\begingroup$ Just letting you now that I have substantially corrected and expanded my answer - although I wasn't able to get full solution, I think I came close, maybe you can make the next step. I am done with editing for the foreseeable future. $\endgroup$ Commented Jul 10 at 18:56
  • $\begingroup$ While I appreciate your response, it is clear that $\psi_i(p)$ cannot be expressed in elementary functions as soon as $n \geq 3$---it corresponds to an elliptical integral in that case. Therefore, I am truly interested in elementary approximations of the multiplicative type expressed in the post. $\endgroup$
    – Drew Brady
    Commented Jul 10 at 18:57
  • $\begingroup$ Just one more idea, barely fleshed out - it seems that if you can bound $p_i$ from below, you can obtain a polynomial approximating $\left(1 + u \left(\frac{1}{p_i} - 1 \right)\right)^{-\frac{1}{2}}$ with constant error. Substituting that polynomial into the final integral in my answer results in nasty but apparently analytically tractable integrals, which than has constant error... $\endgroup$ Commented Jul 10 at 21:13

1 Answer 1


So far only partial answer + some theoretical connections. Hope others can expand.

When $g_i \sim N(0,1)$, we have $g^2_i \sim \text{Gamma}(\frac{1}{2}, 1)$. We then have $p_i g^2_i \sim \text{Gamma}(\frac{1}{2}, p_i^{-1})$ (using the shape-rate parametrization of Gamma)

Similarly to obtaining Dirichlet distribution by normalizing identically scaled Gamma distributions, normalizing general Gamma distributions leads to scaled Dirichlet distribution, i.e. for the vector random variable $Z(p)$

$$ Z(p)_i = \frac{p_i g_i^2}{\sum_{j=1}^n p_j g_j^2} \\ \psi(p) = E(Z(p)) $$

you have

$$ Z(p) \sim \text{ScaledDirichlet}\left(\frac{1}{2}, p^{-1}\right) $$

EDIT: The limitation to Aitchinson measure was previously omitted:

Expectation of the scaled dirichlet with those parameters w.r.t. the Aitchinson measure $a$ is given by:

$$ E_a(Z(p)) = (\ominus \mathcal{C}(p^{-1})) \oplus E_a(X) $$

Where $X \sim \text{Dirichlet}(\frac{1}{2}, ..., \frac{1}{2})$, $\mathcal{C}$ is the closure operator $\mathcal{C}(x) = \frac{x}{\sum x_i}$, $\oplus$ is the perturbation operation over the Aitchinson simplex $x \oplus y = \mathcal{C}((x_1 y_1, \ldots, x_n y_n))$ and $\ominus$ is the related inverse $\ominus x = \mathcal{C}((x_1^{-1}, \ldots, x_n^{-1})$.

Now we may note that $E_a(X) = \left(\frac{1}{n}, \ldots, \frac{1}{n}\right)$ which is the zero element w.r.t. $\oplus$, so we have:

$$ E_a(Z(p)) = p $$

An R gist demonstrating the identity: https://gist.github.com/martinmodrak/14dbaacaffdb99fe2d11cd8a023f573a

EDIT - More hope:

In full generality, there is no closed form for the expectation of the scaled Dirichlet w.r.t. the standard Lebesgue measure.

However, here, we have a special case as the first argument is a vector of all $\frac{1}{2}$s. The joint density is the not-completely-scary:

$$ f_Z(x,p) = \pi^{-\frac{n}{2}} \Gamma\left(\frac{n}{2}\right) \left(\prod_i^n \frac{1}{\sqrt{p_i x_i}} \right) \left(\sum_i^n \frac{x_i}{p_i} \right)^{-\frac{n}{2}} $$

And there is some hope as Wolfram Alpha tells me that for $n = 2$

$$ \psi_1(p) = \frac{1}{1 + \sqrt{\frac{p_2}{p_1}}} $$

Following Dickey 1968, equation 4.4 we can also convert all moments into a 1D integral. For the expectation of the first component (other can be obtained through symmetry) we obtain in your case: $$ \psi_1(p) = \frac{1}{2} \prod_i^n p_i^{-\frac{1}{2}} \int_0^1 u^{\frac{n}{2} - 1} \left(1 + u \left(\frac{1}{p_1} - 1 \right) \right)^{-\frac{3}{2}} \prod_{i = 2}^n \left(1 + u \left(\frac{1}{p_i} - 1 \right) \right)^{-\frac{1}{2}} \text{d}u $$

(the gist linked above now contains a demonstration of this as well)

It might be possible to obtain the bounds you seek by bounding the integrand by analytically integrable function, but I am waaaay out of the time to investigate this.

I also find it possible that actually working within the Aitchinson geometry (where the solution is easy) might prove useful to solving your greater problem for which you seek the bound.

Source: Monti et al. 2011 - The shifted-scaled Dirichlet distribution in the simplex - this also discusses the connections to the Aitchinson geometry more thoroughly.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.