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Any exponential family distribution can be expressed as $p(x|\theta) = g(\theta) f(x) e^{\phi(\theta)^T T(x)} = f(x) e^{\phi(\theta)^T T(x) - A(\theta)}$ where $A(\theta) = -\log{g(\theta)}$.

We know that this integrates to 1 since it's a valid pdf:

$\int f(x) e^{\phi(\theta)^T T(x) - A(\theta)} dx = 1$

The expected value of the sufficient statistic is obtained by differentiating both sides with respect to $\theta$. Assuming this is a parameter vector, we will need a vector derivative:

$\begin{align} \frac{\partial}{\partial \theta_i} \int f(x) e^{\phi(\theta)^T T(x) - A(\theta)} dx &= \frac{\partial}{\partial \theta_i} (1) \\ \int \frac{\partial}{\partial \theta_i} [ f(x) e^{\phi(\theta)^T T(x) - A(\theta)}] dx &= 0 \\ \int p(x|\theta) \left[ \frac{\partial \phi(\theta)^T}{\partial \theta_i} T(x) - \frac{\partial A(\theta)}{\partial \theta_i } \right] dx &= 0 \\ \nabla_i \phi(\theta)^T \int p(x|\theta) T(x) dx - \nabla_i A(\theta) \int p(x|\theta) dx &= 0 \\ \nabla_i \phi(\theta)^T E \left[ T(x) \right] - \nabla_i A(\theta) &= 0\end{align}$

But this is not quite correct since the first term is a scalar (dot product between two vectors?) and the second term is a vector.

The expectation should be the vector i.e. I should have something like $E[T(x)]_i$ with the vector index $i$ on the expected value. What has gone wrong?

I think (but not sure) that the answer should be $E[T(x)]_i = \frac{1}{\phi'(\theta)} \nabla_iA(\theta)$

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Let $\theta \in \Theta \subset \mathbb R^m$ and $\phi : \Theta \to \mathbb R^d$ so $T(x) \in \mathbb R^d$ too. I'll use without proof the fact that differentiation and integration can be exchanged for exponential families.

First I'm going to treat this as a natural exponential family and see what happens (this is like if $\phi$ is just the identity function). I'll use $\theta$ as the parameter so right now I have $p(x|\theta) = \exp(\theta^TT(x) - A(\theta))f(x)$.

In this case the differentiation is easy and we have $$ 0 = \int p(x|\theta) \left[T_j(x) - \nabla _jA(\theta)\right]\,\text dx $$ so $$ \nabla A(\theta) = \text E(T(X)). $$

So if you're content to work with $\phi := \phi(\theta)$ as the parameter, then this is a tidy result.


Now I'll consider what happens if $m = d = 1$ to build our intuition for the general case. We'll have $$ 0 = \int p(x|\theta)\left[T(x) \cdot \phi'(\theta) - A'(\theta) \right]\,\text dx \\ \implies \text E(T(X)) \cdot \phi'(\theta) = A'(\theta) $$ so $$ \text E(T(X)) = \frac{A'(\theta)}{\phi'(\theta)} $$ as you thought.


But the situation is more complicated if $m,d > 1$. The derivates in particular are more complicated because now $\phi$ is vector-valued so we'll get a matrix of first derivatives.

We need $$ \frac{\partial}{\partial \theta_j} \phi(\theta)^TT(x). $$ Writing this as a sum, we have $$ \frac{\partial}{\partial \theta_j} \phi(\theta)^TT(x) = \sum_{i=1}^d T_i(x) \frac{\partial}{\partial \theta_j}\phi_i(\theta). $$ $\phi : \Theta\to\mathbb R^d$ with $\Theta\subset\mathbb R^m$ so for each component we have $\phi_i : \Theta\to\mathbb R$ with $i=1,\dots,d$. Each $\phi_i$ is just a typical scalar-valued function so $\frac{\partial}{\partial \theta_j} \phi_i = (\nabla \phi_i)_j$.

I'll collect these gradients into a matrix $\Phi(\theta) \in \mathbb R^{m\times d}$ where $$ \Phi(\theta) = \left[\begin{array}{c|c|c|c}\nabla \phi_1(\theta) & \nabla \phi_2(\theta) & \cdots & \nabla \phi_d(\theta)\end{array} \right]. $$

This means $$ \frac{\partial}{\partial \theta_j}\phi_i(\theta) = \Phi(\theta)_{ji} $$ so $$ \frac{\partial}{\partial \theta_j} \phi(\theta)^TT(x) = \left(\Phi(\theta)T(x)\right)_j. $$

All together this means $$ \mathbf 0 = \int p(x|\theta) \left[ \Phi(\theta)T(x) - \nabla A(\theta) \right] \,\text dx \\ \implies \Phi(\theta)\text E [T(X)] = \nabla A(\theta) $$ which is a linear system that we're trying to solve for $\text E [T(X)]$. The dimensions in question and the specific properties of $\phi$ now will determine how many solutions this system has.


Here's one example. Suppose $X \sim \mathcal N(\mu, \sigma^2)$ so $$ p(x|\mu,\sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right). $$ Rearranging and expanding the quadratic we have $$ p(x|\mu,\sigma^2) = \frac{1}{\sqrt{2\pi}}\exp\left(\frac{\mu}{\sigma^2}x - \frac{1}{2\sigma^2} x^2 - \left[\frac{\mu^2}{2\sigma^2} + \frac 12 \log \sigma^2\right]\right) $$ which means we'll have $\theta = (\mu,\sigma^2)$, $\Theta = \mathbb R \times (0,\infty)$, $T(x) = {x \choose x^2}$, $$ \phi(\theta) = \left(\begin{array}{c}\frac{\theta_1}{\theta_2} \\ \frac{-1}{2\theta_2}\end{array}\right), $$ and $$ A(\theta) = \frac 12 \theta_1^2\theta_2^{-1} + \frac 12 \log\theta_2. $$

This means $$ \Phi(\theta)= \left(\begin{array}{cc}\theta_2^{-1} & 0 \\ -\theta_1\theta_2^{-2} & \frac 12 \theta_2^{-2}\end{array}\right). $$ Noting that $$ \det \Phi(\theta) = \frac 12 \theta_2^{-3} > 0 $$ we know this matrix is invertible, and it's just $2\times 2$ so the inverse is easy enough to compute: $$ \Phi^{-1}(\theta) = 2\theta_2^3 \left(\begin{array}{cc}\frac 12 \theta_2^{-2} & 0 \\ \theta_1\theta_2^{-2} & \theta_2^{-1}\end{array}\right) = \left(\begin{array}{cc}\theta_2 & 0 \\ 2\theta_1\theta_2 & 2\theta_2^2\end{array}\right). $$ We also have $$ \nabla A(\theta) = \left(\begin{array}{cc}\theta_1\theta_2^{-1} \\ -\frac 12\theta_1^2\theta_2^{-2} + \frac 12 \theta_2^{-1}\end{array}\right) $$ so $$ \text E(T(X)) = \Phi^{-1}(\theta)\nabla A(\theta) \\ = \left(\begin{array}{c}\theta_1 \\ \theta_1^2 + \theta_2\end{array}\right) = {\mu \choose \mu^2 + \sigma^2} $$ which we know to be the first two raw moments of $X$. This was a lot of work to get there but it shows the unavoidable role of the matrix $\Phi$.


Update: here's the multivariate normal case. Let $X \sim \mathcal N_n(\mu, \Omega^{-1})$ where $\Omega$ is the precision matrix. I'll use $\Omega_{(i)}$ for the $i$th row of $\Omega$, and for a vector $v$ and $1\leq i\leq j$ I'll use $v_{i:j}$ to be $(v_i,v_{i+1},\dots,v_j)$.

The density is $$ p(x \mid \mu, \Omega) = \exp\left[-\frac 12 x^T\Omega x + x^T \Omega \mu - \frac 12 \left(\mu^T\Omega\mu - \log\det \Omega\right)\right] \cdot(2\pi)^{-n/2}. $$

The parameters are $\mu$ and $\Omega$ but I want to deal with these as a flat vector so I'll combine $\mu$ with $\Omega$ "unrolled" as $$ \theta = (\mu, \Omega_{11}, \Omega_{12}, \dots, \Omega_{1n}, \Omega_{22}, \dots, \dots, \Omega_{n-1,n}, \Omega_{nn}). $$ I'm only taking the diagonal and one triangle of $\Omega$ since it's symmetric. This means $\Theta\subset \mathbb R^m$ where $m = n + {n+1\choose 2} = 2n + {n\choose 2}$.

I now need to get this as $\phi(\theta)^TT(x) - A(\theta)$. For $\phi$ and $T$ I'll use the fact that $$ x^T\Omega x = 2\sum_{i < j} x_ix_j \Omega_{ij} + \sum_{i=1}^n x_i^2\Omega_{ii} $$ so I'll take $$ \phi(\theta) = \left(\Omega\mu, -\frac 12 \Omega_{11}, -\Omega_{12}, \dots, -\Omega_{1n}, -\frac 12 \Omega_{22}, -\Omega_{23}, \dots, \dots, -\Omega_{n-1,n}, -\frac 12 \Omega_{nn}\right) $$ (I'm writing this in terms of $\mu$ and $\Omega$ instead of $\theta$ just because it's way clearer, but these can all be written in terms of $\theta$ if desired) and $$ T(x) = (x, x_1^2, x_1x_2, \dots, x_1x_n, x_2^2, x_2x_3, \dots,\dots, x_n^2). $$

We can now construct $\Phi$. I'll build it as a $2\times 2$ block matrix where $$ \Phi(\theta) = \left(\begin{array}{c|c}A&B\\\hline C&D\end{array}\right) $$ with $A$ being $n\times n$ and $D$ being ${n+1c\choose 2}\times{n+1\choose 2}$. $A$, $B$, and $D$ are easy to work out.

For $A$, I need $\frac{\partial \phi_i}{\partial \theta_j}$ where $i,j=1\dots,n$. Since $\phi_i(\theta) = \Omega_{(i)}^T\mu$ for this range of $i$ values, and $\theta_j = \mu_j$, we'll get $$ (\nabla \phi_i)_{1:n} = \Omega_{(i)} $$ (using the symmetry of $\Omega$ for the fact that $\Omega_{k\ell} = \Omega_{\ell k}$). This means that $A = \Omega$.

Now for $i > n$, $\phi_i(\theta)$ is proportional to some $\Omega_{j\ell}$ so derivatives w.r.t. $\mu$ are zero. This means $B = \mathbf O$, the appropriately sized zero matrix.

Next, for $i,j>n$ if $i\neq j$ then $\frac{\partial \phi_i}{\partial \theta_j} = 0$ so $D$ is diagonal. If $i$ corresponds to a diagonal entry of $\Omega$ then we have a derivative of $-\frac 12$, otherwise it'll be a derivative of $-1$. The first element of the diagonal of $D$ corresponds to $\Omega_{11}$ so it begins with $-\frac 12$. The next value of $-\frac 12$ is for $\Omega_{22}$ which happens at element $1+n$ since all of the first row of $\Omega$ has been counted and $\Omega_{21}$ is skipped (since it's accounted for by $\Omega_{12}$). There are $n-2$ remaining elements of $\Omega_{(2)}$ to account for, so the 3rd value of $-\frac 12$ in $D$ is at position $1+n+(n-1)$. The next $-\frac 12$ is at $1+n+(n-1)+(n-2)$ and so on, so the locations of $-\frac12$s are given by the sequence $$ 1+\sum_{j=0}^{i-2}(n-j) $$ with $i=1,\dots,n$. As an example, if $n=5$ then the $-\frac 12$s will be at locations $1,6,10,13,15$. As a sanity check, $D$ is supposed to have a diagonal of length ${n+1\choose 2}$, and the final $-\frac 12$ is for $\Omega_{nn}$. We have $$ 1+\sum_{j=0}^{n-2}(n-j) = \sum_{j=0}^{n-1}(n-j) \\ = \sum_{j=1}^n j = {n+1\choose 2} $$ so this agrees with that.

Finally, we need $C$. I'm going to skip $C$ for now until the final step.

For $\Phi^{-1}$, $\Omega$ and $D$ are always invertible so $\Phi$ is too, and we can use the formula for the inverse of a $2\times 2$ block matrix to find $$ \Phi^{-1}(\theta) = \left(\begin{array}{c|c}\Omega^{-1} & \mathbf O \\ \hline -D^{-1}C\Omega^{-1} & D^{-1}\end{array}\right). $$


Now we can get $\nabla A$ to finish this off. I'll also partition it into the first $n$ elements, which lines up the upper two blocks of $\Phi^{-1}$, and the remaining ${n+1\choose 2}$ which I'll call $V$. We have $$ A(\theta) = \frac 12 \mu^T\Omega\mu - \frac 12 \log\det \Omega $$ so $$ \frac{\partial A}{\partial \theta_{1:n}} = \frac{\partial A}{\partial \mu} = \Omega\mu. $$ This means $$ \nabla A(\theta) = {\Omega\mu \choose V} $$ so $$ \Phi^{-1}(\theta)\nabla A(\theta) = {\mu \choose -D^{-1}(C\mu - V)}. $$ This shows that $\text E(T_{1:n}(X)) = \text E(X) = \mu$ which is nice to see.

For the rest, I won't work it out in general but I'll just show what we get for the elements corresponding to diagonal terms of $\Omega$ and off-diagonal terms.

For $V$, I'll need derivatives of $\log\det \Omega$ w.r.t. $\Omega_{ij}$ which I can get via this result: $$ \frac{\partial}{\partial \Omega_{ij}} \log\det \Omega = (2 \Sigma - \text{diag}(\Sigma)) $$ due to symmetry. This means that for the elements of $V$ corresponding to a diagonal element of $\Omega$ we'll have $\frac 12 \mu_i^2 - \frac 12 \Sigma_{ii}$ while the off-diagonal will be $\mu_i\mu_j - \Sigma_{ij}$.

For $C\mu$, the diagonal elements contribute $\mu_i^2$ because for that row the only non-zero element is $\mu_i$ in the $i$th column. $-D^{-1}$ will multiply the result by $2$, and then $V_i$ is $$ 2\left(\mu_i^2 - \frac 12 \mu_i^2 + \frac 12 \Sigma_{ii} \right) \\ = \mu_i^2 + \Sigma_{ii}. $$ For the off-diagonal, you can show that $C\mu$ will contribute $2\mu_i\mu_j$ so we get $$ 2\mu_i\mu_j - \mu_i\mu_j + \Sigma_{ij} = \mu_i\mu_j + \Sigma_{ij}. $$

Thus after all this work we have confirmed that $\text E(X_i) = \mu_i$ and $\text E(X_iX_j) = \mu_i\mu_j + \Sigma_{ij}$.

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  • $\begingroup$ Wow. Thanks for an awesome reply! But what about if I want to consider e.g. a multivariate normal distribution in D dimensions. In this case the matrix of derivatives $\Phi(\theta) $ will be DxD meaning I won't be able to invert it explicitly. And yet we should still be able to obtain algebraic solution for E[T(X)] in this same way? $\endgroup$ – user11128 Sep 7 at 22:28
  • $\begingroup$ @user11128 I think in general $\Phi(\theta)^{-1}\nabla A(\theta)$ might be as explicit as you can get if you take this approach unless $\Phi$ happens to have some special structure. I'll go through the multivariate normal example and update with that. I'm also looking more into general conditions for $\Phi$ to be invertible. $\endgroup$ – jld Sep 8 at 1:01
  • $\begingroup$ @user11128 I just updated with a multivariate normal example, $\Phi$ had enough structure that the inverse was easy to do $\endgroup$ – jld Sep 12 at 16:48

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