# Multivariate Cramer-Rao inequality

The Cramer-Rao inequality addresses the question of how accurately one can estimate a set of parameters $\vec{\theta} = \{\theta_1, \theta_2, \ldots, \theta_m \}$ characterizing a probability distribution $P(x) \equiv P(x; \vec{\theta})$, given only some samples $\{x_1, \ldots, x_n\}$ taken from $P$. Specifically, the inequality provides a rigorous lower bound on the covariance matrix of any unbiased set of estimators to these $\{\theta_i\}$ values. In this post, we review the general, multivariate form of the inequality, including its significance and proof.

### Introduction and theorem statement

The analysis of data very frequently requires one to attempt to characterize a probability distribution. For instance, given some random, stationary process that generates samples $\{x_i\}$, one might wish to estimate the mean $\mu$ of the probability distribution $P$ characterizing this process. To do this, one could construct an estimator function $\hat{\mu}(\{x_i\})$ — a function of some samples taken from $P$ — that is intended to provide an approximation to $\mu$. Given $n$ samples, a natural choice is provided by
$$\hat{\mu}(\{x_i\}) = \frac{1}{n}\sum_{i = 1}^n x_i, \tag{1}$$
the mean of the samples. This particular choice of estimator will always be unbiased given a stationary $P$ — meaning that it will return the correct result, on average. However, each particular sample set realization will return a slightly different mean estimate. This means that $\hat{\mu}$ is itself a random variable having its own distribution and width.

More generally, one might be interested in a distribution characterized by a set of $m$ parameters $\{\theta_i\}$. Consistently good estimates to these values require estimators with distributions that are tightly centered around the true $\{\theta_i\}$ values. The Cramer-Rao inequality tells us that there is a fundamental limit to how tightly centered such estimators can be, given only $n$ samples. We state the result below.

Theorem: The multivariate Cramer-Rao inequality.

Let $P$ be a distribution characterized by a set of $m$ parameters $\{\theta_i\}$, and let $\{\hat{\theta_i}\equiv \hat{\theta_i}(\{x_i\})\}$ be an unbiased set of estimator functions for these parameters. Then, the covariance matrix (see definition below) for the $\hat{\{\theta_i\}}$ satisfies,

$$cov(\hat{\theta}, \hat{\theta}) \geq \frac{1}{n} \times \frac{1}{ cov(\nabla_{\vec{\theta}} \log P(x),\nabla_{\vec{\theta}} \log P(x) )}. \tag{2} \label{CR}$$
Here, the inequality holds in the sense that left side of the above equation, minus the right, is positive semi-definite. We discuss the meaning and significance of this equation in the next section.

### Interpretation of the result

To understand (\ref{CR}), we must first review a couple of definitions. These follow.

Definition 1. Let $\vec{u}$ and $\vec{v}$ be two jointly-distributed vectors of stationary random variables. The covariance matrix of $\vec{u}$ and $\vec{v}$ is defined by
$$cov(\vec{u}, \vec{v})_{ij} = \overline{(u_{i}- \overline{u_i})(v_{j}- \overline{v_j})} \equiv \overline{\delta u_{i} \delta v_{j}}\tag{3} \label{cov},$$
where we use overlines for averages. In words, (\ref{cov}) states that $cov(\vec{u}, \vec{v})_{ij}$ is the correlation function of the fluctuations of $u_i$ and $v_j$.

Definition 2. A real, square matrix $M$ is said to be positive semi-definite if
$$\vec{a}^T\cdot M \cdot \vec{a} \geq 0 \tag{4} \label{pd}$$
for all real vectors $\vec{a}$. It is positive definite if the “$\geq$” above can be replaced by a “$>$”.

The interesting consequences of (\ref{CR}) follow from the following observation:

Observation. For any constant vectors $\vec{a}$ and $\vec{b}$, we have
$$cov(\vec{a}^T\cdot\vec{u}, \vec{b}^T \cdot \vec{v}) = \vec{a}^T \cdot cov(\vec{u}, \vec{v}) \cdot \vec{b}. \tag{5} \label{fact}$$
This follows from the definition (\ref{cov}).

Taking $\vec{a}$ and $\vec{b}$ to both be along $\hat{i}$ in (\ref{fact}), and combining with (\ref{pd}), we see that (\ref{CR}) implies that
$$\sigma^2(\hat{\theta}_i^2) \geq \frac{1}{n} \times \left (\frac{1}{ cov(\nabla_{\vec{\theta}} \log P(x),\nabla_{\vec{\theta}} \log P(x) )} \right)_{ii},\tag{6}\label{CRsimple}$$
where we use $\sigma^2(x)$ to represent the variance of $x$. The left side of (\ref{CRsimple}) is the variance of the estimator function $\hat{\theta}_i$, whereas the right side is a function of $P$ only. This tells us that there is fundamental — distribution-dependent — lower limit on the uncertainty one can achieve when attempting to estimate any parameter characterizing a distribution. In particular, (\ref{CRsimple}) states that the best variance one can achieve scales like $O(1/n)$, where $n$ is the number of samples available$^1$ — very interesting!

Why is there a relationship between the left and right matrices in (\ref{CR})? Basically, the right side relates to the inverse rate at which the probability of a given $x$ changes with $\theta$: If $P(x \vert \theta)$ is highly peaked, the gradient of $P(x \vert \theta)$ will take on large values. In this case, a typical observation $x$ will provide significant information relating to the true $\theta$ value, allowing for unbiased $\hat{\theta}$ estimates that have low variance. In the opposite limit, where typical observations are not very $\theta$-informative, unbiased $\hat{\theta}$ estimates must have large variance$^2$.

We now turn to the proof of (\ref{CR}).

### Theorem proof

Our discussion here expounds on that in the online text of Cízek, Härdle, and Weron. We start by deriving a few simple lemmas. We state and derive these sequentially below.

Lemma 1 Let $T_j(\{x_i\}) \equiv \partial_{\theta_j} \log P(\{x_i\}; \vec{\theta})$ be a function of a set of independent sample values $\{x_i\}$. Then, the average of $T_j(\{x_i\})$ is zero.

Proof: We obtain the average of $T_j(\{x_i\})$ through integration over the $\{x_i\}$, weighted by $P$,
$$\int P(\{x_i\};\vec{\theta}) \partial_{\theta_j} \log P(\{x_i\}; \vec{\theta}) d\vec{x} = \int P \frac{\partial_{\theta_j} P}{P} d\vec{x} = \partial_{\theta_j} \int P d\vec{x} = \partial_{\theta_j} 1 = 0. \tag{7}$$

Lemma 2. The covariance matrix of an unbiased $\hat{\theta}$ and $\vec{T}$ is the identity matrix.

Proof: Using (\ref{cov}), the assumed fact that $\hat{\theta}$ is unbiased, and Lemma 1, we have
\begin{align} cov \left (\hat{\theta}(\{x_i\}), \vec{T}(\{x_i\}) \right)_{jk} &= \int P(\{x_i\}) (\hat{\theta}_j – \theta_j ) \partial_{\theta_k} \log P(\{x_i\}) d\vec{x}\\ & = \int (\hat{\theta}_j – \theta_j ) \partial_{\theta_k} P d\vec{x} \\ &= -\int P \partial_{\theta_k} (\hat{\theta}_j – \theta_j ) d \vec{x} \tag{8} \end{align}
Here, we have integrated by parts in the last line. Now, $\partial_{\theta_k} \theta_j = \delta_{jk}$. Further, $\partial_{\theta_k} \hat{\theta}_j = 0$, since $\hat{\theta}$ is a function of the samples $\{x_i\}$ only. Plugging these results into the last line, we obtain
$$cov \left (\hat{\theta}, \vec{T} \right)_{jk} = \delta_{jk} \int P d\vec{x} = \delta_{jk}. \tag{9}$$

Lemma 3. The covariance matrix of $\vec{T}$ is $n$ times the covariance matrix of $\nabla_{\vec{\theta}} \log P(x_1 ; \vec{\theta})$ — a single-sample version of $\vec{T}$.

Proof: From the definition of $\vec{T}$, we have
$$T_j = \partial_{\theta_j} \log P(\{x_i\}, \vec{\theta}) = \sum_{i=1}^n \partial_{\theta_j} \log P(x_i, \vec{\theta}), \tag{10}$$
where the last line follows from the fact that the $\{x_i\}$ are independent, so that $P(\{x_i\}, \vec{\theta}) = \prod P(x_i; \vec{\theta})$. The sum on the right side of the above equation is a sum of $n$ independent, identically-distributed random variables. If follows that their covariance matrix is $n$ times that for any individual.

Lemma 4. Let $x$ and $y$ be two scalar stationary random variables. Then, their correlation coefficient is defined to be $\rho \equiv \frac{cov(x,y)}{\sigma(x) \sigma(y)}$. This satisfies
$$-1 \leq \rho \leq 1 \label{CC} \tag{11}$$

Proof: Consider the variance of $\frac{x}{\sigma(x)}+\frac{y}{\sigma(y)}$. This is
\begin{align} var \left( \frac{x}{\sigma(x)}+\frac{y}{\sigma(y)} \right) &= \frac{\sigma^2(x)}{\sigma^2(x)} + 2\frac{ cov(x,y)}{\sigma(x) \sigma(y)} + \frac{\sigma^2(y)}{\sigma^2(y)} \\ &= 2 + 2 \frac{ cov(x,y)}{\sigma(x) \sigma(y)} \geq 0. \tag{12} \end{align}
This gives the left side of (\ref{CC}). Similarly, considering the variance of $\frac{x}{\sigma(x)}-\frac{y}{\sigma(y)}$ gives the right side.

We’re now ready to prove the Cramer-Rao result.

Proof of Cramer-Rao inequality. Consider the correlation coefficient of the two scalars $\vec{a} \cdot \hat{\theta}$ and $\vec{b} \cdot \vec{T}$, with $\vec{a}$ and $\vec{b}$ some constant vectors. Using (\ref{fact}) and Lemma 2, this can be written as
\begin{align} \rho & \equiv \frac{cov(\vec{a} \cdot \hat{\theta} ,\vec{b} \cdot \vec{T})}{\sqrt{var(\vec{a} \cdot \hat{\theta})var(\vec{b} \cdot \vec{T})}} \\ &= \frac{\vec{a}^T \cdot \vec{b}}{\left(\vec{a}^T \cdot cov(\hat{\theta}, \hat{\theta}) \cdot \vec{a} \right)^{1/2} \left( \vec{b}^T \cdot cov(\vec{T},\vec{T}) \cdot \vec{b} \right)^{1/2}}\leq 1. \tag{13} \end{align}
The last inequality here follows from Lemma 4. We can find the direction $\hat{b}$ where the bound above is most tight — at fixed $\vec{a}$ — by maximizing the numerator while holding the denominator fixed in value. Using a Lagrange multiplier to hold $\left( \vec{b}^T \cdot cov(\vec{T},\vec{T}) \cdot \vec{b} \right) \equiv 1$, the numerator’s extremum occurs where
$$\vec{a}^T + 2 \lambda \vec{b}^T \cdot cov(\vec{T},\vec{T}) = 0 \ \ \to \ \ \vec{b}^T = – \frac{1}{2 \lambda} \vec{a}^T \cdot cov(\vec{T}, \vec{T})^{-1}. \tag{14}$$
Plugging this form into the prior line, we obtain
$$– \frac{\vec{a}^T \cdot cov(\vec{T},\vec{T})^{-1} \cdot \vec{a}}{\left(\vec{a}^T \cdot cov(\hat{\theta}, \hat{\theta}) \cdot \vec{a} \right)^{1/2} \left(\vec{a}^T \cdot cov(\vec{T},\vec{T})^{-1} \cdot \vec{a} \right)^{1/2}}\leq 1. \tag{15}$$
Squaring and rearranging terms, we obtain
$$\vec{a}^T \cdot \left (cov(\hat{\theta},\hat{\theta}) – cov(\vec{T},\vec{T})^{-1} \right ) \cdot \vec{a} \geq 0. \tag{16}$$
This holds for any $\vec{a}$, implying that $cov(\hat{\theta}, \hat{\theta}) – cov(\vec{T},\vec{T})^{-1}$ is positive semi-definite — see (\ref{pd}). Applying Lemma 3, we obtain the result$^3$. $\blacksquare$

Thank you for reading — we hope you enjoyed.

[1] More generally, (\ref{fact}) tells us that an observation similar to (\ref{CRsimple}) holds for any linear combination of the $\{\theta_i\}$. Notice also that the proof we provide here could also be applied to any individual $\theta_i$, giving $\sigma^2(\hat{\theta}_i) \geq 1/n \times 1/\langle(\partial_{\theta_i} \log P)^2\rangle$. This is easier to apply than (\ref{CR}), but is less stringent.

[2] It might be challenging to intuit the exact function that appears on the right side of $(\ref{CR})$. However, the appearance of $\log P$’s does make some intuitive sense, as it allows the derivatives involved to measure rates of change relative to typical values, $\nabla_{\theta} P / P$.

[3] The discussion here covers the “standard proof” of the Cramer-Rao result. Its brilliance is that it allows one to work with scalars. In contrast, when attempting to find my own proof, I began with the fact that all covariance matrices are positive definite. Applying this result to the covariance matrix of a linear combination of $\hat{\theta}$ and $\vec{T}$, one can quickly get to results similar in form to the Cramer-Rao bound, but not quite identical. After significant work, I was eventually able to show that $\sqrt{cov(\hat{\theta},\hat{\theta})} – 1/\sqrt{cov(\vec{T},\vec{T}) } \geq 0$. However, I have yet to massage my way to the final result using this approach — the difficulty being that the matrices involved don’t commute. By working with scalars from the start, the proof here cleanly avoids all such issues.

• Wonseok Hwang

Although the result is correct, the logic from 1st line of Eq. (8) to 2nd line of Eq. (8) seems to be wrong as no integration by part is possible between theta_k and vec{x}. It is rather due to vanishment of int partial theta_k (hat{theta_j} – theta_j) P(vec{x}) dvec{x} due to unbiasness of estimator hat{theta}_j

• Wonseok Hwang

But thanks for the good post. It helps a lot.

• efavdb

Glad you liked it, WonSeok and thank you very much for your comment. I’ll adjust the text as soon as I get a chance.