Uniformly Minimum Variance Unbiased Estimator (UMVUE)

Published in Winter, 2025

Now we would like to focus on the class of unbiased estimators of a real valued parameter $\tau(\theta)$, $\tau : \Theta \to \mathbb{R}$. We would like to introduce the method of Uniformly Minimum Variance Unbiased Estimator (UMVUE):

Let $\vec X = \begin{pmatrix} X_1 \ X_2 \ \vdots \ X_n \end{pmatrix}$ be a random vector with a joint pdf (pmf) given by $p_{\theta}(x_1,\cdots,x_n) = p_{\theta}(\vec x)$, and the vectoer $\vec x$ given by $\vec x = \begin{pmatrix} x_1 \ x_2 \ \vdots \ x_n \end{pmatrix}$.

Definition. An estimator $T(\vec X)$ of a real valued parameter $\theta \in \Theta \subset \mathbb{R}^d$ where $\tau(\theta) : \Theta \to \mathbb{R}$ is said to be a UMVUE of $\tau(\theta)$ if: (1) $\mathbb{E}_{\theta}\{ T(\vec X)\} = \tau(\theta), \forall \theta \in \Theta$; (2) For any other unbiased estimator of $\tau(\theta)$, call it $T^*(\vec X)$, where $\mathbb{E}_{\theta}\{ T^*(\vec X)\} = \tau(\theta),\forall \theta \in \Theta$, and $$ \mathbf{Var}_{\theta}\{T(\vec X)\} \leq \mathbf{Var}_{\theta}\{ T^*(\vec X)\}, \forall \theta \in \Theta. $$

How do we actually find a UMVUE? There are two ways, one is by Cramér-Rao Lower Bound, one is by Rao-Blackwell \& Lehmann-Scheffé theorem. We first introduce the Cramér-Rao Lower Bound (CRLB). Assume $d=1$, $\Theta \subset \mathbb{R}$, $\vec X \sim p_{\theta}$ (joint pdf / pmf).

  • Assume that the family ${ p_{\theta} : \theta \in \Theta \subset \mathbb{R} }$ has a common support, i.e

\(\mathcal{S} := \left\{ \vec x = \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} : p_{\theta}(\vec x) > 0 \right\}\) and does not depend on $\theta$. (In this case $Uniform(0,\theta)$ is excluded);

  • Furthermore, assume for $\vec x \in \mathcal{S}$, $\theta \in \Theta \subset \mathbb{R}$, $\displaystyle{\frac{d}{d\theta}\log p_{\theta}(\vec x)}$ exists;

  • Also, assume for any statistics $h(\vec X)$ with $\mathbb{E}{ \vert h(\vec X) \vert } < +\infty, \forall \theta \in \Theta$ we have

\[\frac{d}{d\theta} \int_{\mathcal{S}} h(\vec x) p_{\theta}(\vec x) d\vec x = \int_{\mathcal{S}} h(\vec x) \frac{d}{d\theta} p_{\theta}(\vec x)d\vec x\]

whenever the $R.H.S$ is finite.

Theorem. (Cramér-Rao Lower Bound) Assume the above conditions hold, let $T(\vec X)$ be such that $$ \mathbf{Var}_{\theta}(T(\vec X))<+\infty, \mathbb{E}_{\theta}\{ T(\vec X)\} = \tau(\theta),\forall \theta \in \Theta, $$ then if $\displaystyle{0<\mathbb{E}_{\theta}\left\{ \left[ \frac{d}{d\theta} \log p_{\theta}(\vec x) \right]^2 \right\}<+\infty,\forall \theta \in \Theta}$, then $$ \mathbf{Var}_{\theta}(T(\vec X)) \geq \frac{[\tau'(\theta)]^2}{\displaystyle{\mathbb{E}_{\theta}\left\{ \left[ \frac{d}{d\theta} \log p_{\theta}(\vec x) \right]^2 \right\}}}. $$

Here, we denote

\[I(\theta) = \displaystyle{\mathbb{E}_{\theta}\left\{ \left[ \frac{d}{d\theta} \log p(\theta,\vec x) \right]^2 \right\}}\]

it is called the Fisher Information.

Proof. $W.L.O.G$, we will only prove the continuous case. Note $\tau(\theta) = \mathbb{E}_{\theta}\{ T(\vec X)\}, \forall \theta \in \Theta$, so $$ \begin{align*} \frac{d}{d\theta} \tau(\theta) &= \frac{d}{d\theta} \left\{ \mathbb{E}_{\theta}[T(\vec X)] \right\}\\ &=\frac{d}{d\theta} \left\{ \int_{\mathcal{S}} T{(\vec x)} p_{\theta}(\vec x) d\vec x \right\}\\ &= \int_{\mathcal{S}} T(\vec x) \cdot \frac{d}{d\theta} p_{\theta}(\vec x) d \vec x\\ &= \int_{\mathcal{S}} T(\vec x) \frac{d}{d\theta} [ \log p_{\theta}(\vec x) ]p_{\theta}(\vec x) d\vec x. \end{align*} $$ Therefore, $$ \tau'(\theta) = \mathbb{E}_{\theta}\left\{ T(\vec X) \frac{d}{d\theta} [\log p_{\theta}(\vec x)] \right\}, \forall \theta \in \Theta. $$ On the other hand, by (\romannumeral 3), if $h(\vec x) \equiv 1$, then $$ 0 = \int_{\mathcal{S}} \frac{d}{d\theta} p_{\theta}(\vec x) d\vec x= \int_{\mathcal{S}} \left[ \frac{d}{d\theta} \log p_{\theta}(\vec x) \right] p_{\theta}(\vec x) d\vec x, \forall \theta \in \Theta. $$ Thus we have $$ \mathbb{E}_{\theta} \left\{ \frac{d}{d\theta} \log p_{\theta}(\vec x) \right\}=0, \forall \theta \in \Theta. $$ The equation above is also known as the Bartlett Identity. So now we have $$ \tau'(\theta) = \mathbf{Cov}\left( T(\vec X), \frac{d}{d\theta} \log p_{\theta}(\vec x) \right), \forall \theta \in \Theta $$ and hence by Cauchy-Schwarz inequality, we have $$ \begin{align*} [\tau'(\theta)]^2 & = \mathbf{Cov}^2\left( T(\vec X), \frac{d}{d\theta} \log p_\theta (\vec x) \right)\\ & \leq \mathbf{Var}_\theta (T(\vec X)) \cdot \mathbf{Var}_\theta \left( \frac{d}{d\theta} \log p_\theta (\vec x) \right). \end{align*} $$ Now using Bartlett's identity, we have $$ [\tau'(\theta)]^2 \leq \mathbf{Var}_\theta(T(\vec X)) \cdot \mathbb{E}_\theta \left\{ \left[ \frac{d}{d\theta} \log p_\theta(\vec x)\right]^2 \right\}. $$

Note that if we have $X_1,\cdots,X_n \overset{i.i.d}{\sim} f(x,\theta)$, then by definition

\[p_\theta(\vec x) = \prod_{i=1}^n f(\theta,x_i)\]

hence in this case

\[\mathbb{E}\left\{ \left[ \frac{d}{d\theta} \log p_\theta(\vec x) \right]^2\right\} = n \mathbb{E}\left\{ \left[ \frac{d}{d\theta} \log f(\theta,x_i) \right]^2\right\}.\]

and now

\[\mathbf{Var}_\theta(T(\vec X)) \geq \frac{[\tau'(\theta)]^2}{n I_i(\theta)}\]

sometimes it is easier to compute the Fisher information for one sample, we may do so and multiply by $n$ in the end, with the assumption that we are working with random sample (i.i.d).

Corollary. (Second Bartlett's Identity) We have $$ \mathbb{E}\left\{ \left[ \frac{d}{d\theta} \log p_\theta (\vec x) \right]^2\right\} = -\mathbb{E}\left\{ \frac{d^2}{d\theta^2} \log p_\theta (\vec x) \right\}. $$
Proof. Using Chain rule, we have $$ \frac{d^2}{d\theta^2} \log p_\theta(\vec x) = \frac{p^{''}_\theta (\vec x)}{p_\theta (\vec x)} - \left[ \frac{d}{d\theta} \log p_\theta(\vec x) \right]^2 $$ and taking expectation on both sides, we have $$ \mathbb{E}\left\{ \frac{d^2}{d\theta^2} \log p_\theta (\vec x)\right\} = \mathbb{E}\left\{ \frac{p^{''}_\theta(\vec x)}{p_\theta(\vec x)} \right\} - \mathbb{E}\left\{ \left[ \frac{d}{d\theta} \log p_\theta (\vec x) \right]^2\right\} $$ where the first term on the right hand side of the equation equals $0$, because $$ \frac{d}{d\theta} \mathbb{E}\left\{ \frac{d}{d\theta} \log p_\theta(\vec x) \right\} = \int_\mathcal{S} \frac{d}{d\theta} \left\{ \frac{d}{d\theta} \log p_\theta (\vec x)\right\} p_\theta(\vec x) dx. $$

Example: Suppose we have a random sample $X_1,\cdots,X_n \overset{i.i.d}{\sim} Bernoulli(\theta)$, its probability mass function is given by $f(x,\theta) = \theta^x (1-\theta)^{1-x}, x \in {0,1}$ and $\theta \in (0,1)$. Find the CRLB of all unbiased estimator of $\tau(\theta) = \theta$.

Solution: We first find the Fisher information. We investigate the function

\[\log f(\theta,x_i) = x_i \log\theta + (1-x_i) \log(1-\theta)\]

and

\[\frac{d}{d\theta} \log f (\theta,x_i) = \frac{x_i}{\theta} - \frac{1-x_i}{1-\theta}, \hspace{0.3cm}\frac{d^2}{d\theta^2} \log f(\theta,x_i) = -\frac{x_i}{\theta^2} + \frac{1-x_i}{(1-\theta)^2}\]

by Bartlett’s identity, we have

\[I_i(\theta) = -\mathbb{E}\left\{ -\frac{X_i}{\theta^2} + \frac{1-X_i}{(1-\theta)^2}\right\} = \frac{1}{\theta(1-\theta)}.\]

Now $\tau(\theta) = \theta$ so $[\tau’(\theta)]^2 = 1$ hence the CRLB is given by

\[\mathbf{Var}_\theta(T(\vec X)) \geq \frac{1}{n I_i(\theta)} = \frac{\theta(1-\theta)}{n}.\]

An unbiased estimator of $\tau(\theta)$ might be the UMVUE, but its variance could be larger than the CRLB. That is, CRLB is not a sharp lower bound.

Theorem. Suppose that $\vec X =(X_1,\cdots,X_n)$ and we have a joint pdf $p_\theta(\vec x)$, and $T(\vec X)$ be unbiased for $\tau(\theta)$, then $T(\vec X)$ attains the CRLB if and only if $$ a(\theta) \cdot \{ T(\vec X) - \tau(\theta)\} = \frac{d}{d\theta} \log p_\theta(\vec x) $$ for some function $a(\theta)$ and for all $\theta \in \Theta$.

Example:

We say $f_\theta(x)$ belongs to exponential family (note that you may find me use $f_\theta(x), f(x,\theta), f(x \vert \theta)$, they are all the same notation, meaning $\theta$ is our parameter, but the density function is a function of $x$), if its density function takes the form

\[f_\theta(x) = h(x) \cdot c(\theta) \cdot \exp\{ \omega(\theta) \cdot T(x)\}\]

for some non-negative function $h(x)$ of $x$ and $c(\theta)$ of $\theta$, and $T(x)$ is a function of $x$, also its support does not depend on $\theta$. Then show that if a random sample $X_1,\cdots,X_n \overset{i.i.d}{\sim} f_\theta(x)$ where $f_\theta(x)$ belongs to exponential family, then

\[T(\vec x) = \frac{1}{n} \sum_{i=1}^n T(x_i)\]

is the UMVUE of $\tau(\theta) = -c’(\theta) / (c(\theta) \omega’(\theta))$.

Solution: Fist of all, the joint pdf of the sample takes the form

\[p_\theta(\vec x) = \left( \prod_{i=1}^n h(x_i) \right) c(\theta)^n \cdot \exp\left\{ \omega(\theta) \cdot \sum_{i=1}^n T(x_i) \right\}\]

so we have

\[\frac{d}{d\theta} \log p_\theta(\vec x) = n \cdot \frac{c'(\theta)}{c(\theta)} + \omega'(\theta) \cdot \sum_{i=1}^n T(x_i) = \omega'(\theta) \cdot \left\{ \sum_{i=1}^n T(x_i) - \frac{-n c'(\theta)}{c(\theta) \omega'(\theta)}\right\}.\]

By the previous theorem, it suffies to check $T(\vec X)$ is unbiased. Using Bartlett’s identity, we know that

\[\mathbb{E}\left\{ \frac{d}{d\theta} \log p_\theta(\vec x) \right\} = 0\]

which means

\[\mathbb{E}\left\{ \sum_{i=1}^n T(\vec X_i)\right\} = n \tau(\theta)\]

hence $T(\vec X) = \displaystyle{\frac{1}{n} \sum_{i=1}^n T(X_i)}$ is unbiased, and by the previous theorem, it is the UMVUE.

Example: Let a random sample $X_1,\cdots,X_n \overset{i.i.d}{\sim} Poisson(\theta)$ with density $\displaystyle{f(x,\theta) = e^{-\theta} \frac{\theta^x}{x!}}$, $x \in \mathbb{N}_0$.

  • find the UMVUE of $\theta$.

  • Find the CRLB of all estimates of $\theta$.

Solution:

We rewrite the density of Poisson distribution as

\[f(x,\theta) = \frac{e^{-\theta}}{x!} e^{x \log\theta}\]

let $h(x) = 1/x! , c(\theta) = e^{-\theta}, \omega(\theta) = \log(\theta), T(x) = x$, and its support $x \in \mathbb{N}_0$ does not depend on $\theta$, so it belongs to exponential family, then we use the result from the previous example, now $\tau(\theta) = \theta$, so the UMVUE is just the sample mean, given by

\[UMVUE(\theta) = \frac{1}{n} \sum_{i=1}^n X_i.\]

By direct computation, we have

\[\frac{d}{d\theta} \log f(x,\theta) = -1 + \frac{x}{\theta}, \frac{d^2}{d\theta^2} \log f(x,\theta) = -\frac{x}{\theta^2}\]

so using Bartlett’s identity, the Fisher information of one sample is given by

\[I(\theta) = -\mathbb{E}\left\{ \frac{d^2}{d\theta^2} \log f(X,\theta)\right\} = \mathbb{E}\left\{ -\frac{X}{\theta^2}\right\} = \frac{1}{\theta}\]

since $X \sim Poisson(\theta)$, hence the CRLB can be computed by

\[CRLB = \frac{[\tau'(\theta)]^2}{nI(\theta)} = \frac{\theta}{n}.\]

Recommended citation: Jiajun Zhang, (2025) Uniformly Minimum Variance Unbiased Estimator (UMVUE)