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 $\vec x = \begin{pmatrix} x_1 \ x_2 \ \vdots \ x_n \end{pmatrix}$.
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
whenever the $R.H.S$ is finite.
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.
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).
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.
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)
