Estimation theory

Consider a received discrete signal, ${\displaystyle x[n]}$, of ${\displaystyle N}$ independent samples that consists of an unknown constant ${\displaystyle A}$ with additive white Gaussian noise (AWGN) ${\displaystyle w[n]}$ with zero mean and known variance ${\displaystyle \sigma ^{2}}$ (i.e., ${\displaystyle {\mathcal {N}}(0,\sigma ^{2})}$). Since the variance is known then the only unknown parameter is ${\displaystyle A}$.

The model for the signal is then

${\displaystyle x[n]=A+w[n]\quad n=0,1,\dots ,N-1}$

Two possible (of many) estimators for the parameter ${\displaystyle A}$ are:

• ${\displaystyle {\hat {A}}_{1}=x[0]}$
• ${\displaystyle {\hat {A}}_{2}={\frac {1}{N}}\sum _{n=0}^{N-1}x[n]}$ which is the sample mean

Both of these estimators have a mean of ${\displaystyle A}$, which can be shown through taking the expected value of each estimator

${\displaystyle \mathrm {E} \left[{\hat {A}}_{1}\right]=\mathrm {E} \left[x[0]\right]=A}$

and

${\displaystyle \mathrm {E} \left[{\hat {A}}_{2}\right]=\mathrm {E} \left[{\frac {1}{N}}\sum _{n=0}^{N-1}x[n]\right]={\frac {1}{N}}\left[\sum _{n=0}^{N-1}\mathrm {E} \left[x[n]\right]\right]={\frac {1}{N}}\left[NA\right]=A}$

At this point, these two estimators would appear to perform the same. However, the difference between them becomes apparent when comparing the variances.

${\displaystyle \mathrm {var} \left({\hat {A}}_{1}\right)=\mathrm {var} \left(x[0]\right)=\sigma ^{2}}$

and

${\displaystyle \mathrm {var} \left({\hat {A}}_{2}\right)=\mathrm {var} \left({\frac {1}{N}}\sum _{n=0}^{N-1}x[n]\right){\overset {\text{independence}}{=}}{\frac {1}{N^{2}}}\left[\sum _{n=0}^{N-1}\mathrm {var} (x[n])\right]={\frac {1}{N^{2}}}\left[N\sigma ^{2}\right]={\frac {\sigma ^{2}}{N}}}$

It would seem that the sample mean is a better estimator since its variance is lower for every N > 1.

Continuing the example using the maximum likelihood estimator, the probability density function (pdf) of the noise for one sample ${\displaystyle w[n]}$ is

${\displaystyle p(w[n])={\frac {1}{\sigma {\sqrt {2\pi }}}}\exp \left(-{\frac {1}{2\sigma ^{2}}}w[n]^{2}\right)}$

and the probability of ${\displaystyle x[n]}$ becomes (${\displaystyle x[n]}$ can be thought of a ${\displaystyle {\mathcal {N}}(A,\sigma ^{2})}$)

${\displaystyle p(x[n];A)={\frac {1}{\sigma {\sqrt {2\pi }}}}\exp \left(-{\frac {1}{2\sigma ^{2}}}(x[n]-A)^{2}\right)}$

By independence, the probability of ${\displaystyle \mathbf {x} }$ becomes

${\displaystyle p(\mathbf {x} ;A)=\prod _{n=0}^{N-1}p(x[n];A)={\frac {1}{\left(\sigma {\sqrt {2\pi }}\right)^{N}}}\exp \left(-{\frac {1}{2\sigma ^{2}}}\sum _{n=0}^{N-1}(x[n]-A)^{2}\right)}$

Taking the natural logarithm of the pdf

${\displaystyle \ln p(\mathbf {x} ;A)=-N\ln \left(\sigma {\sqrt {2\pi }}\right)-{\frac {1}{2\sigma ^{2}}}\sum _{n=0}^{N-1}(x[n]-A)^{2}}$

and the maximum likelihood estimator is

${\displaystyle {\hat {A}}=\arg \max \ln p(\mathbf {x} ;A)}$

Taking the first derivative of the log-likelihood function

${\displaystyle {\frac {\partial }{\partial A}}\ln p(\mathbf {x} ;A)={\frac {1}{\sigma ^{2}}}\left[\sum _{n=0}^{N-1}(x[n]-A)\right]={\frac {1}{\sigma ^{2}}}\left[\sum _{n=0}^{N-1}x[n]-NA\right]}$

and setting it to zero

${\displaystyle 0={\frac {1}{\sigma ^{2}}}\left[\sum _{n=0}^{N-1}x[n]-NA\right]=\sum _{n=0}^{N-1}x[n]-NA}$

This results in the maximum likelihood estimator

${\displaystyle {\hat {A}}={\frac {1}{N}}\sum _{n=0}^{N-1}x[n]}$

which is simply the sample mean. From this example, it was found that the sample mean is the maximum likelihood estimator for ${\displaystyle N}$ samples of a fixed, unknown parameter corrupted by AWGN.

To find the Cramér–Rao lower bound (CRLB) of the sample mean estimator, it is first necessary to find the Fisher information number

${\displaystyle {\mathcal {I}}(A)=\mathrm {E} \left(\left[{\frac {\partial }{\partial A}}\ln p(\mathbf {x} ;A)\right]^{2}\right)=-\mathrm {E} \left[{\frac {\partial ^{2}}{\partial A^{2}}}\ln p(\mathbf {x} ;A)\right]}$

and copying from above

${\displaystyle {\frac {\partial }{\partial A}}\ln p(\mathbf {x} ;A)={\frac {1}{\sigma ^{2}}}\left[\sum _{n=0}^{N-1}x[n]-NA\right]}$

Taking the second derivative

${\displaystyle {\frac {\partial ^{2}}{\partial A^{2}}}\ln p(\mathbf {x} ;A)={\frac {1}{\sigma ^{2}}}(-N)={\frac {-N}{\sigma ^{2}}}}$

and finding the negative expected value is trivial since it is now a deterministic constant ${\displaystyle -\mathrm {E} \left[{\frac {\partial ^{2}}{\partial A^{2}}}\ln p(\mathbf {x} ;A)\right]={\frac {N}{\sigma ^{2}}}}$

Finally, putting the Fisher information into

${\displaystyle \mathrm {var} \left({\hat {A}}\right)\geq {\frac {1}{\mathcal {I}}}}$

results in

${\displaystyle \mathrm {var} \left({\hat {A}}\right)\geq {\frac {\sigma ^{2}}{N}}}$

Comparing this to the variance of the sample mean (determined previously) shows that the sample mean is equal to the Cramér–Rao lower bound for all values of ${\displaystyle N}$ and ${\displaystyle A}$. In other words, the sample mean is the (necessarily unique) efficient estimator, and thus also the minimum variance unbiased estimator (MVUE), in addition to being the maximum likelihood estimator.

One of the simplest non-trivial examples of estimation is the estimation of the maximum of a uniform distribution. It is used as a hands-on classroom exercise and to illustrate basic principles of estimation theory. Further, in the case of estimation based on a single sample, it demonstrates philosophical issues and possible misunderstandings in the use of maximum likelihood estimators and likelihood functions.

Given a discrete uniform distribution ${\displaystyle 1,2,\dots ,N}$ with unknown maximum, the UMVU estimator for the maximum is given by

${\displaystyle {\frac {k+1}{k}}m-1=m+{\frac {m}{k}}-1}$

where m is the sample maximum and k is the sample size, sampling without replacement.[2][3] This problem is commonly known as the German tank problem, due to application of maximum estimation to estimates of German tank production during World War II.

The formula may be understood intuitively as;

"The sample maximum plus the average gap between observations in the sample",

the gap being added to compensate for the negative bias of the sample maximum as an estimator for the population maximum.[note 1]

This has a variance of[2]

${\displaystyle {\frac {1}{k}}{\frac {(N-k)(N+1)}{(k+2)}}\approx {\frac {N^{2}}{k^{2}}}{\text{ for small samples }}k\ll N}$

so a standard deviation of approximately ${\displaystyle N/k}$, the (population) average size of a gap between samples; compare ${\displaystyle {\frac {m}{k}}}$ above. This can be seen as a very simple case of maximum spacing estimation.

The sample maximum is the maximum likelihood estimator for the population maximum, but, as discussed above, it is biased.