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1. How to use this iBook
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2. Prologue
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3. Introduction
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5. Correlations and Spectra
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6. Filtering of Stochastic Signals
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7. The Langevin Equation – A Case Study
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8. Characterizing Signal-to-Noise Ratios
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9. The Matched Filter
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10. The Wiener filter
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11. Aspects of Estimation
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11. Aspects of Estimation
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Maximum-likelihood estimation
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But is it a “good” estimate?
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Estimating the mean
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Estimating the autocorrelation function
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Problem 11.2
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Problem 11.3
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Problem 11.4
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Problem 11.8
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Problem 11.9
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Problem 11.10
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12. Spectral Estimation
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Maximum-likelihood estimation
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Example: Estimating the Poisson rate
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But is it a “good” estimate?
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Estimating the mean
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Estimating the autocorrelation function
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How good is our estimator?
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Langevin redux
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Trouble in paradise
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Problems
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Problem 11.1
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Problem 11.2
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Problem 11.3
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Problem 11.4
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Problem 11.5
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Problem 11.6
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Problem 11.7
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Problem 11.8
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Problem 11.9
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Problem 11.10
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Laboratory Exercises
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Laboratory Exercise 11.1
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Laboratory Exercise 11.2
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Laboratory Exercise 11.3
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<h1 id="aspects-of-estimation">Aspects of Estimation<a class="headerlink" href="#aspects-of-estimation" title="Permanent link">¶</a></h1>
<p>In this and the next chapter we will look at the problem of estimating the parameters of a stochastic process from experimental data. That is, how to estimate:</p>
<div class="" id="eq:Eeq1">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.1)</td>
<td class="eqTableEq">
<div>$${m_x} = E\left\{ {x[n]} \right\} = \; < x[n] >$$</div>
</td>
</tr>
</table>
</div>
<div class="" id="eq:Eeq2">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.2)</td>
<td class="eqTableEq">
<div>$${\varphi _{xx}}[k] = E\left\{ {x[n]{x^*}[n + k]} \right\} = \; < x[n]{x^*}[n + k] >$$</div>
</td>
</tr>
</table>
</div>
<div class="" id="eq:Eeq3">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.3)</td>
<td class="eqTableEq">
<div>$${S_{xx}}(\Omega ) = {\mathscr{F}}\left\{ {{\varphi _{xx}}[k]} \right\}$$</div>
</td>
</tr>
</table>
</div>
<p>There is a formal theory of estimation that addresses a variety of issues and the development of that theory can be quite complicated. An introduction can be found in van den Bos<sup id="fnref:bos2007"><a class="footnote-ref" href="#fn:bos2007">1</a></sup>. Instead we look at a few simple criteria. To estimate the mean of a process given <span class="arithmatex">\(N\)</span> samples, we might, for example, use one of the following:</p>
<p><em>arithmetic mean:</em></p>
<div class="" id="eq:Eeq4">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.4)</td>
<td class="eqTableEq">
<div>$${\mu _A} = \,\, < x[n]{ > _A}\; = \frac{1}{N}\sum\limits_{n = 0}^{N - 1} {x[n]}$$</div>
</td>
</tr>
</table>
</div>
<p><em>geometric mean:</em></p>
<div class="" id="eq:Eeq5">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.5)</td>
<td class="eqTableEq">
<div>$${\mu _G} = \,\,\, < x[n]{ > _G}\; = {\left( {\prod\limits_{n = 0}^{N - 1} {x[n]} } \right)^{1/N}}$$</div>
</td>
</tr>
</table>
</div>
<p><em>harmonic mean:</em></p>
<div class="" id="eq:Eeq6">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.6)</td>
<td class="eqTableEq">
<div>$${\mu _H} = \,\,\, < x[n]{ > _H}\; = N{\left( {\sum\limits_{n = 0}^{N - 1} {\frac{1}{{x[n]}}} } \right)^{ - 1}}$$</div>
</td>
</tr>
</table>
</div>
<p>These three ways of defining a mean are collectively referred to as the <a href="https://en.wikipedia.org/wiki/Pythagorean_means">Pythagorean means</a> and, as one might expect, there are useful relationships among the three.</p>
<p>From experience we expect that <a href="Chap_11.html#eq:Eeq4">Equation 11.4</a> will be the preferred form and, in fact, this can be shown from “maximum likelihood” estimation theory under many hypotheses. There are, however, numerous examples that can be found in finance, economics, the social sciences, and the physical sciences where one of the other two means is to be preferred. Let us
look at how a choice is made.</p>
<h2 id="maximum-likelihood-estimation">Maximum-likelihood estimation<a class="headerlink" href="#maximum-likelihood-estimation" title="Permanent link">¶</a></h2>
<p>We assume that <span class="arithmatex">\(N\)</span> independent measurements have been made of an ergodic random process that is described by the probability distribution <span class="arithmatex">\(p({x_i}|\theta )\)</span> where <span class="arithmatex">\(\theta\)</span> is a parameter of the distribution. In the case of coin flipping that parameter might be <span class="arithmatex">\(\theta = p = p(Heads)\)</span> as described in <a href="Chap_4.html">Chapter 4</a>.</p>
<p>The joint distribution of these <span class="arithmatex">\(N\)</span> measurements is given by
<span class="arithmatex">\(p\left( {{x_1},{x_2},{x_3},\,...\,,{x_N}|\theta } \right).\)</span> But because of the independence of these <span class="arithmatex">\(N\)</span> measurements we can rewrite this using <a href="Chap_3.html#eq:label14">Equation 3.8</a> as:</p>
<div class="" id="eq:Eeq7">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.7)</td>
<td class="eqTableEq">
<div>$$p({x_1},{x_2},{x_3},...,{x_N}|\theta ) = \prod\limits_{i = 1}^N {p({x_i}|\theta )}$$</div>
</td>
</tr>
</table>
</div>
<p>Given all of these measurements, the question is: How do we estimate <span class="arithmatex">\(\theta?\)</span> There are a variety of estimation criteria and the one we choose is <a href="https://en.wikipedia.org/wiki/Maximum_likelihood">maximum-likelihood estimation</a>, ML-estimation. In this approach we choose the value of <span class="arithmatex">\(\theta = {\theta _{ML}}\)</span> such that </p>
<div class="" id="eq:Eeq8">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.8)</td>
<td class="eqTableEq">
<div>$$p\left( {{x_1},{x_2},{x_3},\,...\,,{x_N}|{\theta _{ML}}} \right) \geqslant p\left( {{x_1},{x_2},{x_3},\,...\,,{x_N}|{\theta _o}} \right)$$</div>
</td>
</tr>
</table>
</div>
<p>where <span class="arithmatex">\({\theta _{ML}} \ne {\theta _o}.\)</span> In words, the ML-estimate of the parameter <span class="arithmatex">\(\theta\)</span> is the estimate that gives the maximum probability (likelihood) of collecting the data that we have just acquired. To see how this is used in practice let us look at a specific example.</p>
<h4 id="example-estimating-the-poisson-rate">Example: Estimating the Poisson rate<a class="headerlink" href="#example-estimating-the-poisson-rate" title="Permanent link">¶</a></h4>
<p>In this example we refer to the Poisson random process defined in <a href="Chap_8.html#snr-for-signals-and-systems-with-poisson-noise">Chapter 8</a>. We repeat the definition of the Poisson probability distribution:</p>
<div class="" id="eq:Eeq9">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.9)</td>
<td class="eqTableEq">
<div>$$p(n|\lambda T) = \frac{{{{\left( {\lambda T} \right)}^n}{e^{ - \lambda T}}}}{{n!}}$$</div>
</td>
</tr>
</table>
</div>
<p>where <span class="arithmatex">\(\lambda\)</span> is the number of events per unit “time” and <span class="arithmatex">\(T\)</span> is the duration of an observation window. We see that <span class="arithmatex">\(\theta = \lambda T.\)</span></p>
<p>To find the ML-estimate of <span class="arithmatex">\(\theta\)</span> given <span class="arithmatex">\(N\)</span> measurements of, say, photon emission <span class="arithmatex">\(\left\{ {{n_1},{n_2},\,...\,,{n_N}|\theta } \right\}\)</span> we use the result in <a href="Chap_11.html#eq:Eeq7">Equation 11.7</a> to write the <em>likelihood function</em> <span class="arithmatex">\(L(\theta )\)</span>:</p>
<div class="" id="eq:Eeq10">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.10)</td>
<td class="eqTableEq">
<div>$$L(\theta ) = \prod\limits_{i = 1}^N {p({n_i}|\theta )} = \prod\limits_{i = 1}^N {\frac{{{\theta ^{{n_i}}}{e^{ - \theta }}}}{{{n_i}!}}}$$</div>
</td>
</tr>
</table>
</div>
<p>We seek the value of <span class="arithmatex">\(\theta\)</span> that maximizes <span class="arithmatex">\(L(\theta ).\)</span> At first glance this might
seem like an intractable problem. One of the standard procedures—a.k.a. “tricks of the trade”—can, however, be applied.</p>
<p>It is not difficult to show that if <span class="arithmatex">\({\theta _{ML}}\)</span> maximizes <span class="arithmatex">\(\ln \left[ {L(\theta )} \right]\)</span>—where <span class="arithmatex">\(\ln\)</span> is the natural logarithm function—then it maximizes <span class="arithmatex">\(L(\theta )\)</span> as well. See <a href="#problem-111">Problem 11.1</a>. Applying this gives:</p>
<div class="" id="eq:Eeq11">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.11)</td>
<td class="eqTableEq">
<div>$$\begin{array}{*{20}{l}}
{\ln \left[ {L(\theta )} \right]}&{ = \ln \left[ {\prod\limits_{i = 1}^N {\frac{{{\theta ^{{n_i}}}{e^{ - \theta }}}}{{{n_i}!}}} } \right] = \sum\limits_{i = 1}^N {\ln } \left[ {\frac{{{\theta ^{{n_i}}}{e^{ - \theta }}}}{{{n_i}!}}} \right]}\\
{\,\,\,}&{ = \sum\limits_{i = 1}^N {{n_i}\ln } \left( \theta \right) - \sum\limits_{i = 1}^N \theta - \sum\limits_{i = 1}^N {n!} }
\end{array}$$</div>
</td>
</tr>
</table>
</div>
<p>To find the maximum (extremum actually) we set the derivative with
respect to <span class="arithmatex">\(\theta\)</span> to zero.</p>
<div class="" id="eq:Eeq12">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.12)</td>
<td class="eqTableEq">
<div>$$\begin{array}{*{20}{l}}
{\frac{{\partial \ln \left[ {L(\theta )} \right]}}{{\partial \theta }}}&{ = \frac{\partial }{{\partial \theta }}\left( {\sum\limits_{i = 1}^N {{n_i}\ln } \left( \theta \right) - \sum\limits_{i = 1}^N \theta - \sum\limits_{i = 1}^N {n!} } \right)}\\
{\,\,\,}&{ = \frac{1}{\theta }\sum\limits_{i = 1}^N {{n_i}} - N = 0}
\end{array}$$</div>
</td>
</tr>
</table>
</div>
<p>(How would you show that this extremum is a maximum as opposed to a minimum or saddle point?) Solving for <span class="arithmatex">\(\theta\)</span> gives the ML-estimate:</p>
<div class="" id="eq:Eeq13">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.13)</td>
<td class="eqTableEq">
<div>$$\frac{1}{\theta }\sum\limits_{i = 1}^N {{n_i}} - N = 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\theta _{ML}} = \frac{1}{N}\sum\limits_{i = 1}^N {{n_i}}$$</div>
</td>
</tr>
</table>
</div>
<p>The ML-estimate of the parameter <span class="arithmatex">\(\theta = \lambda T\)</span> is simply the <em>arithmetic</em> average
(<span class="arithmatex">\({\mu _A}\)</span>) of the measurements. Other examples can be found in <a href="#problem-112">Problem 11.2</a>.</p>
<h2 id="but-is-it-a-good-estimate">But is it a “good” estimate?<a class="headerlink" href="#but-is-it-a-good-estimate" title="Permanent link">¶</a></h2>
<p>In a given formula such as <a href="Chap_11.html#eq:Eeq4">Equation 11.4</a> we might also ask:</p>
<ol>
<li>On the average does the formula give the “right” answer?</li>
<li>If we take a very large number of samples will our estimation error
become smaller?<a class="indentlist" href=""></a> </li>
</ol>
<p>With regard to the first question we introduce a concept that is central to obtaining good estimates: the concept of <em>bias</em>. Formally if <span class="arithmatex">\(a\)</span> is a (deterministic) parameter of a random process and <span class="arithmatex">\(\hat a\)</span> is the estimate of <span class="arithmatex">\(a\)</span> then the bias <span class="arithmatex">\(B\)</span> is:</p>
<div class="" id="eq:Eeq14">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.14)</td>
<td class="eqTableEq">
<div>$$B = E\left\{ {\hat a} \right\} - a$$</div>
</td>
</tr>
</table>
</div>
<p>We should remember that <span class="arithmatex">\(\hat a\)</span> is a random variable because it is a function of random data. If <span class="arithmatex">\(B > 0\)</span> we have an overestimate of <span class="arithmatex">\(a\)</span> and if <span class="arithmatex">\(B < 0\)</span> we have an underestimate. We define an <em>unbiased estimate</em> as one that has <span class="arithmatex">\(B = 0.\)</span> An unbiased estimate is desirable because it means that the estimate will be neither consistently too high nor
consistently too low. Continuing, we define the mean-square error between the parameter and its estimate as</p>
<div class="" id="eq:Eeq15">
<table class="eqTable">
<tr>
<td class="eqTableTag">(11.15)</td>
<td class="eqTableEq">
<div>$$e = E\left\{ {{{\left( {\hat a - a} \right)}^2}} \right\}$$</div>
</td>
</tr>
</table>
</div>
<p>We will show in <a href="#problem-115">Problem 11.5</a> that choosing an unbiased estimate (<span class="arithmatex">\(B = 0\)</span>) can, under certain circumstances, minimize the mean-square error measure. In <a href="#problem-116">Problem 11.6</a> and <a href="#problem-117">Problem 11.7</a> we will show that this is not always the case. When it is the case, the estimators are referred to as <em>minimum-variance unbiased estimators</em> (<a href="https://en.wikipedia.org/wiki/Minimum-variance_unbiased_estimator">MVUE</a>).</p>
<p>An unbiased estimate—whether it is minimum-variance or not—is frequently referred to as an <em>accurate</em> estimate.</p>