# 1.1 The bayesian paradigm  (Page 2/2)

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$\forall n, n=\{1, , N\}\colon {x}_{n}=A+{W}_{n}$ $({W}_{n}, (0, ^{2}))$ iid. $(A)=\frac{1}{N}\sum_{n=1}^{N} {x}_{n}$ MVUB and MLE estimator. Now suppose that we have prior knowledge that $-{A}_{0}\le A\le {A}_{0}$ . We might incorporate this by forming a new estimator

$\stackrel{}{A}=\begin{cases}-{A}_{0} & \text{if (A)< -{A}_{0}}\\ (A) & \text{if -{A}_{0}\le (A)\le {A}_{0}}\\ {A}_{0} & \text{if (A)> {A}_{0}}\end{cases}$
This is called a truncated sample mean estimator of $A$ . Is $\stackrel{}{A}$ a better estimator of $A$ than the sample mean $(A)$ ?

Let $p(a)$ denote the density of $(A)$ . Since $(A)=\frac{1}{N}\sum {x}_{n}$ , $p(a)=(A, \frac{^{2}}{N})$ . The density of $\stackrel{}{A}$ is given by

$\stackrel{}{p}(a)=((A)\le -{A}_{0})(a+{A}_{0})+p(a){I}_{\left\{-{A}_{0}{A}_{0}\right\}}+((A)\ge {A}_{0})(a-{A}_{0})$
Now consider the MSE of the sample mean $(A)$ .
$\mathrm{MSE}((A))=\int_{()} \,d a$ a A 2 p a

## Note

• $\stackrel{}{A}$ is biased ( ).
• Although $(A)$ is MVUB, $\stackrel{}{A}$ is better in the MSE sense.
• Prior information is aptly described by regarding $A$ as a random variable with a prior distribution $U(-{A}_{0}, {A}_{0})$ , which implies that we know $-{A}_{0}\le A\le {A}_{0}$ , but otherwise $A$ is abitrary.

## The bayesian approach to statistical modeling

$\forall n, n=\{1, , N\}\colon {x}_{n}=A+{W}_{n}$

Prior distribution allows us to incorporate prior information regarding unknown paremter--probable values of parameter aresupported by prior. Basically, the prior reflects what we believe "Nature" will probably throw at us.

• ## (a)

joint distribution $p(x,)=p(, x)p()$
• ## (b)

marginal distributions $p(x)=\int p(, x)p()\,d$ $p()=\int p(, x)p()\,d x$ where $p()$ is a prior .
• ## (c)

posterior distribution $p(x, )=\frac{p(x,)}{p(x)}=\frac{p(, x)p()}{\int p(, x)p()\,d x}$

$\forall , \in \left[0 , 1\right]\colon p(, x)=(n, x)^{x}(1-)^{(n-x)}$ which is the Binomial likelihood. $p()=\frac{1}{B(, )}^{(-1)}(1-)^{(-1)}$ which is the Beta prior distriubtion and $B(, )=\frac{()()}{(+)}$

## Joint density

$p(x,)=\frac{(n, x)}{B(, )}^{(+x-1)}(1-)^{(n-x+-1)}$

## Marginal density

$p(x)=(n, x)\frac{(+)}{()()}\frac{(+x)(n-x+)}{(++n)}$

## Posterior density

$p(x, )=\frac{^{(+x-1)}^{(+n-x-1)}}{B(+x, +n-x)}$ where $B(+x, +n-x)$ is the Beta density with parameters ${}^{}=+x$ and ${}^{}=+n-x$

## Selecting an informative prior

Clearly, the most important objective is to choose the prior $p()$ that best reflects the prior knowledge available to us. In general, however, our prior knowledge is imprecise andany number of prior densities may aptly capture this information. Moreover, usually the optimal estimator can't beobtained in closed-form.

Therefore, sometimes it is desirable to choose a prior density that models prior knowledge and is nicely matched in functional form to $p(, x)$ so that the optimal esitmator (and posterior density) can be expressed in a simple fashion.

## 1. informative priors

• design/choose priors that are compatible with prior knowledge of unknown parameters

## 2. non-informative priors

• attempt to remove subjectiveness from Bayesian procedures
• designs are often based on invariance arguments

Suppose we want to estimate the variance of a process, incorporating a prior that is amplitude-scaleinvariant (so that we are invariant to arbitrary amplitude rescaling of data). $p(s)=\frac{1}{s}$ satisifies this condition. $(^{2}, p(s))\implies (A^{2}, p(s))$ where $p(s)$ is non-informative since it is invariant to amplitude-scale.

## Idea

Given $p(, x)$ , choose $p()$ so that $(p(x, ), p(, x)p())$ has a simple functional form.

## Conjugate priors

Choose $p()\in$ , where  is a family of densities ( e.g. , Gaussian family) so that the posterior density also belongsto that family.

conjugate prior
$p()$ is a conjugate prior for $p(, x)$ if $p()\in \implies p(x, )\in$

$\forall n, n=\{1, , N\}\colon {x}_{n}=A+{W}_{n}$ $({W}_{n}, (0, ^{2}))$ iid. Rather than modeling $(A, U(-{A}_{0}, {A}_{0}))$ (which did not yield a closed-form estimator) consider $p(A)=\frac{1}{\sqrt{2\pi {}_{A}^{2}}}e^{\frac{-1}{2{}_{A}^{2}}(A-)^{2}}$

With $=0$ and ${}_{A}=\frac{1}{3}{A}_{0}$ this Gaussian prior also reflects prior knowledge that it is unlikely for $\left|A\right|\ge {A}_{0}$ .

The Gaussian prior is also conjugate to the Gaussian likelihood $p(A, x)=\frac{1}{(2\pi ^{2})^{\left(\frac{N}{2}\right)}}e^{\frac{-1}{2^{2}}\sum_{n=1}^{N} ({x}_{n}-A)^{2}}$ so that the resulting posterior density is also a simple Gaussian, as shown next.

First note that $p(A, x)=\frac{1}{(2\pi ^{2})^{\left(\frac{N}{2}\right)}}e^{\frac{-1}{2^{2}}\sum_{n=1}^{N} {x}_{n}}e^{\frac{-1}{2^{2}}(NA^{2}-2NA\stackrel{-}{x})}$ where $\stackrel{-}{x}=\frac{1}{N}\sum_{n=1}^{N} {x}_{n}$ .

$p(x, A)=\frac{p(A, x)p(A)}{\int p(A, x)p(A)\,d A}=\frac{e^{\frac{-1}{2}(\frac{1}{^{2}}(NA^{2}-2NA\stackrel{-}{x})+\frac{1}{{}_{A}^{2}}(A-)^{2})()}}{\int_{()} \,d A}$ -1 2 1 2 N A 2 2 N A x - 1 A 2 A 2 -1 2 Q A A -1 2 Q A
where $Q(A)=\frac{N}{^{2}}A^{2}-\frac{2NA\stackrel{-}{x}}{^{2}}+\frac{A^{2}}{{}_{A}^{2}}-\frac{2A}{{}_{A}^{2}}+\frac{^{2}}{{}_{A}^{2}}$ . Now let ${}_{A|x}^{2}\equiv \frac{1}{\frac{N}{^{2}}+\frac{1}{{}_{A}^{2}}}$ ${}_{A|x}^{2}\equiv (\frac{N}{^{2}}\stackrel{-}{x}+\frac{}{{}_{A}^{2}}){}_{A|x}^{2}$ Then by "completing the square" we have
$Q(A)=\frac{1}{{}_{A|x}^{2}}(A^{2}-2{}_{A|x}A+{}_{A|x}^{2})-\frac{{}_{A|x}^{2}}{{}_{A|x}^{2}}+\frac{^{2}}{{}_{A}^{2}}=\frac{1}{{}_{A|x}^{2}}(A-{}_{A|x})^{2}-\frac{{}_{A|x}^{2}}{{}_{A|x}^{2}}+\frac{^{2}}{{}_{A}^{2}}$
Hence, $p(x, A)=\frac{e^{\frac{-1}{2{}_{A|x}\times 2}(A-{}_{A|x})^{2}}e^{\frac{-1}{2}(\frac{^{2}}{{}_{A}^{2}}-\frac{{}_{A|x}^{2}}{{}_{A|x}^{2}})}}{\int_{()} \,d A}$ -1 2 A | x 2 A A | x 2 -1 2 2 A 2 A | x 2 A | x 2 where $\frac{-1}{2{}_{A|x}\times 2}(A-{}_{A|x})^{2}$ is the "unnormalized" Gaussian density and $\frac{-1}{2}(\frac{^{2}}{{}_{A}^{2}}-\frac{{}_{A|x}^{2}}{{}_{A|x}^{2}})$ is a constant, independent of $A$ . This implies that $p(x, A)=\frac{1}{\sqrt{2\pi {}_{A|x}^{2}}}e^{\frac{-1}{2{}_{A|x}^{2}}(A-{}_{A|x})^{2}}$ where $(A|x, ({}_{A|x}, {}_{A|x}^{2}))$ . Now
$(A)=(x, A)=\int Ap(x, A)\,d A={}_{A|x}=\frac{\frac{N}{^{2}}\stackrel{-}{x}+\frac{}{{}_{A}^{2}}}{\frac{N}{^{2}}+\frac{1}{{}_{A}^{2}}}=\frac{{}_{A}^{2}}{{}_{A}^{2}+\frac{^{2}}{N}}\stackrel{-}{x}+\frac{\frac{^{2}}{N}}{{}_{A}^{2}+\frac{^{2}}{N}}=\stackrel{-}{x}-1$
Where $0< =\frac{{}_{A}^{2}}{{}_{A}^{2}+\frac{^{2}}{N}}< 1$

## Interpretation

• When there is little data $({}_{A}^{2}, \frac{^{2}}{N})$  is small and $(A)=$ .
• When there is a lot of data $({}_{A}^{2}, \frac{^{2}}{N})$ , $\approx 1$ and $(A)=\stackrel{-}{x}$ .

## Interplay between data and prior knowledge

Small $N\to (A)$ favors prior.

Large $N\to (A)$ favors data.

## The multivariate gaussian model

The multivariate Gaussian model is the most important Bayesian tool in signal processing. It leads directly tothe celebrated Wiener and Kalman filters.

Assume that we are dealing with random vectors $x$ and $y$ . We will regard $y$ as a signal vector that is to be estimated from an observation vector $x$ .

$y$ plays the same role as  did in earlier discussions. We will assume that $y$ is p1 and $x$ is N1. Furthermore, assume that $x$ and $y$ are jointly Gaussian distributed $(\begin{pmatrix}x\\ y\\ \end{pmatrix}, (\begin{pmatrix}0\\ 0\\ \end{pmatrix}, \begin{pmatrix}{R}_{\mathrm{xx}} & {R}_{\mathrm{xy}}\\ {R}_{\mathrm{yx}} & {R}_{\mathrm{yy}}\\ \end{pmatrix}))$ $(x)=0$ , $(y)=0$ , $(xx^T)={R}_{\mathrm{xx}}$ , $(xy^T)={R}_{\mathrm{xy}}$ , $(yx^T)={R}_{\mathrm{yx}}$ , $(yy^T)={R}_{\mathrm{yy}}$ . $R\equiv \begin{pmatrix}{R}_{\mathrm{xx}} & {R}_{\mathrm{xy}}\\ {R}_{\mathrm{yx}} & {R}_{\mathrm{yy}}\\ \end{pmatrix}$

$x=y+W$ , $(W, (0, ^{2}I))$ $p(y)=(0, {R}_{\mathrm{yy}})$ which is independent of $W$ . $(x)=(y)+(W)=0$ , $(xx^T)=(yy^T)+(yW^T)+(Wy^T)+(WW^T)={R}_{\mathrm{yy}}+^{2}I$ , $(xy^T)=(yy^T)+(Wy^T)={R}_{\mathrm{yy}}=(yx^T)$ . $(\begin{pmatrix}x\\ y\\ \end{pmatrix}, (\begin{pmatrix}0\\ 0\\ \end{pmatrix}, \begin{pmatrix}{R}_{\mathrm{yy}}+^{2}I & {R}_{\mathrm{yy}}\\ {R}_{\mathrm{yy}} & {R}_{\mathrm{yy}}\\ \end{pmatrix}))$ From our Bayesian perpsective, we are interested in $p(x, y)$ .

$p(x, y)=\frac{p(x,y)}{p(x)}=\frac{(2\pi )^{-\left(\frac{N}{2}\right)}(2\pi )^{-\left(\frac{p}{2}\right)}\det R^{\left(\frac{-1}{2}\right)}e^{\frac{-1}{2}\begin{pmatrix}x^T & y^T\\ \end{pmatrix}R^{(-1)}\begin{pmatrix}x\\ y\\ \end{pmatrix}}}{(2\pi )^{-\left(\frac{N}{2}\right)}\det {R}_{\mathrm{xx}}^{\left(\frac{-1}{2}\right)}e^{\frac{-1}{2}x^T{R}_{\mathrm{xx}}^{(-1)}x}}$
In this formula we are faced with $R^{(-1)}=\begin{pmatrix}{R}_{\mathrm{xx}} & {R}_{\mathrm{xy}}\\ {R}_{\mathrm{yx}} & {R}_{\mathrm{yy}}\\ \end{pmatrix}^{(-1)}$ The inverse of this covariance matrix can be written as $\begin{pmatrix}{R}_{\mathrm{xx}} & {R}_{\mathrm{xy}}\\ {R}_{\mathrm{yx}} & {R}_{\mathrm{yy}}\\ \end{pmatrix}^{(-1)}=\begin{pmatrix}{R}_{\mathrm{xx}}^{(-1)} & 0\\ 0 & 0\\ \end{pmatrix}+\begin{pmatrix}-{R}_{\mathrm{xx}}^{(-1)}{R}_{\mathrm{xy}}\\ I\\ \end{pmatrix}Q^{(-1)}\begin{pmatrix}-{R}_{\mathrm{yx}}{R}_{\mathrm{xx}} & I\\ \end{pmatrix}$ where $Q\equiv {R}_{\mathrm{yy}}-{R}_{\mathrm{yx}}{R}_{\mathrm{xx}}{R}_{\mathrm{xy}}$ . (Verify this formula by applying the right hand side above to $R$ to get $I$ .)

#### Questions & Answers

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Source:  OpenStax, Statistical signal processing. OpenStax CNX. Jun 14, 2004 Download for free at http://cnx.org/content/col10232/1.1
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