# 5.1 Lms algorithm analysis

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## Objective

Minimize instantaneous squared error

${e}_{k}(w)^{2}=({y}_{k}-{x}_{k}^Tw)^{2}$

## Lms algorithm

$({w}_{k})=({w}_{k-1})+\mu {x}_{k}{e}_{k}$
Where ${w}_{k}$ is the new weight vector, ${w}_{k-1}$ is the old weight vector, and $\mu {x}_{k}{e}_{k}$ is a small step in the instantaneous error gradient direction.

## Interpretation in terms of weight error vector

Define

${v}_{k}={w}_{k}-{w}_{\mathrm{opt}}$
Where ${w}_{\mathrm{opt}}$ is the optimal weight vector and
${\epsilon }_{k}={y}_{k}-{x}_{k}^T{w}_{\mathrm{opt}}$
where ${\epsilon }_{k}$ is the minimum error. The stochastic difference equation is:
${v}_{k}=I{v}_{k-1}+\mu {x}_{k}{\epsilon }_{k}$

## Convergence/stability analysis

Show that (tightness)

$\lim_{B\to }B\to$ v k B 0
With probability 1, the weight error vector is bounded for all $k$ .

Chebyshev's inequality is

$(({v}_{k})\ge B)\le \frac{(({v}_{k})^{2})}{B^{2}}$
and
$(({v}_{k})\ge B)=\frac{1}{B^{2}}((({v}_{k}))^{2}+\sigma({v}_{k})^2)$
where $(({v}_{k}))^{2}$ is the squared bias. If $(({v}_{k}))^{2}+\sigma({v}_{k})^2$ is finite for all $k$ , then $\lim_{B\to }B\to$ v k B 0 for all $k$ .

Also,

$\sigma({v}_{k})^2=\mathrm{tr}(({v}_{k}{v}_{k}^T))$
Therefore $\sigma({v}_{k})^2$ is finite if the diagonal elements of ${\Gamma }_{k}\equiv ({v}_{k}{v}_{k}^T)$ are bounded.

## Convergence in mean

$(({v}_{k}))\to 0$ as $k\to$ . Take expectation of using smoothing property to simplify the calculation. We haveconvergence in mean if

• ${R}_{\mathrm{xx}}$ is positive definite (invertible).
• $\mu < \frac{2}{{\lambda }_{\mathrm{max}}({R}_{\mathrm{xx}})}$ .

## Bounded variance

Show that ${\Gamma }_{k}=({v}_{k}{v}_{k}^T)$ , the weight vector error covariance is bounded for all $k$ .

We could have $({v}_{k})\to 0$ , but $\sigma({v}_{k})^2\to$ ; in which case the algorithm would not be stable.
Recall that it is fairly straightforward to show that the diagonal elements of the transformed covariance ${C}_{k}=U{\Gamma }_{k}U^T$ tend to zero if $\mu < \frac{1}{{\lambda }_{\mathrm{max}}({R}_{\mathrm{xx}})}$ ( $U$ is the eigenvector matrix of ${R}_{\mathrm{xx}}$ ; ${R}_{\mathrm{xx}}=UDU^T$ ). The diagonal elements of ${C}_{k}$ were denoted by ${\gamma }_{k,i}\forall i, i=\{1, \dots , p\}$ .
$\sigma({v}_{k})^2=\mathrm{tr}({\Gamma }_{k})=\mathrm{tr}(U^T{C}_{k}U)=\mathrm{tr}({C}_{k})=\sum_{i=1}^{p} {\gamma }_{k,i}$
Thus, to guarantee boundedness of $\sigma({v}_{k})^2$ we need to show that the "steady-state" values ${\gamma }_{k,i}\to ({\gamma }_{i})$ .

We showed that

${\gamma }_{i}=\frac{\mu (\alpha +{\sigma }_{\epsilon }^{2})}{2(1-\mu {\lambda }_{i})}$
where ${\sigma }_{\epsilon }^{2}=({\epsilon }_{k}^{2})$ , ${\lambda }_{i}$ is the ${i}^{\mathrm{th}}$ eigenvalue of ${R}_{\mathrm{xx}}$ ( ${R}_{\mathrm{xx}}=U\begin{pmatrix}{\lambda }_{1} & \dots & 0\\ ⋮ & \ddots & ⋮\\ 0 & \dots & {\lambda }_{p}\\ \end{pmatrix}U^T$ ), and $\alpha =\frac{c{\sigma }_{\epsilon }^{2}}{1-c}$ .
$0< c=\frac{1}{2}\sum_{i=1}^{p} \frac{\mu {\lambda }_{i}}{1-\mu {\lambda }_{i}}< 1$
We found a sufficient condition for $\mu$ that guaranteed that the steady-state ${\gamma }_{i}$ 's (and hence $\sigma({v}_{k})^2$ ) are bounded: $\mu < \frac{\frac{2}{3}}{\sum_{i=1}^{p} {\lambda }_{i}}$ Where $\sum_{i=1}^{p} {\lambda }_{i}=\mathrm{tr}({R}_{\mathrm{xx}})$ is the input vector energy.

With this choice of $\mu$ we have:

• convergence in mean
This implies
$\lim_{B\to }B\to$ v k B 0
In other words, the LMS algorithm is stable about the optimumweight vector ${w}_{\mathrm{opt}}$ .

## Learning curve

Recall that

${e}_{k}={y}_{k}-{x}_{k}^T{w}_{k-1}$
and . These imply
${e}_{k}={\epsilon }_{k}-{x}_{k}^T{v}_{k-1}$
where ${v}_{k}={w}_{k}-{w}_{\mathrm{opt}}$ . So the MSE
$({e}_{k}^{2})={\sigma }_{\epsilon }^{2}+({v}_{k-1}^T{x}_{k}{x}_{k}^T{v}_{k-1})={\sigma }_{\epsilon }^{2}+(({x}_{n}{\epsilon }_{n}\forall n, n< k, {v}_{k-1}^T{x}_{k}{x}_{k}^T{v}_{k-1}))={\sigma }_{\epsilon }^{2}+({v}_{k-1}^T{R}_{\mathrm{xx}}{v}_{k-1})={\sigma }_{\epsilon }^{2}+(\mathrm{tr}({R}_{\mathrm{xx}}{v}_{k-1}{v}_{k-1}^T))={\sigma }_{\epsilon }^{2}+\mathrm{tr}({R}_{\mathrm{xx}}{\Gamma }_{k-1})$
Where $(\mathrm{tr}({R}_{\mathrm{xx}}{\Gamma }_{k-1})\equiv {\alpha }_{k-1})\to (\alpha =\frac{c{\sigma }_{\epsilon }^{2}}{1-c})$ . So the limiting MSE is
${\epsilon }_{}$ k e k 2 σ ε 2 c σ ε 2 1 c σ ε 2 1 c
Since $0< c< 1$ was required for convergence, ${\epsilon }_{}$ σ ε 2 so that we see noisy adaptation leads to an MSE larger than the optimal
$({\epsilon }_{k}^{2})=(({y}_{k}-{x}_{k}^T{w}_{\mathrm{opt}})^{2})={\sigma }_{\epsilon }^{2}$
To quantify the increase in the MSE, define the so-called misadjustment :
$M=\frac{-{\epsilon }_{}}{}$ σ ε 2 σ ε 2 ε σ ε 2 1 α σ ε 2 c 1 c
We would of course like to keep $M$ as small as possible.

Fast adaptation and quick convergence require that we take steps as large as possible. In other words,learning speed is proportional to $\mu$ ; larger $\mu$ means faster convergence. How does $\mu$ affect the misadjustment?

To guarantee convergence/stability we require $\mu < \frac{\frac{2}{3}}{\sum_{i=1}^{p} {\lambda }_{i}({R}_{\mathrm{xx}})}$ Let's assume that in fact $\ll (\mu , \frac{1}{\sum_{i=1}^{p} {\lambda }_{i}})$ so that there is no problem with convergence. This condition implies $\ll (\mu , \frac{1}{{\lambda }_{i}})$ or $\ll (\mu {\lambda }_{i}, 1)\forall i, i=\{1, \dots , p\}$ . From here we see that

$\ll (c=\frac{1}{2}\sum_{i=1}^{p} \frac{\mu {\lambda }_{i}}{1-\mu {\lambda }_{i}}\approx \frac{1}{2}\mu \sum_{i=1}^{p} {\lambda }_{i}, 1)$
$M=\frac{c}{1-c}\approx c=\frac{1}{2}\mu \sum_{i=1}^{p} {\lambda }_{i}$
This shows that larger step size $\mu$ leads to larger misadjustment.

Since we still have convergence in mean, this essentially means that with a larger step size we "converge"faster but have a larger variance (rattling) about ${w}_{\mathrm{opt}}$ .

## Summary

small $\mu$ implies

large $\mu$ implies

${w}_{\mathrm{opt}}=\left(\begin{array}{c}1\\ 1\end{array}\right)$ $({x}_{k}, (0, \begin{pmatrix}1 & 0\\ 0 & 1\\ \end{pmatrix}))$ ${y}_{k}={x}_{k}^T{w}_{\mathrm{opt}}+{\epsilon }_{k}$ $({\epsilon }_{k}, (0, 0.01))$

## Lms algorithm

initialization ${w}_{0}=\left(\begin{array}{c}0\\ 0\end{array}\right)$ and ${w}_{k}={w}_{k-1}+\mu {x}_{k}{e}_{k}\forall k, k\ge 1$ , where ${e}_{k}={y}_{k}-{x}_{k}^T{w}_{k-1}$

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