HW2. Math 499. Spring 2007 Independent studies course Supervisor: Dr Angel R. Pineda, Assistant Professor Mathematics Department California State University, Fullerton

Student: Nasser Abbasi


Problem

question: Consider the solution of MATH where $A$ is $m\times n$ matrix, MATH and $\QTR{bf}{\vec{n}}$ is vector of i.i.d. Gaussian MATH noise vector. (i.e. noise is white Gaussian noise). Determine the best solution $\vec{x}$

Answer:

Let us refer to the observed output (which includes the noise $\QTR{bf}{\vec {n}}$) as $\QTR{bf}{\vec{z}}$, hence we write MATH where $\QTR{bf}{\vec{b}}$ is the uncontaminated output (what the observed output would be if there is no noise).

Since the noise $\QTR{bf}{\vec{n}}$ is an additive noise to the output $\QTR{bf}{\vec{b}}$ of the system, the since the noise has zero mean, then the mean of $\QTR{bf}{\vec{z}}$ will be the same as the mean of $\QTR{bf}{\vec{b}}$. But $\QTR{bf}{\vec{b}}$ is a deterministic signal which does not change, hence its mean is its value, hence the mean of MATHis MATH

Now, $\QTR{bf}{\vec{z}}$ is described by a probability density function PDF as follows ( $\QTR{bf}{\vec{z}}$ is in $R^{m}$, hence it is $m$ long vector)

MATH

But since MATH, then the above can be written as

MATH

Since $A$ is a constant matrix (system is assumed to be time-invariant), hence from the above we see that the expression gives the probability of observing $\QTR{bf}{\vec{z}}$ for a given $\QTR{bf}{\vec{x}}$. Hence the best estimate of $\QTR{bf}{\vec{x}}$ would be the one which maximizes this probability. Instead of maximizing the PDF directly, we maximize its natural logarithm (a mathematical convenience trick, no more).

Now find the natural logarithm of the above quantity, and find where the result is maximum.

MATH

Start with $m=n=1$, hence the above becomes Note_1

MATH

Set the above to zero and solve for $x_{1}$

MATH

Hence MATH

This matches the least squares solution MATH

Now I need to do this for $m=n=2,$ and assuming that MATH. We can start from equation (1) above, shown again below

MATH

Let MATH

Then

MATH

so MATH

andMATH

Hence we obtain that

MATH

Hence MATH

and MATH

so, solve for $x_{1},x_{2}$. Write as $Ax=b$ and solve:

MATH

Solve for MATH

so

MATH

and

MATH

Hence

MATH

is the least squares error. To validate

MATH

Compare equations (2) and (3) above, they are the same. Ok, confirmed.