ECE 641 Fall 2008 Professor Bouman
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Errors in the notes
Dear ECE 641 Student, It would GREAT if you could post errors you find in my notes here.
Thanks.
Prof. Bouman
Errors version dated September 3 2008
- Please swap the exponents in (Bernoulli)PMF equation on page 12.
- Please correct the posterior variance on page 19 (middle of the page): Assignment for "a" after the line "More specifically ...".
Dalton
Errors version dated September 5 2008
- I believe the definition of the conditional expectation on page 8 is not true, possibly what was meant was: $ E[X|Y=y]=\int_\mathbb{R} x dF(x|Y=y) $. (Also, less significantly, the pdf/cdf definitions are swapped.)
- On page 15, it may be worthwhile to note that an estimator with minimum MSE is not necessarily efficient (since we are omitting the fact that efficiency is actually defined in terms of achieving the CRLB).
- I believe the expected cost (page 17) should be $ \hat{C}(x)=E[C(x,T(Y))|X=x]=C\left(x,\int_{\mathbb{R}^n} T(y)p_{Y|X}(y|x)dy\right) $
Josh
Errors version dated September 7 2008
- Since the exponents in the PMF equation on Page 12 has been swapped, I believe the equation on the next page (Example 2.1.2) needs to have its exponents swapped also since it's the same PMF.
Alfa
Errors Lab 3
- Note: the following discussion uses prefix notation instead of infix (so everything 'reads' like functions). It may be overkill, but I want to be notationally explicit.
- In case anyone was listening to my comments regarding the lab, I want to correct my stated results. To recap, the conditional from which $ W $ was drawn should indeed depend on the neighborhood using vales from $ X^{(k+1)} $ and $ X^{(k)} $ (the last pixel of the ordering will depend solely on a neighborhood of $ X^{(k+1)} $). This conclusion still holds (and it darn well better!).
- However, I noticed that I made a mistake based on the bad use of dirac functions. To clarify, if boolean operators are a function of two reals, ie $ \ne:\mathbb{R}^2\rightarrow \{0,1\} $ (with the typical interpretation) and
$ \delta(x)=\begin{cases}0, & x\ne 0\\ 1, & x=0\end{cases} $
then $ \delta(\ne\!(x,y)) \text{ iff } =\!(x,y) $. Anyway, my point is that as it is written, and under my interpretation, it is the complement of what it should be. If you replace all the $ \delta $ of the lab with indicator function $ I $ then things are less ambiguous. One could equivalently use: $ 1-\delta(x-y) $.
- Bottom-line: the particular choice of parameters listed in the questions results in all zeros or all ones with high probability.
Apologies for any confusion!
Josh
