Nearly every formula in this course is a sum over a finite list of
numbers. An average brightness is one. So is a histogram count, a
convolution sum, and a discrete Fourier coefficient. This section defines
sigma notation as the bookkeeping tool for that kind of sum. It then
extends the idea to sums over image pixels. There the counting runs over
two directions at once, a row and a column. (linear-algebra's readiness
chapter works the one-direction case in more depth. The practice
questions below draw from its bank.)
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
\sum_{i=1}^{4} a_i and \sum_{j=1}^{4} a_j are...
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The index is a placeholder, not a label with meaning of its own — renaming it from i to j doesn't change which terms get added or in what order.
This condition would only make sense if i and j were separate quantities being compared; here they're just two different names for the same loop variable.
No numeric values are needed — the two sums are identical purely because the index is just a placeholder name, for any sequence a_1,\ldots,a_4.
Why
The summation index is a bound variable: its name has no effect on the sum's value. Both expressions add a_1+a_2+a_3+a_4.
Write out what each sum expands to term by term and compare -- the name of the index letter doesn't appear in the expanded sum.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Evaluate \sum_{k=2}^{5} (2k - 3).
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Why
Terms for k=2,3,4,5: 1, 3, 5, 7. Sum: 1+3+5+7 = 16.
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Why
List out the value of 2k-3 for each k from 2 to 5, then add the four results.
A double sum is just an ordinary sum done twice. First add up one row. Then do that for every row. Then add all the row totals together. Nothing about counting in two directions instead of one changes what "sum" means. It just says where to start and stop counting on two separate directions at once. Compare it to "every seat in every row of the theater": that needs both a row number and a seat number to be unambiguous.
The image-mean formula above is exactly "add up every pixel's brightness, then divide by how many pixels there are." That is the everyday meaning of "average." It only carries extra bookkeeping, the two sigma signs and their counting ranges, so a computer can carry it out without ever being told what a picture even is.
For a single list of numbers a_0, a_1, \dots, a_{n-1}, \sum_{i=0}^{n-1} a_i = a_0 + a_1 + \cdots + a_{n-1}. An image sum runs over two counters at once, row y and column x. So the sigma notation nests two sums, one per direction: \sum_{y=0}^{H-1}\sum_{x=0}^{W-1} I(x, y) is the sum of every pixel value in an H\times W image. It is computed by summing across each row, then adding the row totals together. The order of the two sums does not matter here, since addition never cares what order its terms arrive in. Dividing that double sum by the pixel count HW gives the image's mean brightness, \bar I = \frac{1}{HW}\sum_{y=0}^{H-1}\sum_{x=0}^{W-1} I(x, y), a quantity this course uses constantly, starting in Chapter 2.
Summing across one row of pixels
This bar chart is one row of an image, treated as a plain sequence of numbers. Watch the running total build as the sum adds one pixel at a time -- this is exactly the inner sum in the double-sigma formula above, before the outer sum adds row totals together.
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Worked example
Computing a mean brightness by hand
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1
Find the mean brightness of the 2\times 3 image I = \begin{bmatrix} 100 & 150 & 200 \\ 50 & 100 & 150 \end{bmatrix}. Set up the double sum: H=2 rows, W=3 columns, so HW = 6 pixels total.
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2
Sum row y=0: 100+150+200 = 450. Sum row y=1: 50+100+150=300. Add the row totals: \sum_{y=0}^{1}\sum_{x=0}^{2} I(x,y) = 450+300 = 750
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3
Divide by the pixel count: \bar I = \frac{750}{6} = 125
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Worked example
A second mean, on a bigger patch
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1
Find the mean brightness of the 2\times 4 image above -- half its pixels are 0, half are 255. Set up the double sum: H=2, W=4, so HW=8 pixels.
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2
Four pixels are 0, four are 255: total =4(0)+4(255)=1020.
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3\bar I = \frac{1020}{8} = 127.5
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Where this shows up
Manufacturing inspection
A camera checks a part on the line for a flaw: a scratch, a missing piece, an off color. Its first call often comes from nothing more than a handful of double-sum totals like \bar I. This mean gets figured over a small region of the image. A mean brightness far outside the normal range flags the part for closer review, or for pulling it off the line, before any fancier technique ever runs. The double sum is not a warm-up exercise here. It is the real math a factory line runs, millions of times a day, because it is cheap and exact.
Check your understanding (counts toward your progress)
Expand \sum_{y=0}^{1}\sum_{x=0}^{1} I(x,y) into a sum of four individual terms, with no sigma signs remaining.
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Why
Why
I(0,0)+I(1,0)+I(0,1)+I(1,1) -- the inner sum over x runs for each fixed y in turn, so all four (x, y) combinations from the two ranges appear exactly once.
In \sum_{y=0}^{H-1}\sum_{x=0}^{W-1} I(x,y), what does the order of the two sigma signs affect?
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Addition is commutative and associative -- summing the same set of numbers in a different order never changes the total.
Both orderings touch every single pixel exactly once -- no pixel is included or excluded by swapping sum order.
The dimensions H and W are fixed by the image itself, unrelated to which sigma is written first.
Why
Since every term is added exactly once regardless of order, swapping the two sigma signs changes nothing about the result.
Find the mean brightness of \begin{bmatrix}0&255\\255&0\end{bmatrix}.
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Why
The sum of all four entries is 0+255+255+0=510; dividing by the pixel count 4 gives \bar I = 127.5.
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Why
What does HW represent in the mean-brightness formula?
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Diagonal length would involve a square root of a sum of squares, not a plain product.
The brightest possible value (255 for 8-bit images) is unrelated to the image's height and width.
Channel count is a separate property from spatial dimensions entirely.
Why
Height times width counts every distinct (x,y) pair exactly once -- the total pixel count.
Practice: Double sums by hand (3 problems · doesn’t affect your progress)
This 2\times 2 image has one black pixel and one white pixel on each diagonal. \begin{bmatrix}0&255\\255&0\end{bmatrix}. What is its mean brightness?
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Why
Sum is 0+255+255+0=510; dividing by 4 pixels gives 127.5.
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Why
Every one of the nine pixels in a 3\times 3 image equals 80. Compute its mean brightness.
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Why
Averaging n copies of the same value returns that value.
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Why
Expand \sum_{y=0}^{1}\sum_{x=0}^{1} I(x,y) into a sum of four individual terms, with no sigma signs left.
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First mistake
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Why
Why
I(0,0)+I(1,0)+I(0,1)+I(1,1).