Question

Let \(f(x,y)\) be an image with the histogram as shown in the figure. X-axis denotes grey levels (r) and Y-axis denotes the probability (p) of occurrence of the grey levels.

What is the variance of the equalized image?

• A. 2/27
• B. 3/27
• C. 2/3
• D. 1/27

Hint:

Equalization maps the three equal-probability grey levels to cumulative-probability values 1/3, 2/3, 1; find the variance of this new set with each value equally likely.

Answer 1

The Correct Option is A

Step 1: Read the histogram.
The histogram shows three discrete grey levels \(r_1, r_2, r_3\), each occurring with the same probability, since the three impulses in the figure are drawn to the same height on the dashed line. Because probabilities must sum to 1, \[ p(r_1) = p(r_2) = p(r_3) = \frac{1}{3} \]

Step 2: Recall histogram equalization.
Histogram equalization maps each input grey level \(r_k\) to an output level \(s_k\) using the cumulative distribution function (CDF) of the input histogram: \[ s_k = \sum_{i=1}^{k} p(r_i) \] (working on the normalized \([0,1]\) grey-level scale).

Step 3: Compute the transformed (equalized) grey levels.
\[ s_1 = \frac{1}{3}, \qquad s_2 = \frac{2}{3}, \qquad s_3 = 1 \] Each transformed level keeps the same probability as its corresponding \(r_k\), i.e. \(p(s_1)=p(s_2)=p(s_3)=\frac{1}{3}\), since equalization here is a one-to-one relabeling of grey levels, not a redistribution of probability mass.

Step 4: Compute the mean of the equalized image.
\[ E[s] = \frac{1}{3}\left(\frac{1}{3} + \frac{2}{3} + 1\right) = \frac{2}{3} \]

Step 5: Compute the mean of the squares.
\[ E[s^2] = \frac{1}{3}\left(\frac{1}{9}+\frac{4}{9}+1\right) = \frac{1}{3}\times\frac{14}{9} = \frac{14}{27} \]

Step 6: Apply the variance formula.
\[ \text{Var}(s) = E[s^2] - (E[s])^2 = \frac{14}{27} - \frac{4}{9} = \frac{14}{27} - \frac{12}{27} = \frac{2}{27} \] \[ \boxed{\text{Variance of the equalized image} = \frac{2}{27}} \]