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?
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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.
Step 1: Establish the equal-probability property. Since the three stems in the histogram are drawn at the same height on the dashed reference line, the input image has exactly three grey levels, each with equal likelihood $p = \frac{1}{3}$.
Step 2: Build the equalized levels using the cumulative sum rule. Histogram equalization transforms each grey level to the running cumulative probability up to and including that level (on a normalized $[0,1]$ scale): $$ s_1 = \frac13,\quad s_2 = \frac13+\frac13 = \frac23,\quad s_3 = \frac13+\frac13+\frac13 = 1 $$ Because the transform is strictly increasing, the three levels map to three distinct output levels, each inheriting probability $\frac13$.
Step 3: Compute the mean directly. \[ \mu = \frac{s_1+s_2+s_3}{3} = \frac{\frac13+\frac23+1}{3} = \frac{2}{3} \]
Step 4: Compute variance using the sum-of-squared-deviations definition instead of E[s^2]. \[ \text{Var}(s) = \frac{1}{3}\Big[(s_1-\mu)^2 + (s_2-\mu)^2 + (s_3-\mu)^2\Big] \] Deviations: $s_1-\mu = -\frac13$ so squared $=\frac19$. $s_2-\mu = 0$ so squared $=0$. $s_3-\mu = \frac13$ so squared $=\frac19$.
Step 5: Sum and average the squared deviations. \[ \text{Var}(s) = \frac{1}{3}\left(\frac19 + 0 + \frac19\right) = \frac{1}{3}\times\frac{2}{9} = \frac{2}{27} \] This matches the value obtained by the $E[s^2]-\mu^2$ route, confirming the result. \[ \boxed{\text{Variance} = \frac{2}{27}} \]
