1. Which of the information (R or B or G) is better to use for naïve bays classi

Question

1. Which of the information (R or B or G) is better to use for naïve bays classification if we
only have to use one of them? i.e. which one classifies better plant vs non-plant. Explain
why.
2. Which one (R or B or G) is the worst one? Explain why.
3. I have an unclassified pixel (we don’t know if it belongs to plant or the background). But I
know its RGB information. R = 5, G = 8 and B = 9. Does this pixel belong to plant or nonplant?
Write the Bayes equation for this question.
4. What about a pixel with R = 30, G = 15, and B = 24? Write the Bayes rule and explant
your solution.

Explanation / Answer

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ANS:

1) Green is the best variable for naive bayes classification as there is the least amount of common area shared between plant and non-plant PDGs among all the channels.

Let's define P(plant| x < G < y) as probability that the segment is plant given the green color channel is between (x,y) and P(non-plant| x < G < y)  as probability that the segment is non-plant given the green color channel is between (x,y) . Then ,from the diagram we see that that for the green channel,we have the smallest segment (x,y) which is (10,12) on the channel axis where both P(plant| x<G <y) > 0 and P(non-plant| x<G<y) > 0 which will give more unambiguos classification.

2. Blue pixel is the worst one for classification the there is greater length of segment (14,20) for which PDG of both plant and non-plant are greater than zero (which increases the likelihood for ambiguos classification.

// For the next two question since we don't know whether the pixel is plant or non-plant , i assume the // prior probability as 0.5 each that is, P(plant)prior = P(Non-plant)prior = 0.5   

3. From the PDG, we can easily see that P(Non-plant| R <=5,G <= 8 and B <= 9) = 0 Hence, the pixel is a plant. Since, R, G and B channels are independent of each other, we write  

P(Plant| R=5, G=8, B=9) = P(R=5|Plant) *P(G=8|Plant )*P(B=9|Plant)*P(plant)prior/P(R=5,G=8,B=9)

  P(Non-Plant| R=5, G=8, B=9) =

P(R=5|Non-Plant)*P(G=8|Non-plant) * P(B=9|Non-Plant) *P(Non-plant)prior//P(R=5,G=8,B=9)

Now P(R= 5|plant) > 0 , P(G =8|plant) >0 and P(B=9|plant) >0 while

P(R=5|non-plant) = P(G=8|non-plant) = P(B=9|non-plant) = 0

Hence ,the pixel is plant

4. With R = 30, G = 15 and B= 24 ,from the PDG , we can infer that the segment belongs to the non- plant .

More Formally,we write

P(Plant| R=30, G=15, B=24) =

P(R=30|Plant) *P(G=15|Plant )*P(B=24|Plant)*P(plant)prior/P(R=30,G=15,B=24)

  P(Non-Plant| R=30, G=15, B=24) =   

P(R=30|Non-Plant)*P(G=15|Non-plant) * P(B=24|Non-Plant) *P(Non-plant)prior//P(R=30,G=15,B=24)

Now P(R= 30|non-plant) > 0 , P(G=15|non-plant) >0 and P(B=24|non-plant) >0 while

P(R=30|plant) = P(G=15|plant) = P(B=24|plant) = 0

Hence, the pixel is non-plant

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