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Quantitative Aptitude · Correlation and Regression

Concurrent Deviation Method for Correlation (CA Foundation)

Updated 1 October 2026 · Fact-checked

The concurrent deviation method estimates correlation using only the direction of change (up or down) in paired values. Count pairs where both series move the same way (c), let m = n − 1, then r = ±√(±(2c − m) ÷ m). It is quick but rough.

Understand Concurrent Deviation Method

Most correlation methods use the actual size of the values. The concurrent deviation method ignores size. It asks only one thing: when X goes up or down, does Y move the same way?

A deviation here means the change from the previous value in the series. If a value is higher than the one before it, mark +. If lower, mark −. If equal, mark 0. A concurrent deviation is a pair of changes where X and Y have the same sign, both + or both −.

If most pairs move together, correlation is positive. If most move in opposite directions, it is negative. If it is about half and half, correlation is near zero. The formula turns this count into a number between −1 and +1.

The method is fast and needs no squares or products of values. The cost is accuracy. It throws away the size of changes, so the answer is only a rough estimate. It suits long series where you want a quick feel for the relationship, such as prices that rise and fall together.

Key formulas to remember

Coefficient of concurrent deviation
r_c = ± √( ± (2c − m) ÷ m )
c = number of concurrent deviations (pairs with the same sign). m = number of pairs of deviations = n − 1, where n is the number of pairs of observations.
Number of pairs of deviations
m = n − 1
The first value has no previous value, so it gives no deviation. Never use m = n.
Sign rule
If (2c − m) > 0: r_c = + √((2c − m) ÷ m). If (2c − m) < 0: r_c = − √((m − 2c) ÷ m).
The sign outside the root and inside it is the same as the sign of (2c − m). If 2c = m, r_c = 0.
Direction of each change
+ if value rises, − if value falls, 0 if no change
A pair counts in c only when both signs are + or both are −. A pair with a 0 is not counted in c.

How to solve Concurrent Deviation Method questions

Use this method for any question that gives paired series and asks for the coefficient of concurrent deviation.

  1. 1Write down n, the number of pairs of observations, and find m = n − 1.
  2. 2For series X, compare each value with the one before it. Write + for a rise, − for a fall, 0 for no change. Do the same for Y.
  3. 3Compare the signs of X and Y in each row. Mark the row as concurrent if both are + or both are −.
  4. 4Count the concurrent rows. This is c.
  5. 5Compute 2c − m. Note whether it is positive, negative or zero.
  6. 6If 2c − m is positive, r = +√((2c − m) ÷ m). If negative, r = −√((m − 2c) ÷ m). If zero, r = 0.
  7. 7Simplify the fraction first, then take the square root and match it with the options.

Quickest way: Sign-pairing and option elimination

When to use it: Use in the MCQ paper when two short series are given and options differ in sign or size.

  1. Find m = n − 1 first. Many wrong options come from using n.
  2. Write only the signs for X and Y as two rows of + and − symbols. Do not write the differences.
  3. Count matching columns for c. Count mismatches too; c plus mismatches plus zero-pairs must equal m.
  4. Decide the sign of r straight away: more matches than half of m means positive, fewer means negative.
  5. Eliminate options with the wrong sign. Then compute √(|2c − m| ÷ m) only for the remaining options.
  6. If c = m, r = +1. If c = 0 and no zeros, r = −1. Use these without calculation.

Common mistakes in Concurrent Deviation Method

  • Using m = n instead of m = n − 1.

    Students count the number of observations and forget the first value has no previous value.

    Fix: Always count the signs you actually wrote. The number of signs in one series is m.

  • Taking the sign of the root wrongly when 2c − m is negative.

    Students put a minus outside the root but leave a negative number inside it.

    Fix: Use −√((m − 2c) ÷ m). The number inside the root must be positive, and the sign outside follows 2c − m.

  • Counting pairs with a 0 sign as concurrent.

    Students feel that no change in both series means they move together.

    Fix: Count only pairs where both signs are + or both are −. Pairs involving 0 are not counted in c, but they still stay in m.

  • Comparing each value with the first value or with the average.

    Students mix this method with deviations from the mean in Karl Pearson's method.

    Fix: Compare each value only with the one just before it.

  • Counting c as the number of + signs in X.

    Students count signs in one series instead of matching signs across both series.

    Fix: Put the two sign rows one under the other and count only the columns that match.

  • Treating the answer as an exact correlation.

    The result looks like Pearson's r, so students compare the two as equal.

    Fix: Remember this is a rough estimate based on direction only. It can differ from Pearson's r for the same data.

Worked examples

Example 1

For X: 10, 12, 15, 14, 18, 20 and Y: 5, 8, 9, 7, 12, 15, the coefficient of concurrent deviation is: (a) 0 (b) 0.50 (c) 1 (d) −1

Show the solution
  1. n = 6, so m = 5.
  2. Signs of X: 12>10 (+), 15>12 (+), 14<15 (−), 18>14 (+), 20>18 (+).
  3. Signs of Y: 8>5 (+), 9>8 (+), 7<9 (−), 12>7 (+), 15>12 (+).
  4. Every column matches, so c = 5.
  5. 2c − m = 10 − 5 = 5, which is positive.
  6. r = +√(5 ÷ 5) = +√1 = 1.

Answer: (c) 1

Example 2

For X: 2, 4, 3, 6, 5, 8, 7 and Y: 9, 7, 8, 4, 6, 7, 5, the coefficient of concurrent deviation is closest to: (a) 0.33 (b) −0.33 (c) 0.58 (d) −0.58

Show the solution
  1. n = 7, so m = 6.
  2. Signs of X: 4>2 (+), 3<4 (−), 6>3 (+), 5<6 (−), 8>5 (+), 7<8 (−).
  3. Signs of Y: 7<9 (−), 8>7 (+), 4<8 (−), 6>4 (+), 7>6 (+), 5<7 (−).
  4. Compare columns: 1 differs, 2 differs, 3 differs, 4 differs, 5 matches (+,+), 6 matches (−,−). So c = 2.
  5. 2c − m = 4 − 6 = −2, which is negative.
  6. r = −√((6 − 4) ÷ 6) = −√(1/3) = −0.577, which rounds to −0.58.

Answer: (d) −0.58

Example 3

From 9 pairs of observations, 6 pairs of deviations are concurrent. The coefficient of concurrent deviation is closest to: (a) 0.50 (b) 0.58 (c) 0.71 (d) 0.87

Show the solution
  1. n = 9, so m = n − 1 = 8.
  2. c = 6, so 2c − m = 12 − 8 = 4, which is positive.
  3. r = +√(4 ÷ 8) = √0.5 = 0.707.
  4. This rounds to 0.71. Option (b) 0.58 comes from wrongly using m = 9, and option (a) is 0.50, the value inside the root before taking the square root.

Answer: (c) 0.71

Exam tips

  • Check the question for n. If it gives the number of pairs of observations, use m = n − 1. If it already says 'number of pairs of deviations', use that number as m.
  • Decide the sign of r first by comparing c with m ÷ 2. This often removes two options at once.
  • Write only + and − signs, never full differences. This saves time and avoids arithmetic slips.
  • Use a quick total check: matches plus mismatches plus zero-pairs must equal m. If not, you missed a sign.
  • Questions are often direct: c and n are given and you only apply the formula. Do these first and quickly.

Practice questions from Correlation and Regression

Concurrent Deviation Method: frequently asked questions

What is the formula for the coefficient of concurrent deviation?

r_c = ± √(±(2c − m) ÷ m). Here c is the number of pairs of deviations with the same sign and m = n − 1. The sign outside and inside the root follows the sign of (2c − m).

Why do we use n − 1 and not n?

Each deviation is a change from the previous value. The first observation has no previous value, so n observations give only n − 1 changes.

What if there is no change between two values?

Mark it 0. A pair with a 0 sign is not counted as concurrent, but it is still part of m.

Is the concurrent deviation result the same as Pearson's r?

No. It uses only the direction of change, not the size. It gives a rough estimate and can differ from Pearson's r for the same data.