Correlation and regression are two sides of the same coin — correlation tells you how strongly two variables move together, while regression tells you by how much one variable changes when the other changes by one unit.
Here is the core distinction you must lock in before reading further:
Correlation is symmetric. The correlation between X and Y is identical to the correlation between Y and X. It is a pure number between -1 and +1 with no units.
Regression is directional. The regression of Y on X (predicting Y using X) gives a different equation from the regression of X on Y (predicting X using Y). These are two distinct lines — they only coincide when |r| = 1 (perfect correlation).
Think of it this way: imagine you are tracking daily temperature (X) and ice-cream sales (Y). Correlation answers "do they move together?" Regression answers "if temperature rises by 1 degree, by how many units do sales rise?" — and that answer depends on which variable you treat as the predictor.
Why does this matter for SSC CGL? Because questions routinely exploit the confusion between the two regression coefficients b_yx and b_xy. Candidates who do not know which coefficient belongs to which line consistently mark the wrong answer. The exam does not reward vague understanding — it rewards the ability to slot numbers into the right formula in under 90 seconds.
One useful analogy: correlation is like the R-squared of a relationship expressed as a simpler number, while regression is the actual slope of the line you would draw on a scatter diagram. The scatter diagram itself is the visual entry point — a tight cluster along a diagonal means high correlation; a wide cloud means low correlation.
The most tested formula:
Where:
Cov(X, Y) = \frac\{1\}{n} \Sigma (x - \bar{x})(y - \bar{y})\sigma_X = \sqrt{\text{Var}(X)}, \sigma_Y = \sqrt{\text{Var}(Y)}The raw-score (computational) version — the one you will actually use in exam numericals:
Key properties you must memorize:
$-1 \leq r \leq +1$ always.r = +1: perfect positive linear relationship.r = -1: perfect negative linear relationship.r = 0: no linear correlation (variables may still be non-linearly related).r is unit-free and unaffected by change of origin or scale.Used when data is in ranks or the distribution is not normal. Formula:
Where d_i is the difference between the ranks of the i-th observation on X and Y, and n is the number of pairs.
When there are tied ranks, the average rank is assigned and a correction factor is added to \Sigma d^2. SSC CGL rarely tests the tied-ranks correction, but know that it exists.
Regression of Y on X (used to predict Y given X):
Where the regression coefficient:
Regression of X on Y (used to predict X given Y):
Where the regression coefficient:
This is the single most important identity in this chapter for SSC CGL:
The sign of r equals the common sign of both regression coefficients (if both are positive, r is positive; if both are negative, r is negative). You will never see one regression coefficient positive and the other negative — that would be a contradiction.
Also note: r, b_yx, and b_xy always carry the same sign.
Both regression lines always pass through (\bar{x}, \bar{y}). This is a theorem with a direct exam application: if you are given the means and one regression equation, you can find an unknown coefficient by substituting the means.
The regression line is derived by minimizing the sum of squared residuals — that is, \Sigma(y_i - \hat{y}_i)^2 is minimized. This gives the "best-fit" line. You do not need to derive this for SSC CGL, but you should know that the formulas for b_{yx} and the intercept a come from setting up and solving the normal equations:
From these: b = \frac{n\Sigma xy - \Sigma x \cdot \Sigma y}{n\Sigma x^2 - (\Sigma x)^2} and a = \bar{y} - b\bar{x}.
r^2 \leq 1 implies b_{yx} \cdot b_{xy} \leq 1. So the product of the two regression coefficients can never exceed 1.b_{yx} > 1, then b_{xy} < 1 (and vice versa). Both cannot simultaneously exceed 1.This bound is a quick sanity check when options look suspicious.
When both regression lines are given in the form y = a + bx and x = c + dy, extract b_yx = b (coefficient of x in the y-on-x line) and b_xy = d (coefficient of y in the x-on-y line), then compute r = √(b × d).
Worked example: y = 2x + 3 and x = 0.4y + 1. Here b_yx = 2, b_xy = 0.4. So r = √(2 × 0.4) = √0.8 ≈ 0.894.
Standard method (computing covariance, both SDs): 4-5 steps. This shortcut: 2 steps — multiply, take square root. Time saved: roughly 60 seconds versus 20 seconds.
Both regression lines pass through (\bar{x}, \bar{y}). If you are given the means and a regression equation with one unknown coefficient, simply plug in the means and solve.
Worked example: \bar{x} = 4, \bar{y} = 6, equation Y = 2 + bX. Substitute: 6 = 2 + b(4) → b = 1. Done in 10 seconds flat versus setting up the full normal equations (45+ seconds).
Given Var(X), Var(Y), and Cov(X,Y): (1) take square roots of variances to get \sigma_X and \sigma_Y, (2) multiply them, (3) divide Cov by that product.
Worked example: Var(X) = 9, Var(Y) = 16, Cov = 6. Step 1: \sigma_X = 3, \sigma_Y = 4. Step 2: 3 × 4 = 12. Step 3: 6 / 12 = 0.5. Three arithmetic operations, no formula hunting needed. Saves 2 formula-lookup steps compared to writing out the full Pearson formula each time.
In any valid data set, b_yx × b_xy ≤ 1. Scan the answer options for r: if an option implies r > 1 or r < -1, eliminate immediately without calculation.
Also: if one regression coefficient exceeds 1 (say b_yx = 2), the other must be less than 0.5 (since 2 × b_xy ≤ 1 → b_xy ≤ 0.5). Use this to cross-check answer choices and eliminate in 5 seconds versus computing from scratch (30-40 seconds).
Rule: in the line "Y = a + bX", the dependent variable is Y — so this is the regression of Y on X, meaning b_yx = b. In the line "X = c + dY", the dependent variable is X — so this is the regression of X on Y, meaning b_xy = d.
The error candidates make: treating the coefficient in the second line as b_yx. Keeping the dependent variable = left-hand side as your anchor, you will never confuse the two coefficients. This single pattern stops roughly 30% of wrong answers on this topic.
When a correlation or regression question appears in the exam hall, run this decision tree:
Step 1 — What type of question is it?
b_yx and b_xy → use r = √(b_yx × b_xy).r = Cov / (σ_X × σ_Y).b_yx or r.Step 2 — Sign check. Both b_yx and b_xy positive → r is positive. Both negative → r is negative.
Step 3 — Bounds check. Is |r| ≤ 1? Is b_yx × b_xy ≤ 1? If your answer violates these, recheck your coefficient identification.
Step 4 — Eliminate impossible options. Any option with |r| > 1 is wrong. Any option where both regression coefficients are given with opposite signs is wrong.
If you spend more than 90 seconds on a straightforward numericals-type question here, you have misidentified which formula applies — go back to Step 1.
Why this question: Tests the foundational property that both regression lines pass through the mean point — this property appears in multiple forms across SSC CGL papers.
Solving path: Both regression lines pass through (\bar{x}, \bar{y}) = (4, 6). The line is Y = 2 + bX. Substitute directly: 6 = 2 + b(4) → 4 = 4b → b = 1. No formula needed — pure substitution.
Why this question: The range of r is a definitional question that appears almost every year. Get this right in 5 seconds.
Solving path: By definition of Pearson's coefficient, r is bounded: -1 ≤ r ≤ +1. Option B. Eliminate A (range -2 to +2 is wrong), C (unbounded is wrong), D (only non-negative is wrong — negative correlation exists).
Why this question: Classic two-regression-line problem. Tests whether you correctly identify b_yx versus b_xy and apply the product-root rule.
Solving path: Line 1 (y on x): y = 2x + 3, so b_yx = 2. Line 2 (x on y): x = 0.4y + 1, so b_xy = 0.4. Then r = √(2 × 0.4) = √0.8. Compute: √0.8 = √(4/5) = 2/√5 ≈ 2/2.236 ≈ 0.894. Both coefficients positive → r is positive. Answer: 0.894.
Why this question: Direct application of the covariance formula for r. Tests whether you remember to take square roots of variances before dividing.
Solving path: σ_X = √9 = 3, σ_Y = √16 = 4. Then r = Cov(X,Y) / (σ_X × σ_Y) = 6 / (3 × 4) = 6/12 = 0.5. Answer: 0.5.
Why this question: Tests the least-squares formula for b_yx using deviation scores. Most direct regression coefficient calculation type.
Solving path: b_{yx} = \Sigma(x - \bar{x})(y - \bar{y}) / \Sigma(x - \bar{x})^2 = 60 / 30 = 2. Answer: 2. This is a one-step division — if you spend more than 20 seconds here, you are overthinking it.
Why this question: Tests product-root rule with non-trivial numbers — √0.48 requires quick approximation.
Solving path: r = √(b_yx × b_xy) = √(0.8 × 0.6) = √0.48. Now √0.49 = 0.7, so √0.48 is just below 0.7 — approximately 0.6928, i.e., 0.69. Both coefficients positive → r is positive. Answer: 0.69 (approximately).
Why this question: Another two-regression-line problem — tests the same product-root rule but with smaller coefficients, giving a less obvious square root.
Solving path: Line 1 (y on x): y = 0.4x + 3, so b_yx = 0.4. Line 2 (x on y): x = 0.3y + 2, so b_xy = 0.3. Then r = √(0.4 × 0.3) = √0.12. Compute: √0.12 = √(12/100) = (2√3)/10 ≈ (2 × 1.732)/10 ≈ 3.464/10 ≈ 0.346. Answer: 0.346.
Swapping b_yx and b_xy. The regression coefficient of Y on X comes from the equation where Y is on the left-hand side. The equation "X = c + dY" gives b_xy = d, not b_yx. This swap causes wrong answers on at least one in three regression PYQs.
Forgetting to take square roots of variances. The Pearson formula uses σ_X and σ_Y (standard deviations), not Var(X) and Var(Y). Plugging in variances directly gives a completely wrong answer that still "looks reasonable" — the most dangerous type of error.
Treating r as having units or scale-dependence. Candidates sometimes think changing the unit of measurement (e.g., from kilograms to grams) changes r. It does not. r is dimensionless and invariant under linear transformations.
Assuming r = 0 means the variables are independent. Zero Pearson correlation means no linear relationship. A perfect U-shaped (quadratic) relationship would give r = 0 but the variables are clearly dependent. SSC CGL occasionally tests this conceptually.
Taking r as the geometric mean of the two regression coefficients rather than the square root of their product. These are the same thing mathematically, but candidates sometimes compute (b_yx + b_xy)/2 (the arithmetic mean) instead — a completely different value.
Sign error on r. After computing √(b_yx × b_xy), always assign the sign based on the common sign of the regression coefficients — not based on which value is larger. If both b's are negative (e.g., b_yx = -0.8, b_xy = -0.6), then r is negative, even though the square root is always non-negative by convention.