Joint, marginal & conditional distributions
A joint distribution gives P(X = x, Y = y) for every pair, usually as a table. Summing a row or column gives a marginal distribution — the name comes from writing these totals in the margins. Dividing a row by its total gives a conditional distribution, and X and Y are independent exactly when every joint entry equals the product of its margins.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Check the table sums to 1All joint probabilities together must total 1. This catches transcription errors instantly.
- Sum across for marginalsP(X = x) is the row total; P(Y = y) is the column total.
- Divide for conditionalsP(Y = y | X = x) = joint / row total. Each conditional distribution sums to 1 on its own.
- Test independence entry by entryEvery cell must equal the product of its margins. One failure means dependence.
Worked example
A joint table has P(0,0) = 0.2, P(0,1) = 0.3, P(1,0) = 0.1, P(1,1) = 0.4. Find the marginals and test independence.
- Total: 0.2 + 0.3 + 0.1 + 0.4 = 1.0 ✓.
- Marginal of X: P(X=0) = 0.5, P(X=1) = 0.5. Marginal of Y: P(Y=0) = 0.3, P(Y=1) = 0.7.
- Independence would require P(0,0) = 0.5 × 0.3 = 0.15.
- But the table gives 0.2.
Answer. Marginals X: (0.5, 0.5), Y: (0.3, 0.7). Not independent, since 0.2 ≠ 0.15.
Where marks get dropped
These are the specific errors that cost credit on joint, marginal & conditional distributions questions — QED's rubric penalises each of them separately.
- Reading a marginal as a conditional. The row total is P(X = x); the conditional needs dividing by it.
- Testing independence on a single cell that happens to match. EVERY cell must factor.
- Forgetting that conditional distributions must each sum to 1 — a useful check on the division.
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Joint, marginal & conditional distributions — frequently asked questions
Can I recover the joint from the marginals?
Only if the variables are independent, when the joint is the product. In general the marginals lose the dependence structure entirely.
What is covariance?
Cov(X,Y) = E[XY] − E[X]E[Y], measuring linear association. Independence implies zero covariance, but zero covariance does not imply independence.
How do I compute E[XY] from a table?
Sum xy·P(x,y) over every cell. Comparing it with E[X]E[Y] gives the covariance directly.
The rest of Probability
Events, conditional probability, random variables. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Sample spaces & events
- 2Conditional probability & independence
- 3Bayes’ theorem
- 4Random variables & expected value
- 5Variance & standard deviation
- 6Binomial & uniform distributions
- 7Law of total probability & tree diagrams
- 8Geometric & Poisson distributions
- 9Joint, marginal & conditional distributions
- 10Markov & Chebyshev bounds
- 11Equally likely outcomes & counting
- 12Linearity of expectation
- 13Indicator random variables
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