23-Ind-B1 Reliability and Maintainability · December 2019

Question 4 of 9: Joint Probability Function — Correlation and Independence

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2019 — 17-Ind-B1 Applied Probability & Statistics. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides), statistical tables supplied. Format: three sections — Section A: do 2 of 4 (20 marks); Section B: do 2 of 3 (40 marks); Section C: do 1 of 2 (20 marks) — a 9-question, 80-mark paper as actually structured by the section table and each question's own printed marks (the front page's own summary line states “Exam: 5 Questions. Total marks: 100”, which is internally inconsistent with 20+40+20=80 from its own section breakdown and every question's printed mark value). All nine questions across the three sections are solved below for completeness.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — probability density functions and moments (ch. 4), point/interval estimation (ch. 8), two-sample hypothesis testing and the sign test (ch. 9–10, 16), simple linear regression (ch. 11), single-factor ANOVA and multiple comparisons (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — Bartlett's test, Tukey's HSD, $2^k$ factorial designs (ch. 3, 6).

Question 4 (Section A.4): Joint Probability Function — Correlation and Independence (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Joint pmf $p(x,y)$:

$x\backslash y$	0	1	2	$p_X(x)$
0	0.05	0.10	0.20	0.35
1	0.05	0.15	0.05	0.25
2	0.25	0.10	0.05	0.40
$p_Y(y)$	0.35	0.35	0.30	1.00

Find. (a) The correlation coefficient $\rho_{xy}$; (b) whether $X$ and $Y$ are independent.

Approach. Compute the marginals directly from the table, then $E[X],E[Y],E[XY]$ (the last via the full nine-term sum over the joint table), then $\mathrm{Cov}(X,Y)$ and $\rho_{xy}=\mathrm{Cov}/\sqrt{\mathrm{Var}(X)\mathrm{Var}(Y)}$; check independence by comparing every joint cell against the product of its marginals.

Marginals and their moments. $p_X=(0.35,0.25,0.40)$ for $x=0,1,2$; $p_Y=(0.35,0.35,0.30)$ for $y=0,1,2$ (each sums to 1, and the table's own row/column sums confirm them). $$E[X]=0(.35)+1(.25)+2(.40)=1.05,\qquad E[X^2]=0(.35)+1(.25)+4(.40)=1.85,\qquad \mathrm{Var}(X)=1.85-1.05^2=0.7475$$ $$E[Y]=0(.35)+1(.35)+2(.30)=0.95,\qquad E[Y^2]=0(.35)+1(.35)+4(.30)=1.55,\qquad \mathrm{Var}(Y)=1.55-0.95^2=0.6475$$
Joint moment and covariance. Summing $xy\cdot p(x,y)$ over all nine cells (only cells with $x,y\ne0$ contribute): $$E[XY]=1(1)(.15)+1(2)(.05)+2(1)(.10)+2(2)(.05)=0.15+0.10+0.20+0.20=0.65$$ $$\mathrm{Cov}(X,Y)=E[XY]-E[X]E[Y]=0.65-(1.05)(0.95)=\boxed{-0.3475}$$
(a) Correlation coefficient. $$\rho_{xy}=\frac{\mathrm{Cov}(X,Y)}{\sqrt{\mathrm{Var}(X)\,\mathrm{Var}(Y)}}=\frac{-0.3475}{\sqrt{(0.7475)(0.6475)}}=\boxed{-0.499}$$ A moderate negative linear association: larger $X$ tends to pair with smaller $Y$ and vice versa.
(b) Independence check (essay + verification). $X,Y$ are independent iff $p(x,y)=p_X(x)\,p_Y(y)$ for every cell. Testing just one cell is enough to disprove it: $p(0,0)=0.05$ but $p_X(0)p_Y(0)=(0.35)(0.35)=0.1225\ne0.05$. $X$ and $Y$ are NOT independent. This is also guaranteed by part (a) alone: independent random variables always have $\rho=0$, and here $\rho_{xy}=-0.499\ne0$, so a nonzero correlation is already sufficient (though not the only possible route) to rule out independence.

Final Results
Quantity	Value
$E[X],\ \mathrm{Var}(X)$	$1.05,\ 0.7475$
$E[Y],\ \mathrm{Var}(Y)$	$0.95,\ 0.6475$
$\mathrm{Cov}(X,Y)$	$-0.3475$
$\rho_{xy}$	$-0.499$
Independent?	No ($\rho\ne0$; joint table $\ne$ product of marginals)

Topic: Covariance and correlation from a discrete joint probability table; the independence test.

Key relations: $\mathrm{Cov}(X,Y)=E[XY]-E[X]E[Y]$; $\rho_{xy}=\mathrm{Cov}(X,Y)/\sqrt{\mathrm{Var}(X)\mathrm{Var}(Y)}$; independence requires $p(x,y)=p_X(x)p_Y(y)$ for every $(x,y)$.

Why this works: Independence is a much stronger condition than zero correlation ($\rho=0$ does not imply independence in general), but the converse always holds — independence forces $\rho=0$, so any nonzero $\rho$ is immediate proof of dependence.

Common pitfall: Checking only the marginals sum to 1 and stopping, without comparing individual joint cells against the marginal product — marginals alone never determine the joint distribution.

Source: Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th ed., §5.2–5.3 (joint probability distributions, covariance and correlation).

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