Question 4 of 8: Continuous Probability Density Function
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, December 2015 — 04-BS-2 Probability and Statistics (2 hours, closed book except one hand-written information sheet and an approved calculator). Statistical tables of the Normal, t, chi-square and F distributions are supplied with the exam. The rubric states "Any 5 questions constitute a complete paper"; every question is answered.
Reference texts: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed. (Pearson) — used throughout for normal/binomial/Poisson/hypergeometric probability, confidence intervals, hypothesis tests, and simple linear regression.
Check — source defect in Question 4. As literally printed, K/y cannot integrate
to a finite value over (1, ∞), so the density as given is not valid. This solution
reconstructs the intended density as f(y) = 1/y² for y > 1 — the unique
minimal correction (recovering one dropped exponent) that makes the function integrate to 1 on the stated
domain, and it is the standard "infinite mean" example from this course's own textbook (Walpole). This choice is disclosed here rather than presented as certain.
Question 4: Continuous Probability Density Function (20 marks)
Given. As reconstructed (see the check callout above), $f(y)=K/y^2$ for $y>1$, $0$ otherwise.
Find. (a) K; (b) E(Y); (c) Var(Y); (d) F(y).
Approach. Normalize the density to find K, then apply the definitions of expectation, variance, and the CDF directly by integration.
(a) Find K. Normalization requires $\int_1^\infty f(y)\,dy=1$:
$$\int_1^\infty \dfrac{K}{y^2}\,dy = K\Big[-\dfrac1y\Big]_1^\infty = K(0-(-1))=K=1.$$
So $\boxed{K=1}$, i.e. $f(y)=1/y^2$ for $y>1$.
Fig. 4 — f(y)=1/y² for y>1 (open circle at y=1, where f(1)=1).
(b) Find E(Y).
$$E(Y)=\int_1^\infty y\cdot\dfrac{1}{y^2}\,dy=\int_1^\infty \dfrac1y\,dy=\big[\ln y\big]_1^\infty.$$
Since $\ln y\to\infty$ as $y\to\infty$, this integral diverges: $\boxed{E(Y)=\infty}$ (the mean does not exist). This is not an arithmetic slip — $f(y)=1/y^2$ is a genuine, valid density (it integrates to exactly 1, part (a)), but its right tail is heavy enough that the mean is infinite. This is the classic textbook example used to show that "integrates to 1" does not guarantee a finite mean.
(c) Find Var(Y). Variance requires $E(Y^2)-[E(Y)]^2$, and since $E(Y)$ itself is already infinite (part b), $\mathrm{Var}(Y)$ is undefined (equivalently, $\boxed{\mathrm{Var}(Y)=\infty}$, since $E(Y^2)=\int_1^\infty y^2\cdot y^{-2}\,dy=\int_1^\infty 1\,dy$ diverges too, even more strongly).
(d) Find F(y). For $y>1$,
$$F(y)=\int_1^y \dfrac{1}{t^2}\,dt=\Big[-\dfrac1t\Big]_1^y=1-\dfrac1y.$$
$$F(y)=\begin{cases}0 & y\le1\\[4pt] 1-\dfrac1y & y>1\end{cases}$$
Unlike the mean, $F(y)$ is perfectly well-behaved: $F(1)=0$, $F(y)\to1$ as $y\to\infty$, confirming $f$ really is a valid density despite its undefined moments.
Fig. 5 — F(y)=1−1/y for y>1, rising smoothly from 0 to 1.
