23-Ind-B1 Reliability and Maintainability · December 2019

Question 1 of 9: Validity of a Piecewise Probability Function

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

Question 1 (Section A.1): Validity of a Piecewise Probability Function (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. $f(x)=\alpha x$ on $0\lt x\lt3$; $f(x)=3-2x$ on $3\le x\le5$; $f(x)=0$ elsewhere.

Find. (a) The value of $\alpha$ that makes $f$ a valid pdf, and whether one exists; (b) $E[X]$; (c) $\mathrm{Var}(X)$, both under the instruction to proceed “assuming it is a valid pdf.”

Approach. A valid pdf must satisfy two independent conditions everywhere on its support: $f(x)\ge0$ and $\int f(x)\,dx=1$. Check non-negativity on each piece first (it does not depend on $\alpha$ where $\alpha$ does not appear), then solve the normalizing condition for $\alpha$, then compute the moments as instructed.

1. (a) Check non-negativity on $[3,5]$. The second piece, $3-2x$, contains no $\alpha$ term at all — it is fully determined by $x$ alone. On its stated domain, $3-2x$ is strictly decreasing, taking the value $3-2(3)=-3$ at the left end and $3-2(5)=-7$ at the right end: $$f(x)=3-2x\lt0\quad\text{for every }x\in[3,5]$$ Since a valid probability density function requires $f(x)\ge0$ for all $x$, and this piece is negative throughout its entire domain regardless of what $\alpha$ is chosen (there is no $\alpha$ present to rescale or flip its sign), no value of $\alpha$ can make $f(x)$ a valid pdf. This is decisive on its own; the normalizing condition below cannot rescue it.
2. (a, continued) Solve the normalizing condition anyway, to see what “$\alpha$” the question is pointing at. Total probability must integrate to 1: $$\int_0^3 \alpha x\,dx+\int_3^5(3-2x)\,dx=1$$ $$\int_0^3\alpha x\,dx=\frac{9\alpha}{2},\qquad \int_3^5(3-2x)\,dx=\Big[3x-x^2\Big]_3^5=(15-25)-(9-9)=-10$$ $$\frac{9\alpha}{2}-10=1\ \Rightarrow\ \alpha=\boxed{\dfrac{22}{9}\approx2.444}$$ This $\alpha$ satisfies the total-probability-equals-1 condition only — it does nothing to fix the sign problem found in step 1, since $\alpha$ never multiplies the second piece.
3. (b) Mean, proceeding formally with $\alpha=22/9$ as instructed. $$E[X]=\int_0^3 x(\alpha x)\,dx+\int_3^5 x(3-2x)\,dx=\alpha\int_0^3x^2\,dx+\int_3^5(3x-2x^2)\,dx$$ $$=\frac{22}{9}(9)+\Big[\tfrac{3}{2}x^2-\tfrac{2}{3}x^3\Big]_3^5=22+\big(-\tfrac{275}{6}\big)-\big(-\tfrac{9}{2}\big)=22-\frac{124}{3}=\boxed{-\dfrac{58}{3}\approx-19.33}$$
4. (c) Variance, same formal basis. First $E[X^2]$: $$E[X^2]=\alpha\int_0^3x^3\,dx+\int_3^5(3x^2-2x^3)\,dx=\frac{22}{9}\Big(\frac{81}{4}\Big)+\Big[x^3-\tfrac{x^4}{2}\Big]_3^5=49.5+(-174)=\boxed{-124.5}$$ Then $$\mathrm{Var}(X)=E[X^2]-(E[X])^2=-124.5-\left(-\frac{58}{3}\right)^2=-124.5-373.78=\boxed{-\dfrac{8969}{18}\approx-498.3}$$ A variance can never be negative for a genuine random variable ($\mathrm{Var}(X)=E[(X-\mu)^2]\ge0$ always, being an average of squares) — this impossible result is a second, independent confirmation of step 1's conclusion: $f(x)$ as printed is not a valid probability density function for any $\alpha$, and the numbers in (b)/(c) are formal bookkeeping only, not meaningful moments of a real distribution.