Exercises — Continuous random variables — PDF, CDF, percentiles

A PDF needs two things: $f(x)\ge 0$ everywhere, and total area $=1$.

(a) NO. On $[0,1]$, $-x$ is negative (e.g. at $x=0.5$, $f=-0.5<0$). A density can never be negative — density is "how thickly probability is packed," and you can't pack a negative amount. Fails condition 1.

(b) NO. It is non-negative, good. But the area is a rectangle of height $3$, width $1$, so area $=3\ne 1$. Too much total probability. Fails condition 2.

(c) YES. Non-negative ✓. Area = rectangle height $2$, width $0.5$, so $2\times 0.5=1$ ✓. Note the height is $2>1$ — that is allowed, because only the area must equal 1.

$F(-\infty )=0$ (no area accumulated yet from the far left), $F(+\infty )=1$ (all the area collected), and $F$ is always non-decreasing — since $f\ge 0$, adding more of the number line can only add area, never remove it.

$P(X=3)=0$. A single point is an interval of width zero, and ${\int }_3^{3}f=0$. The value $f(3)$ is a density, not a probability — do not confuse the two.

(a) $3x^2\ge 0$ on $[0,1]$ ✓. Area: ${\int }_0^{1}3x^{2}dx=[x^3{]}_0^1=1$ ✓. Why: those are exactly the two PDF conditions.

(b) Accumulate area from the left edge $x=0$: for $0\le x\le 1$, $F(x)={\int }_0^{x}3t^{2}dt=[t^3{]}_0^x=x^3.$ Below the support nothing has accumulated; above it everything has: $F(x)=\{\begin{array}{ll}0,& x<0\\ x^3,& 0\le x\le 1\\ 1,& x>1.\end{array}$ Why the three pieces: the parent note's "restrict to the support" rule — $F$ must be $0$ before the support and $1$ after.

(c) Use $P(a\le X\le b)=F(b)-F(a)$: $P(0.5\le X\le 1)=F(1)-F(0.5)=1-0.125=\mathrm{0.875.}$

The PDF is the rate of accumulation of area, so by the Fundamental Theorem of Calculus, $f=F^{\prime }$ (see Fundamental Theorem of Calculus): $f(x)=\frac{d}{dx}\left(1-e^{-3x}\right)=3e^{-3x},x\ge 0,$ and $f(x)=0$ for $x<0$. This is the Exponential distribution with rate $\lambda =3$.

Median $m$ solves $F(m)=0.5$: $1-e^{-3m}=0.5\Rightarrow e^{-3m}=0.5\Rightarrow -3m=\ln 0.5\Rightarrow m=\frac{\ln 2}{3}\approx \mathrm{0.231.}$ Why: the median is the 50th percentile; we invert the CDF, never the PDF.

(a) Area must be $1$. The support has width $6-2=4$, so a rectangle of height $c$ over it has area $4c$. Set $4c=1\Rightarrow c=\frac{1}{4}$.

(b) For $2\le x\le 6$, accumulate area from the left edge $x=2$: $F(x)={\int }_2^x\frac{1}{4}dt=\frac{1}{4}(x-2).$ Full definition: $F(x)=\{\begin{array}{ll}0,& x<2\\ \frac{x-2}{4},& 2\le x\le 6\\ 1,& x>6.\end{array}$

Look at Figure s01. The accent-red curve is the CDF: a straight ramp rising from height $0$ at $x=2$ to height $1$ at $x=6$. The dashed guide-lines drop from the ramp down to the $x$-axis at the three quartile values $3,4,5$ — each is where the ramp crosses height $0.25$, $0.5$, $0.75$ respectively. A constant density gives a constant slope, hence a straight ramp.

(c) Set $F(x_p)=p$ and solve the linear equation $\frac{x-2}{4}=p\Rightarrow x=2+4p$:

• $Q_1$: $p=0.25\Rightarrow x=2+1=3$.
• Median: $p=0.5\Rightarrow x=2+2=4$.
• $Q_3$: $p=0.75\Rightarrow x=2+3=5$. They are evenly spaced — the signature of a uniform distribution.

(a) The proposed density is NOT valid — the exercise is deliberately baited. Split at the corner $x=0$ since $|x|=-x$ for $x<0$ and $|x|=x$ for $x\ge 0$: ${\int }_{-1}^1\frac{1}{2}|x|dx=2{\int }_0^1\frac{1}{2}tdt=2\cdot \frac{1}{4}=\frac{1}{2}\ne 1.$ So $f(x)=\frac{1}{2}|x|$ has total area $\frac{1}{2}$ — it fails PDF condition 2. The task is to repair it, not to pretend it was correct. Write $f(x)=k|x|$ and choose $k$ so the area is $1$: ${\int }_{-1}^{1}k|x|dx=2k{\int }_0^{1}tdt=2k\cdot \frac{1}{2}=k\stackrel{!}{=}1\Rightarrow k=1.$ So the valid density is $f(x)=|x|$ (height $1$ at the ends). We use this corrected $f(x)=|x|$ for parts (b) and (c). (Symmetry lets us double the right half in every step below.)

(b) Look at Figure s02. The black V is the corrected PDF $f(x)=|x|$. The accent-red dot marks the corner at $x=0$ where the formula switches from $-t$ (left) to $+t$ (right); the shaded left wedge has area $\frac{1}{2}$, so by the time we reach $x=0$ we have already banked $F(0)=\frac{1}{2}$. The right wedge (lighter shading) supplies the remaining $\frac{1}{2}$.

For $-1\le x\le 0$, accumulate from the left edge $-1$, where $|t|=-t$: $F(x)={\int }_{-1}^x(-t)dt={\left[-\frac{t^2}{2}\right]}_{-1}^x=-\frac{x^2}{2}+\frac{1}{2}=\frac{1-x^2}{2}.$ Sanity: $F(-1)=0$ ✓, $F(0)=\frac{1}{2}$ ✓ (half the area sits on the left half). For $0\le x\le 1$, continue from the $F(0)=\frac{1}{2}$ already banked, now with $|t|=t$: $F(x)=\frac{1}{2}+{\int }_0^{x}tdt=\frac{1}{2}+\frac{x^2}{2}=\frac{1+x^2}{2}.$ Sanity: $F(1)=\frac{2}{2}=1$ ✓.

(c) By symmetry the middle band is $F(0.5)-F(-0.5)$: $F(0.5)=\frac{1+0.25}{2}=\frac{1.25}{2}=0.625,F(-0.5)=\frac{1-0.25}{2}=\frac{0.75}{2}=0.375,$ $P(-0.5\le X\le 0.5)=0.625-0.375=\mathrm{0.25.}$

(a) Accumulate from the left edge $x=1$: $F(x)={\int }_1^x\frac{2}{t^3}dt={\left[-\frac{1}{t^2}\right]}_1^x=1-\frac{1}{x^2},x\ge 1,$ and $0$ for $x<1$. Why accumulating the PDF gives the CDF: by the definition $F(x)=P(X\le x)={\int }_{-\infty }^{x}f(t)dt$ — the CDF at $x$ is the total area under $f$ up to $x$. Because the support starts at $1$, there is no area below $1$, so the accumulation begins at the left edge $t=1$; integrating $2/t^3$ from $1$ to $x$ literally sums up all the density collected so far, which is exactly the meaning of "probability of being at most $x$."

(b) Median: $F(m)=0.5\Rightarrow 1-\frac{1}{m^2}=0.5\Rightarrow \frac{1}{m^2}=0.5\Rightarrow m^2=2\Rightarrow m=\sqrt{2}\approx 1.414$.

(c) 90th percentile: $1-\frac{1}{x^2}=0.9\Rightarrow \frac{1}{x^2}=0.1\Rightarrow x^2=10\Rightarrow x=\sqrt{10}\approx 3.162$.

(d) See Expectation and Variance of continuous RVs: $E[X]={\int }_1^{\infty }x\cdot \frac{2}{x^3}dx={\int }_1^{\infty }\frac{2}{x^2}dx={\left[-\frac{2}{x}\right]}_1^{\infty }=0-(-2)=2.$ It exists and equals $2$. (Heavy tails can make the mean diverge, but here the $1/x^2$ integrand decays fast enough.)

(a) By definition the 90th percentile solves $\Phi (x_{0.9})=0.90$, so $x_{0.9}=1.2816$.

(b) The 10th percentile solves $\Phi (x_{0.1})=0.10$. Using symmetry, $\Phi (-1.2816)=1-\Phi (1.2816)=1-0.90=0.10$, so $x_{0.1}=-1.2816$. Why: the standard normal is symmetric about $0$, so the 10th and 90th percentiles are mirror images.

(c) $Q_3=x_{0.75}=0.6745$ and $Q_1=x_{0.25}$. By symmetry $x_{0.25}=-x_{0.75}=-0.6745$. So $Q_3-Q_1=0.6745-(-0.6745)=\mathrm{1.349.}$

(a) Why we invert $F$: the recipe says $X=F^{-1}(U)$, so we need the rule that, given a target cumulative probability $u$, returns the $x$ whose CDF equals $u$. That is precisely "solve $F(x)=u$ for $x$." Set $u=1-e^{-2x}$ and unwind step by step: $e^{-2x}=1-u\Rightarrow -2x=\ln (1-u)\Rightarrow x=-\frac{1}{2}\ln (1-u).$ So $F^{-1}(u)=-\frac{1}{2}\ln (1-u)$. Each manipulation just peels back one operation of $F$ (subtract from 1, take log, divide by $-2$) so that $x$ is expressed in terms of the desired probability $u$ — that is what "inverting the CDF" means.

(b) Why plug in $u=0.5$: the generator's output $U$ is the cumulative probability we hand to $F^{-1}$. Feeding $u=0.5$ asks "which $x$ has half the probability below it?" — i.e. the median: $X=-\frac{1}{2}\ln (1-0.5)=-\frac{1}{2}\ln (0.5)=\frac{\ln 2}{2}\approx \mathrm{0.3466.}$ As expected, the middle of $[0,1]$ maps to the median of $X$.

(c) This step needs $F$ to be strictly increasing on its support so that $F^{-1}$ exists as a genuine (single-valued, invertible) function. Here $f(x)=2e^{-2x}>0$ for all $x\ge 0$, so $F$ is strictly increasing on $[0,\infty )$ and is genuinely invertible — that is what licenses the equivalence $F^{-1}(U)\le x⟺U\le F(x)$. With that guaranteed: $P(X\le x)=P(F^{-1}(U)\le x)=P(U\le F(x))=F(x),$ where the last step uses that $U$ is uniform, so $P(U\le p)=p$. The output's CDF is exactly $F$, as desired.

Two unknowns $k,b$ need two equations.

Equation 1 (valid PDF — total area $=1$): ${\int }_0^{b}kxdx=\frac{k{b}^2}{2}=1\Rightarrow k=\frac{2}{b^2}.$

CDF: for $0\le x\le b$, accumulate from the left edge and substitute $k=2/b^2$: $F(x)={\int }_0^{x}ktdt=\frac{k{x}^2}{2}=\frac{2}{b^2}\cdot \frac{x^2}{2}=\frac{x^2}{b^2}.$ (Note: $F(x)=x^2/b^2$, not $x^2$ — the $b^2$ must not be dropped.)

Equation 2 (median at $2$): set $F(2)=0.5$: $\frac{2^2}{b^2}=\frac{4}{b^2}=0.5\Rightarrow b^2=8\Rightarrow b=2\sqrt{2}\approx \mathrm{2.828.}$

Then from Equation 1, $k=\frac{2}{b^2}=\frac{2}{8}=\frac{1}{4}.$

Answer: $k=\frac{1}{4}$, $b=2\sqrt{2}$, i.e. $f(x)=\frac{1}{4}x$ on $[0,2\sqrt{2}]$. Check: median where $\frac{x^2}{8}=0.5\Rightarrow x^2=4\Rightarrow x=2$ ✓; total area $\frac{1}{4}\cdot \frac{(2\sqrt{2}{)}^2}{2}=\frac{1}{4}\cdot \frac{8}{2}=\frac{1}{4}\cdot 4=1$ ✓.

(a) Two rectangles: $0.5\times 1+0.25\times 2=0.5+0.5=1$ ✓.

(b) Look at Figure s03. The left panel is the PDF — two flat levels (height $0.5$ then $0.25$). The right panel is the accent-red CDF: it climbs as a steep ramp (slope $0.5$) on $[0,1]$, then bends to a gentler ramp (slope $0.25$) on $[1,3]$; the ramp's slope is exactly the local density. The red dot at $(1,0.5)$ marks the seam where $F$ has just reached $0.5$ — the median.

• $x<0$: $F=0$.
• $0\le x\le 1$: $F(x)={\int }_0^{x}0.5dt=0.5x$. At $x=1$, $F=0.5$.
• $1<x\le 3$: carry the banked $0.5$, then add the second rectangle: $F(x)=0.5+{\int }_1^{x}0.25dt=0.5+0.25(x-1)$. At $x=3$, $F=0.5+0.5=1$ ✓.
• $x>3$: $F=1$.

(c) We want $F(m)=0.5$. But $F$ reaches $0.5$ exactly at $x=1$ (end of the first ramp). So the median is $m=1$. (Half the area sits in $[0,1]$, half in $(1,3]$ — the corner is precisely the balance point.)