Corrected exercises: the continuous random variable and its density function (201-337)

The interval $[2, 5]$ straddles the two steps. From 2 to 3: $1\times 0.2=0.2$. From 3 to 5: $2\times 0.1=0.2$. So $P(2\leq X\leq 5)=0.4$. Pitfall: using a single height for the whole width of 3.

A rectangle of width $10-2=8$ and height $c$: $8c=1$, so $c=\frac{1}{8}=0.125$. Pitfall: dividing by the endpoint $10$ instead of the width $8$.

$c=\frac{1}{16}$ and the relevant width is $7-4=3$: $P(X\leq 7)=\frac{3}{16}=0.1875$. Pitfall: taking the width $7$, as if the support started at 0, which gives $\frac{7}{16}$.

The area under $f$ is a triangle with base 5 and height $f(5)=5k$: $\frac{1}{2}\times 5\times 5k=12.5k=1$, so $k=\frac{1}{12.5}=0.08$. Pitfall: forgetting the $\frac{1}{2}$ of the triangle, which gives $k=0.04$.

A triangle with base 3 and height $f(3)=0.24$: $P(X\leq 3)=\frac{1}{2}\times 3\times 0.24=0.36$. Pitfall: answering $f(3)=0.24$, the height instead of the area.

The density is symmetric about the midpoint of the support: $E(X)=\frac{3+15}{2}=9$.

Density $\frac{1}{20}$ on $[5, 25]$. We want $\frac{t-5}{20}=0.3$, so $t-5=6$ and $t=11$. Pitfall: writing $0.3\times 20=6$ and forgetting that the support starts at 5.

$P(X>3)=e^{-0.4\times 3}=e^{-1.2}\approx 0.3012$. Pitfall: computing $1-e^{-1.2}=0.6988$, which is $P(X\leq 3)$.

Mean 2, so $\lambda=\frac{1}{2}=0.5$. $P(1\leq X\leq 4)=P(X>1)-P(X>4)=e^{-0.5}-e^{-2}=0.606531-0.135335\approx 0.4712$. Pitfall: taking $\lambda=2$, the mean itself, which gives $e^{-2}-e^{-8}\approx 0.1350$.

$\lambda=\frac{1}{8}$. The median satisfies $P(X>m)=0.5$, that is $e^{-m/8}=0.5$, so $-\frac{m}{8}=\ln 0.5$ and $m=8\ln 2\approx 5.55$. Pitfall: answering with the mean, $8$; the median of an exponential is smaller than its mean.

$P(X\leq t)=0.75$ means $P(X>t)=0.25$, that is $e^{-0.3t}=0.25$. So $-0.3t=\ln 0.25$ and $t=\frac{\ln 4}{0.3}\approx 4.62$. Pitfall: writing $e^{-0.3t}=0.75$, which gives the quantile of order $0.25$, about $0.96$.

Two conditions, and not one more. NONNEGATIVITY: $f$ only takes the values $0.10$, $0.25$, $0.15$ and $0$, all positive or zero. TOTAL AREA EQUAL TO 1: the curve bounds three rectangles, all of width 2. The first has area $2\times 0.10=0.20$, the second $2\times 0.25=0.50$, the third $2\times 0.15=0.30$. The sum is $0.20+0.50+0.30=1.00$, so $f$ is a density. Note in passing that $0.20$, $0.50$ and $0.30$ are the probabilities that the outage falls in each of the three two-hour slices: the long outage, between 2 h and 4 h, is the most frequent. The pitfall here is to add up the HEIGHTS, $0.10+0.25+0.15=0.50$, and conclude that something is wrong. Heights never add up: they are probabilities PER HOUR, and only multiplying them by a width gives a probability.

The interval $[1, 3]$ straddles two steps. Split it: from 1 to 2, the height is $0.10$ and the width 1, so the area is $1\times 0.10=0.10$; from 2 to 3, the height is $0.25$ and the width 1, so the area is $1\times 0.25=0.25$. In total $P(1\leq X\leq 3)=0.10+0.25=0.35$. Roughly one outage in three lasts between one and three hours. The check is immediate: $0.35$ lies between 0 and 1, and it is smaller than $P(0\leq X\leq 4)=0.70$, which is consistent since $[1, 3]$ is contained in $[0, 4]$.

$P(X=2)=0$. The event “the outage lasts exactly two hours, to the billionth of a second” corresponds to a segment of ZERO width under the curve: its area is zero, whatever the height of the curve at that point. Going from $[1, 3]$ to $(1, 3)$ amounts to removing the two values 1 and 3, so two zero areas: $P(1<X<3)=0.35-0-0=0.35$, exactly as in b). This is the second surprise of the chapter: strict or non-strict, the result is the same for a continuous variable. In the discrete case this is false, and the gap $P(X\leq 3)-P(X<3)$ is precisely $P(X=3)$, which is not zero.

$P(X>4)$ is the area of the third rectangle alone: $2\times 0.15=0.30$. $P(X\leq 2)$ is the area of the first: $2\times 0.10=0.20$. Check with the complement: $P(X\leq 4)=0.20+0.50=0.70$ and $1-0.70=0.30$, which gives $P(X>4)$ back. And since $P(X=4)=0$, writing $P(X\geq 4)$ or $P(X>4)$ changes nothing.

Kolmogorov's three axioms, on which the whole theory rests. FIRST, nonnegativity: for every event $A$, $P(A)\geq 0$. SECOND, normalization: $P(\Omega)=1$, the whole sample space is certain. THIRD, additivity: if $A$ and $B$ are mutually exclusive, $P(A\cup B)=P(A)+P(B)$, and this extends to a sequence of pairwise mutually exclusive events. It is the SECOND that requires the total area to equal 1: the area under the density represents $P(\Omega)$, since the outage necessarily lasts some amount of time. The result of c) contradicts no axiom. An event of probability zero is not an impossible event: the outage does take a value, and each of those values has probability zero. That is the price of moving to the continuous case, and it is exactly what countable additivity allows, since the values of an interval do not form a sequence.

The graph of $f$ is the segment joining the origin to the point $(4, 4k)$. The area under this graph, between 0 and 4, is that of a right TRIANGLE with base 4 and height $4k$, namely $\frac{1}{2}\times 4\times 4k=8k$. The total-area condition requires $8k=1$, so $k=\frac{1}{8}=0.125$ and $f(x)=\frac{x}{8}$ on $[0, 4]$, zero elsewhere. No integration is needed: the density is linear, so the area is that of a triangle. The classic pitfall is to write $f(4)=1$ “so that it adds up to 1”: it is the AREA that equals 1, never the height at the end.

$f(2)=\frac{2}{8}=0.25$ and $f(4)=\frac{4}{8}=0.50$. On $[0, 4]$, $x$ is nonnegative so $\frac{x}{8}$ is too, and $f$ is zero elsewhere: nonnegativity holds everywhere. Both conditions are therefore met, and $f$ is a density. Be careful about what these numbers say: $f(4)=0.50$ is not a 50 percent probability of finding the flaw at the end of the bar, it is a probability PER METRE near 4 m. Over one centimetre around 4 m, the probability is about $0.50\times 0.01=0.005$.

$P(X\leq 2)$ is the area of the triangle with base 2 and height $f(2)=0.25$, namely $\frac{1}{2}\times 2\times 0.25=0.25$. $P(2\leq X\leq 4)$ is the area of the trapezoid of width 2 with parallel sides $f(2)=0.25$ and $f(4)=0.50$, namely $\frac{0.25+0.50}{2}\times 2=0.75$. These two numbers must add up to 1, since the two intervals are mutually exclusive and cover the whole support: $0.25+0.75=1.00$, and the check passes. The result reads clearly: three flaws out of four are in the right half of the bar, which is what the model was meant to say.

With the trapezoid: parallel sides $f(1)=0.125$ and $f(3)=0.375$, width 2, so $P(1\leq X\leq 3)=\frac{0.125+0.375}{2}\times 2=0.50$. As a difference of triangles: the area up to 3 is $\frac{1}{2}\times 3\times 0.375=0.5625$, the area up to 1 is $\frac{1}{2}\times 1\times 0.125=0.0625$, and $0.5625-0.0625=0.50$. The two methods agree, as they should: a trapezoid is a difference of two similar triangles. A striking result: the interval $[1, 3]$ covers only half the bar, and yet it carries exactly half the probability.

Yes, without the slightest difficulty. The uniform distribution on $[0, 0.25]$ has a constant density $f(x)=\frac{1}{0.25}=4$, four times greater than 1, and yet its area is $0.25\times 4=1$. What is capped at 1 is the AREA, never the height: a density is a probability per unit, and if the unit is small, the density is large. The same thing happens in physics, where a density in kilograms per cubic metre is not a mass. Here $f$ stays below 1 because the bar is long; if it were 1 metre long with the same shape, we would have $f(x)=2x$ and $f(1)=2$, with an area still equal to $\frac{1}{2}\times 1\times 2=1$. Confusing height with probability costs marks on every question of the chapter, and it is the error that then leads students to write $P(X=a)=f(a)$.

For $x$ between 0 and 4, the accumulated area is that of the right triangle with base $x$ and height $f(x)=\frac{x}{8}$, namely $F(x)=\frac{1}{2}\times x\times\frac{x}{8}=\frac{x^{2}}{16}$. No integration: a triangle is enough. We get $F(2)=\frac{4}{16}=0.25$ and $F(4)=\frac{16}{16}=1$, which is the mandatory check: the cumulative distribution function must reach 1 at the end of the support. Outside, $F(x)=0$ for $x<0$, since no area has been accumulated yet, and $F(x)=1$ for $x>4$, since all of it has. Compare with part c) of exercise 2: $F(2)=0.25$ indeed gives $P(X\leq 2)$ back.

$P(X\leq 1.5)=F(1.5)=\frac{2.25}{16}=0.140625$, about $0.1406$. The complement gives $P(X>1.5)=1-0.140625=0.859375$, about $0.8594$. Finally $P(0.5\leq X\leq 1.5)=F(1.5)-F(0.5)=0.140625-\frac{0.25}{16}=0.140625-0.015625=0.125$. Exactly one eighth. That is the whole point of $F$: a subtraction replaces cutting the region into trapezoids, and you can no longer pick the wrong figure. A check with the trapezoid, just for fun: parallel sides $f(0.5)=0.0625$ and $f(1.5)=0.1875$, width 1, area $\frac{0.0625+0.1875}{2}=0.125$. The two routes agree.

The median splits the area into two equal parts, so it solves $F(m)=0.5$, that is $\frac{m^{2}}{16}=0.5$, so $m^{2}=8$ and $m=\sqrt{8}=2\sqrt{2}\approx 2.83$. The negative root is rejected, since it lies outside the support $[0, 4]$. A common-sense check: the density increases, so the right half of the bar is more heavily loaded than the left, and the median must fall to the RIGHT of the midpoint 2, which it does.

Same method. First quartile: $\frac{q_{1}^{2}}{16}=0.25$, so $q_{1}^{2}=4$ and $q_{1}=2$. Third quartile: $\frac{q_{3}^{2}}{16}=0.75$, so $q_{3}^{2}=12$ and $q_{3}=2\sqrt{3}\approx 3.46$. The three numbers $2$, $2.83$ and $3.46$ cut the bar into four zones of equal probability, of widths $2$, $0.83$, $0.63$ and $0.54$: the zones get narrower towards the right, since the density is higher there. The interquartile range is $3.46-2=1.46$.

Three properties of $F$: it is INCREASING, since area keeps being added; it goes from 0 to 1, $F(-\infty)=0$ and $F(+\infty)=1$; and it satisfies $P(a\leq X\leq b)=F(b)-F(a)$. Add a fourth, the decisive one: here $F$ is CONTINUOUS, without any jump. In the discrete case, on the contrary, $F$ is a STEP function, and the height of the jump at $a$ is exactly $P(X=a)$: for a fair die, $F$ jumps by $\frac{1}{6}$ at each integer from 1 to 6. That is the complete explanation of $P(X=a)=0$: the probability of an isolated value is the height of the jump of $F$ at that value, and a continuous function does not jump. The two worlds part ways right there, and nowhere else.

The area under a constant density $c$ on $[0, 12]$ is that of a RECTANGLE of width 12 and height $c$, namely $12c$. The condition $12c=1$ gives $c=\frac{1}{12}\approx 0.0833$. The general formula follows: on $[a, b]$, the uniform density equals $\frac{1}{b-a}$. Note that this number has a unit, reciprocal minutes, and that is precisely what forbids reading it as a probability.

Everything is read as a ratio of lengths, since the height is the same everywhere. $P(X\leq 5)$ is the area of the rectangle of width 5, namely $5\times\frac{1}{12}=\frac{5}{12}\approx 0.4167$. $P(3\leq X\leq 7)$ is the area of the rectangle of width $7-3=4$, namely $\frac{4}{12}=\frac{1}{3}\approx 0.3333$. Check: the interval $[3, 7]$ takes up one third of $[0, 12]$, and the probability is indeed one third. For a uniform distribution, a probability is nothing more than the proportion of the length covered, and it is the only distribution in the chapter where this is true.

The density is symmetric about the midpoint, so $E(X)=\frac{0+12}{2}=6$ minutes: on average, the rider waits six minutes, half the interval between two buses. The variance of a uniform distribution on $[a, b]$ is $\frac{(b-a)^{2}}{12}$, here $\frac{144}{12}=12$, and the standard deviation is $\sqrt{12}=2\sqrt{3}\approx 3.46$ minutes. The plausibility check passes: a standard deviation of 3.46 minutes over an interval of 12 minutes means a large spread, which is what we expect from a distribution in which all waiting times are equally likely. Beware of the 12 in the denominator, which here has nothing to do with the 12 minutes: the formula would read $\frac{(b-a)^{2}}{12}$ even on an interval of 5 units.

Conditioning is read in areas, like everything else: $P(X\geq 9\mid X\geq 6)=\frac{P(9\leq X\leq 12)}{P(6\leq X\leq 12)}=\frac{3/12}{6/12}=\frac{3}{6}=0.50$. Compare with the probability of waiting at least 3 minutes from the moment of arrival, $P(X\geq 3)=\frac{9}{12}=0.75$. The two numbers DIFFER, and that is the whole point of the question: the waiting time already spent changes what follows. Here it improves things, since $0.50<0.75$: the probability of waiting three more minutes has gone down. In other words, the uniform model AGES: the longer you have waited, the closer the bus is. This matches intuition, and it is exactly what is about to break down in the next exercise.

The sentence confuses height and area. $\frac{1}{12}$ is the DENSITY, a probability per minute; the probability of waiting exactly 5 minutes is $P(X=5)=0$, because the corresponding segment has zero width. What $\frac{1}{12}$ measures is, for example, $P(4.5\leq X\leq 5.5)=1\times\frac{1}{12}\approx 0.0833$, the probability of a wait within a one-minute window around 5. On $[0, 0.5]$, the uniform density equals $\frac{1}{0.5}=2$, greater than 1 without any contradiction: the area is $0.5\times 2=1$. That would happen if the bus came every thirty seconds, and the density would double again if the interval were fifteen seconds. Writing $P(X=5)=\frac{1}{12}$ costs the whole question and, worse, leads to the same mistake with the normal distribution, where students then start reading probabilities off the height of the bell curve.