Corrected exercises: probability density functions (MATH 141)

a) Two conditions, checked one at a time. Sign: $3x^2 \ge 0$ on $[0, 1]$ and $f = 0$ elsewhere, so $f \ge 0$ everywhere. Area: $\int_{-\infty}^{\infty} f(x)\,dx = \int_0^1 3x^2\,dx = \left[x^3\right]_0^1 = 1$. Both hold, so $f$ is a density, and the value $f(1) = 3$ plays no role in either condition. Then $P(X \le \frac12) = \int_0^{1/2} 3x^2\,dx = \left[x^3\right]_0^{1/2} = \frac18$, the shaded area of the figure, and $P(X \ge \frac12) = 1 - \frac18 = \frac78$. Half of the interval carries only one eighth of the probability: the mass sits near $1$, where the curve is high. That is how the height of $f$ matters: it says WHERE the probability is concentrated, not how large a probability is.

b) $\int_{-1}^{\sqrt3} x\,dx = \left[\frac{x^2}{2}\right]_{-1}^{\sqrt3} = \frac32 - \frac12 = 1$. The area condition holds, and it is a trap: the integral is a SIGNED area, and the negative part over $[-1, 0)$ has been cancelled by extra positive area on the right. Since $g(x) = x < 0$ for $-1 \le x < 0$, the sign condition fails and $g$ is NOT a density. Used as one, it would give $P(-1 \le X \le 0) = \int_{-1}^0 x\,dx = -\frac12$, a negative probability. Checking only that the total is $1$ is the classic half-answer, and it costs the whole question.

c) Sign: $(1 + x)^3 > 0$ for $x \ge 0$, so $h > 0$ there. Area: the interval is infinite, so the integral is improper and is written as a limit. An antiderivative of $2(1 + x)^{-3}$ is $-(1 + x)^{-2}$; check by differentiating: $\frac{d}{dx}\left[-(1+x)^{-2}\right] = 2(1+x)^{-3}$. So $\int_0^t h(x)\,dx = \left[-\frac{1}{(1+x)^2}\right]_0^t = 1 - \frac{1}{(1+t)^2}$, which tends to $1$ as $t \to \infty$. $h$ is a density. Then $P(X > 1) = \lim_{t \to \infty} \int_1^t h(x)\,dx = \lim_{t \to \infty}\left(\frac14 - \frac{1}{(1+t)^2}\right) = \frac14$. Writing $\left[-\frac{1}{(1+x)^2}\right]_1^{\infty}$ with $\infty$ plugged in as a number is what markers penalize: the limit is the definition, and it must appear.

d) The sign condition forces $c \ge 0$, and $c = 0$ gives area $0$, not $1$. For $c > 0$: $\int_1^t \frac{c}{x}\,dx = c \ln t \to \infty$ as $t \to \infty$. The area under $\frac{1}{x}$ on $[1, \infty)$ is INFINITE, and no positive multiple of an infinite area equals $1$. Normalizing means dividing by the total area, which is only possible when that area is finite and non-zero. Compare with part c): $\frac{2}{(1+x)^3}$ decreases fast enough for the area to converge, $\frac{1}{x}$ does not.

e) The student confuses a height with an area. The value $f(x)$ is a probability per unit length, like a density in kilograms per metre: only its product with a length, and in general its integral, is a probability. For a short interval, $P(1 - \Delta x \le X \le 1) \approx f(1)\,\Delta x = 3\,\Delta x$. With $\Delta x = 0.01$ this is about $0.03$, and the exact value is $1 - 0.99^3 = 0.029701$. As $\Delta x \to 0$ the probability tends to $0$: $P(X = 1) = \int_1^1 f(x)\,dx = 0$, whatever the height $f(1)$ is. So heights can be any non-negative number, including $3$ or $1000$; what can never exceed $1$ is an area.

a) Sign first: $x(2 - x) \ge 0$ on $[0, 2]$, so $f \ge 0$ forces $k \ge 0$. Area: $\int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac83 = \frac43$. The condition $k \cdot \frac43 = 1$ gives $k = \frac34$. The frequent slip is $k = \frac43$: the constant is the RECIPROCAL of the area, since it has to scale the area down to $1$. The quick test is the one you should always run: with $k = \frac43$ the total area would be $\frac{16}{9} > 1$.

b) With $F$ written as the area from $0$: $P(X \le \frac12) = \frac34\left[x^2 - \frac{x^3}{3}\right]_0^{1/2} = \frac34\left(\frac14 - \frac{1}{24}\right) = \frac34 \cdot \frac{5}{24} = \frac{5}{32}$. Now $f(2 - x) = k(2 - x)x = f(x)$: the graph is symmetric about $x = 1$, so the right-hand strip $[\frac32, 2]$ has the same area $\frac{5}{32}$. The middle strip is what remains: $1 - 2 \cdot \frac{5}{32} = \frac{22}{32} = \frac{11}{16}$. Direct check: $\frac34\left[x^2 - \frac{x^3}{3}\right]_{1/2}^{3/2} = \frac34\left(\frac98 - \frac{5}{24}\right) = \frac34 \cdot \frac{22}{24} = \frac{11}{16}$. Symmetry is worth a line on an exam only if you STATE the identity $f(2 - x) = f(x)$ that justifies it.

c) $P(X = 1) = \int_1^1 f(x)\,dx = 0$, even though $f(1) = \frac34$ is the maximum of $f$: a single point has no width, hence no area. Consequently $P(X < 1) = P(X \le 1) - P(X = 1) = P(X \le 1)$, and by the symmetry of part b) this is $\frac12$. For a continuous random variable, strict and non-strict inequalities give the same probability; the distinction matters only for discrete variables, which are not in this chapter.

d) Here the domain is the whole line, so the integral is improper at BOTH ends and must be split at a point, say $0$, each half being a limit: $\int_0^t \frac{dx}{1 + x^2} = \arctan t \to \frac{\pi}{2}$ as $t \to \infty$, and likewise $\int_{-t}^0 \frac{dx}{1+x^2} = \arctan t \to \frac{\pi}{2}$. The total is $\pi$, so $k = \frac{1}{\pi}$. Then $P(X > 1) = \frac{1}{\pi}\lim_{t \to \infty}(\arctan t - \arctan 1) = \frac{1}{\pi}\left(\frac{\pi}{2} - \frac{\pi}{4}\right) = \frac14$, and $P(|X| \le \sqrt3) = \frac{1}{\pi}\left(\arctan\sqrt3 - \arctan(-\sqrt3)\right) = \frac{1}{\pi} \cdot \frac{2\pi}{3} = \frac23$. Every value is exact: $\arctan 1 = \frac{\pi}{4}$ and $\arctan\sqrt3 = \frac{\pi}{3}$ are the ones to know.

e) The mean is $\mu = \int_{-\infty}^{\infty} x\,g(x)\,dx$, an improper integral that exists only if BOTH halves converge. On the right, with $u = 1 + x^2$: $\int_0^t \frac{x}{\pi(1 + x^2)}\,dx = \frac{1}{2\pi}\ln(1 + t^2)$, which tends to $\infty$. The right half diverges, so the integral that defines the mean diverges and $X$ HAS NO MEAN. The student computed, in effect, $\lim_{t \to \infty}\int_{-t}^{t} x\,g(x)\,dx = 0$, the symmetric limit: that is not the definition of an integral over $(-\infty, \infty)$, and here it hides two infinite halves. Symmetry about $0$ gives the mean only once the mean is known to exist. This density, called the Cauchy density, is the standard counterexample.

a) The graph is never below the axis, so $f \ge 0$. The region under it is a triangle with base $2$ and height $1$, of area $\frac12 \cdot 2 \cdot 1 = 1$. Both conditions hold: $f$ is a density. Reading an area by geometry is legitimate and fast; it is exactly what the integral would give.

b) For $x < 0$ no area has accumulated: $F(x) = 0$. For $0 \le x \le 1$: $F(x) = \int_0^x t\,dt = \frac{x^2}{2}$. For $1 \le x \le 2$ the area accumulated on $[0, 1]$ must be CARRIED: $F(x) = F(1) + \int_1^x (2 - t)\,dt = \frac12 + \left[2t - \frac{t^2}{2}\right]_1^x = \frac12 + 2x - \frac{x^2}{2} - \frac32 = 2x - \frac{x^2}{2} - 1$. For $x > 2$ all the area is behind: $F(x) = 1$. Checks at the joints: the middle formula gives $F(1) = 2 - \frac12 - 1 = \frac12$, as the first one does, and $F(2) = 4 - 2 - 1 = 1$. A cumulative distribution function is continuous, increasing, starts at $0$ and ends at $1$; the solution figure shows it next to $f$.

c) $F(\frac32) = 3 - \frac98 - 1 = \frac78$ and $F(\frac12) = \frac18$, so $P(\frac12 \le X \le \frac32) = \frac78 - \frac18 = \frac34$. Then $P(X > \frac32) = 1 - F(\frac32) = \frac18$. On the figure, the region to the right of $x = \frac32$ is a small triangle with base $\frac12$ and height $f(\frac32) = \frac12$, of area $\frac12 \cdot \frac12 \cdot \frac12 = \frac18$. The two methods agree, and on an exam the geometric one is a ten-second check of the algebraic one.

d) Going back from $F$ to $f$ is FTC 1: $f_Y(y) = F_Y'(y) = \frac{3}{y^4}$ for $y > 1$, and $f_Y(y) = 0$ for $y < 1$. Check: $\int_1^t 3y^{-4}\,dy = 1 - t^{-3} \to 1$. Then $P(Y > 2) = 1 - F_Y(2) = \frac{1}{8}$, read directly from $F$ without any integral. The median solves $F_Y(m) = \frac12$: $1 - \frac{1}{m^3} = \frac12$, so $m^3 = 2$ and $m = \sqrt[3]{2}$. With $F$ in hand, every probability is an evaluation and the median is an equation; this is why the question often asks for $F$ first.

e) The student started the area at $x = 1$ instead of at $-\infty$: the term $F(1) = \frac12$, the probability already accumulated on $[0, 1]$, is missing, and the formula is too small by exactly $\frac12$. Two tests catch it at once. Continuity: the student's formula gives $2 - \frac12 - \frac32 = 0$ at $x = 1$, while the first piece gives $\frac12$; a cumulative distribution function never jumps when $X$ has a density. The end value: at $x = 2$ the student's formula gives $4 - 2 - \frac32 = \frac12$, but $F(2)$ must equal $1$ since all the area is behind. The correct formula is $2x - \frac{x^2}{2} - 1$.

a) $\mu = \int_0^1 x \cdot 2x\,dx = \left[\frac{2x^3}{3}\right]_0^1 = \frac23$. For the median, first $F(x) = \int_0^x 2t\,dt = x^2$ on $[0, 1]$, then $F(m) = \frac12$ gives $m^2 = \frac12$, so $m = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}$, the positive root since $m$ lies in $[0, 1]$. The mean multiplies by $x$ inside the integral; the median does not, it solves an equation in the upper bound. Mixing the two, for example solving $\int_0^m x f(x)\,dx = \frac12$, is a frequent slip.

b) Both numbers are positive, so compare their squares: $m^2 = \frac12 = \frac{9}{18}$ and $\mu^2 = \frac49 = \frac{8}{18}$. Hence $m > \mu$, with $m \approx 0.71$ and $\mu \approx 0.67$. The shape explains it: the density $2x$ is heavy near $1$ and thins out towards $0$, so its TAIL is on the left. The median only counts area, while the mean weighs each piece of area by its position, like a torque: the thin left tail has little area but a long lever arm, and it pulls the balance point to the left of the half-area cut. Rule to remember: the mean lies on the side of the tail.

c) $\mu = \int_0^1 3x(1 - x)^2\,dx = 3\int_0^1 (x - 2x^2 + x^3)\,dx = 3\left(\frac12 - \frac23 + \frac14\right) = 3 \cdot \frac{1}{12} = \frac14$. For $F$, with the substitution $u = 1 - t$: $F(x) = \int_0^x 3(1 - t)^2\,dt = \left[-(1 - t)^3\right]_0^x = 1 - (1 - x)^3$. Then $F(m) = \frac12$ gives $(1 - m)^3 = \frac12$, so $1 - m = \frac{1}{\sqrt[3]2}$ and $m = 1 - \frac{1}{\sqrt[3]{2}}$. That is the exact answer, and on a MATH 141 paper it is the expected one.

d) $m < \frac14$ is equivalent to $1 - \frac{1}{\sqrt[3]2} < \frac14$, that is $\frac{1}{\sqrt[3]2} > \frac34$. Cubing, a legitimate step between positive numbers since $t \mapsto t^3$ is increasing, it becomes $\frac12 > \frac{27}{64}$, which is true because $\frac12 = \frac{32}{64}$. Hence $m < \mu = \frac14$, with $m \approx 0.21$. This time the mass is piled near $0$ and the tail runs to the right, so the mean is pulled to the right of the median: the same rule as in part b), mirrored. A comparison of two exact numbers is settled by an algebraic chain like this one, never by a decimal from a machine.

e) The median is defined by an AREA condition, $F(m) = \frac12$, and nothing forces the solution to be the middle of the interval: part a) gives $\frac{\sqrt2}{2}$ and part c) gives $1 - \frac{1}{\sqrt[3]2}$, neither of which is $\frac12$. The student is right when the density is symmetric about $\frac12$, that is $f(1 - x) = f(x)$, as for the uniform density $f = 1$ or for $f(x) = 6x(1 - x)$: then the two halves of the region are mirror images and have equal areas. In that case the mean is $\frac12$ too, and it certainly exists because the interval is bounded.

a) $f \ge 0$ since an exponential is positive. For the area: $\int_0^t \frac14 e^{-s/4}\,ds = \left[-e^{-s/4}\right]_0^t = 1 - e^{-t/4}$, and $e^{-t/4} \to 0$ as $t \to \infty$, so the total area is $1$ and $f$ is a density. The same computation IS the cumulative distribution function: $F(t) = 1 - e^{-t/4}$ for $t \ge 0$, and $F(t) = 0$ for $t < 0$. From now on every probability is a subtraction of values of $F$, which is the efficient way to work with this density.

b) $P(T > 8) = 1 - F(8) = e^{-8/4} = e^{-2}$, the shaded tail of the figure. Order of magnitude: $e^2 \approx 2.72^2 \approx 7.4$, so $e^{-2} \approx 0.14$, about one customer in seven. Then $P(2 \le T \le 4) = F(4) - F(2) = (1 - e^{-1}) - (1 - e^{-1/2}) = e^{-1/2} - e^{-1}$. With $\frac1e \approx 0.37$ and $\frac{1}{\sqrt e} \approx \frac{1}{1.65} \approx 0.61$, this is about $0.24$. The exact forms $e^{-2}$ and $e^{-1/2} - e^{-1}$ are the answers; the decimals only tell the reader that the numbers make sense.

c) $\mu = \int_0^\infty t \cdot \frac14 e^{-t/4}\,dt = \lim_{s \to \infty}\int_0^s \frac{t}{4} e^{-t/4}\,dt$. Integration by parts with $u = t$ and $dv = \frac14 e^{-t/4}\,dt$, so $du = dt$ and $v = -e^{-t/4}$; the choice puts the power of $t$ in $u$, where differentiating removes it. Then $\int_0^s \frac{t}{4}e^{-t/4}\,dt = \left[-te^{-t/4}\right]_0^s + \int_0^s e^{-t/4}\,dt = -se^{-s/4} + 4\left(1 - e^{-s/4}\right)$. The limit $\lim_{s \to \infty} se^{-s/4} = \lim_{s \to \infty}\frac{s}{e^{s/4}}$ is of the form $\frac{\infty}{\infty}$; by L'Hôpital's rule it equals $\lim_{s \to \infty}\frac{1}{\frac14 e^{s/4}} = 0$. Hence $\mu = 4$ minutes. In general the exponential density $\frac1\mu e^{-t/\mu}$ has mean $\mu$, which is why its parameter is written that way.

d) The median solves $F(m) = \frac12$: $1 - e^{-m/4} = \frac12$, so $e^{-m/4} = \frac12$, $-\frac{m}{4} = -\ln 2$ and $m = 4\ln 2 \approx 2.8$ minutes. It is SMALLER than the mean $4$. The fraction of customers who wait less than the mean is $P(T < 4) = F(4) = 1 - e^{-1} \approx 0.63$: almost two customers out of three wait less than the average wait. There is no contradiction: a few very long waits, the long right tail of the figure, pull the mean up, while the median only counts customers. Saying that half the customers wait more than the mean is the classic error on this density.

e) The customers who have already waited $2$ minutes are those with $T > 2$, a proportion $P(T > 2) = e^{-1/2}$ of all customers. Among them, those who wait more than $6$ minutes in total form the proportion $P(T > 6) = e^{-3/2}$ of all customers. The fraction asked for is the ratio of the two tail areas: $\frac{e^{-3/2}}{e^{-1/2}} = e^{-1}$. That is exactly $P(T > 4) = e^{-1}$, the probability that a customer who just arrived waits more than $4$ minutes. Having already waited $2$ minutes changes nothing about the $4$ extra minutes: the exponential density has no memory. It is a good model for a queue fed by random arrivals, and a poor one for a machine that wears out, which exercise 9 treats.