Continuous random variables

Worked example 1 — A density can be greater than 1

Show that \(f(x)=\begin{cases}3x^2,&0\leq x\leq1,\\0,&\text{otherwise}\end{cases}\) is a valid pdf.

Solution

On \([0,1]\), \(3x^2\geq0\). Outside this interval the density is zero. Also, \[\int_0^1 3x^2\dd x=[x^3]_0^1=1.\] Both conditions hold, so this is a valid density, even though \(f(1)=3\).

Worked example 2 — Find a normalising constant

A normalising constant is chosen so that the total area under the density is 1. Find \(k\) for \(f(x)=\begin{cases}kx^2,&0\leq x\leq2,\\0,&\text{otherwise}.\end{cases}\)

Solution

\[\begin{aligned} \int_0^2 kx^2\dd x&=1\\ k\left[\frac{x^3}{3}\right]_0^2&=\frac{8k}{3}=1\\ k&=\frac38. \end{aligned}\] Since \(\frac38x^2\geq0\) on \([0,2]\), the non-negativity condition is also satisfied.

Worked example 3 — Left tail, interval and right tail

Let \(f(x)=\begin{cases}\frac34x(2-x),&0\leq x\leq2,\\0,&\text{otherwise}.\end{cases}\) Find:

1. \(\Prob(X\leq1.5)\);

2. \(\Prob(0.5\leq X\leq1)\);

3. \(\Prob(X\geq1)\).

Solution

An antiderivative on \([0,2]\) is \(\frac34x^2-\frac14x^3\). Use only the part of the interval where the density is non-zero.

1. Left tail: \[\Prob(X\leq1.5)=\int_0^{3/2}\frac34x(2-x)\dd x =\left[\frac34x^2-\frac14x^3\right]_0^{3/2} =\frac{27}{32}\approx0.844.\]

2. Between two values: \[\Prob(0.5\leq X\leq1)=\int_{1/2}^1\frac34x(2-x)\dd x =\left[\frac34x^2-\frac14x^3\right]_{1/2}^{1} =\frac12-\frac5{32}=\frac{11}{32}\approx0.344.\]

3. Right tail: \[\Prob(X\geq1)=\int_1^2\frac34x(2-x)\dd x =1-\int_0^1\frac34x(2-x)\dd x=1-\frac12=\frac12.\] We subtracted the left-hand area from the total area 1. Symmetry about \(x=1\) also gives equal areas on either side.

Worked example 4 — A triangular piecewise model

Let \[f(x)=\begin{cases} \frac23(x-1),&1\leq x\leq2,\\[2pt] \frac13(4-x),&2<x\leq4,\\[2pt] 0,&\text{otherwise}. \end{cases}\]

1. Show that \(f\) is a valid density.

2. Find \(\Prob(1.5\leq X\leq3)\).

3. Find \(\Prob(X>3\mid X>2)\).

Solution

1. Both formulas are non-negative on their stated intervals. The total area is \[\begin{aligned} \int_1^2\frac23(x-1)\dd x+\int_2^4\frac13(4-x)\dd x &=\left[\frac{(x-1)^2}{3}\right]_1^2 +\left[-\frac{(4-x)^2}{6}\right]_2^4\\ &=\frac13+\frac23=1. \end{aligned}\] Equivalently, use the two triangle areas: \(\frac12(1)(\frac23)+\frac12(2)(\frac23)=1\).

2. The interval crosses \(x=2\), so \[\begin{aligned} \Prob(1.5\leq X\leq3) &=\int_{3/2}^2\frac23(x-1)\dd x+\int_2^3\frac13(4-x)\dd x\\ &=\left(\frac13-\frac1{12}\right)+\left(-\frac16+\frac46\right) =\frac14+\frac12=\frac34. \end{aligned}\]

3. The event \(X>3\) is contained in \(X>2\). Hence \[\Prob(X>3\mid X>2)=\frac{\Prob(X>3)}{\Prob(X>2)} =\frac{\int_3^4(4-x)/3\dd x}{2/3}=\frac{1/6}{2/3}=\frac14.\]

Worked example 5 — A linear density

Let \(f(x)=\begin{cases}\frac{x}{2},&0\leq x\leq2,\\0,&\text{otherwise}.\end{cases}\) Find \(\E(X)\).

Solution

First, the model is normalised: \(\int_0^2\frac{x}{2}\dd x=[x^2/4]_0^2=1\). \[\begin{aligned} \E(X)&=\int_0^2x\left(\frac{x}{2}\right)\dd x\\ &=\frac12\left[\frac{x^3}{3}\right]_0^2 =\frac43. \end{aligned}\]

Worked example 6 — Mean of a piecewise distribution

Find the mean of the triangular model from worked example 4.

Solution

\[\begin{aligned} \E(X)&=\int_1^2x\,\frac23(x-1)\dd x+\int_2^4x\,\frac13(4-x)\dd x\\ &=\frac23\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_1^2 +\frac13\left[2x^2-\frac{x^3}{3}\right]_2^4\\ &=\frac59+\frac{16}{9}=\frac73. \end{aligned}\]

Worked example 7 — A median found by integration

Let \(f(x)=\frac38x^2\) for \(0\leq x\leq2\), and zero otherwise. Find the median.

Solution

\[\begin{aligned} \int_0^m\frac38x^2\dd x&=\frac12\\ \left[\frac{x^3}{8}\right]_0^m&=\frac12\\ \frac{m^3}{8}&=\frac12\quad\Longrightarrow\quad m=\sqrt[3]{4}\approx1.59. \end{aligned}\]

Worked example 8 — Locate a piecewise median first

Find the median of the triangular model in worked example 4.

Solution

For the triangular model in worked example 4, \(\Prob(X\leq2)=\frac13<\frac12\). Therefore the median lies in the second interval, \(2<m<4\).

Using the definition, the area to the left of \(m\) gives \[\underbrace{\frac13}_{\text{area from 1 to 2}}+\int_2^m\frac{4-x}{3}\dd x=\frac12.\] An equivalent, shorter calculation uses the area to the right of \(m\), which must also be \(\frac12\): \[\begin{aligned} \int_m^4\frac{4-x}{3}\dd x&=\frac12\\ \frac{(4-m)^2}{6}&=\frac12\\ (4-m)^2&=3\quad\Longrightarrow\quad m=4\pm\sqrt3. \end{aligned}\] Only \(4-\sqrt3\) lies in \((2,4)\).

Worked example 9 — Numerical boundaries — [CALCULATOR]

Let \(X\) have density \(f(x)=12x^2(1-x)\) for \(0\leq x\leq1\), and zero otherwise. Find the median \(m\) and the value \(q\) for which \(\Prob(X\leq q)=0.8\). Give both to three significant figures.

Solution

1. Median: write and evaluate the area integral. \[\begin{aligned} \int_0^m12x^2(1-x)\dd x&=0.5\\ \left[4x^3-3x^4\right]_0^m&=0.5\\ 4m^3-3m^4&=0.5,\qquad 0<m<1. \end{aligned}\] TI-Nspire CX: in Scratchpad Calculate, insert nSolve from the Catalog (or type it). Using an unassigned variable \(x\), enter

   nSolve(4*x^3-3*x^4=0.5,x,0,1)

   The last two entries restrict the search to \([0,1]\). The result is \(0.614272\ldots\), so \(m\approx0.614\).

2. Boundary for probability 0.8. \[\int_0^q12x^2(1-x)\dd x=0.8 \quad\Longrightarrow\quad 4q^3-3q^4=0.8.\] Change the right-hand side in the calculator command:

   nSolve(4*x^3-3*x^4=0.8,x,0,1)

   This gives \(q=0.787682\ldots\approx0.788\).

Check: use the unrounded values in \(4x^3-3x^4\) to recover 0.5 and 0.8 respectively. Round only the final answers.

Worked example 10 — An interior mode

Find the mode for \(f(x)=\begin{cases}\frac{\pi}{4}\sin\left(\frac{\pi(x-2)}2\right),&2\leq x\leq4,\\0,&\text{otherwise}.\end{cases}\)

Solution

This is non-negative on \([2,4]\) and its total integral is 1. On \(2<x<4\), \[f'(x)=\frac{\pi^2}{8}\cos\left(\frac{\pi(x-2)}2\right)=0 \quad\Longrightarrow\quad x=3.\] The angle lies between 0 and \(\pi\), so \(\pi/2\) is the only relevant zero of cosine. Since \(f'\) changes from positive to negative and both endpoint densities are zero, \(x=3\) gives the maximum.

Worked example 11 — A mode at an endpoint

For \(f(x)=3x^2\) on \([0,1]\) (zero otherwise), find the mode.

Solution

The density \(f\) increases throughout the interval. Its maximum occurs at \(x=1\), so the mode is \(1\). There is no interior maximum to find by solving \(f'(x)=0\).

Worked example 12 — How variable is an app’s launch time?

The launch time \(X\), in seconds, has density \(f(x)=\begin{cases}\frac{(x-3)^2}{9},&0\leq x\leq3,\\0,&\text{otherwise}.\end{cases}\) Find the mean, variance and standard deviation.

Solution

1. Mean. Multiply the density by \(x\): \[\begin{aligned} \mu&=\frac19\int_0^3x(x-3)^2\dd x =\frac19\int_0^3(x^3-6x^2+9x)\dd x\\ &=\frac19\left[\frac{x^4}{4}-2x^3+\frac92x^2\right]_0^3 =\frac34\text{ seconds}. \end{aligned}\]

2. Second moment and variance. Multiply the density \(f(x)\) by \(x^2\), then integrate: \[\begin{aligned} \E(X^2)&=\int_0^3 x^2\left(\frac{(x-3)^2}{9}\right)\dd x =\frac19\int_0^3(x^4-6x^3+9x^2)\dd x\\ &=\frac19\left[\frac{x^5}{5}-\frac32x^4+3x^3\right]_0^3 =\frac9{10}\text{ seconds}^2. \end{aligned}\] Now subtract the square of the mean: \[\begin{aligned} \bigl[\E(X)\bigr]^2&=\left(\frac34\right)^2=\frac9{16},\\[3pt] \Var(X)&=\frac9{10}-\frac9{16}=\frac{27}{80}\text{ seconds}^2. \end{aligned}\]

3. Standard deviation.

   \[\sigma=\sqrt{\frac{27}{80}}=\frac{3\sqrt{15}}{20}\approx0.581\text{ seconds}.\]

Worked example 13 — A fixed amount plus a variable amount

For app launch time \(X\) in seconds, \(\E(X)=\frac34\) and \(\Var(X)=\frac{27}{80}\). A simulator’s launch time \(Y=5+2X\) seconds doubles \(X\) and adds a fixed 5-second delay. Find the mean, variance and standard deviation of \(Y\).

Solution

1. \(\E(Y)=5+2\left(\frac34\right)=\frac{13}{2}=6.5\) seconds.

2. \(\Var(Y)=2^2\left(\frac{27}{80}\right)=\frac{27}{20}\) seconds\(^2\).

3. \(\sigma_Y=2\sqrt{\frac{27}{80}}=\sqrt{\frac{27}{20}}\approx1.16\) seconds.

Worked example 14 — A negative multiplier

For the same launch time \(X\), a performance score starts at 10 points and loses 2 points per second of launch time. Let \(Z=10-2X\) be this score, in points.

Solution

1. \(\E(Z)=10-2\left(\frac34\right)=\frac{17}{2}\) points.

2. \(\Var(Z)=(-2)^2\left(\frac{27}{80}\right)=\frac{27}{20}\) points\(^2\).

3. \(\sigma_Z=|-2|\sqrt{\frac{27}{80}}=\sqrt{\frac{27}{20}}\) points.

Worked example 15 — Set a fair entry price

On each play, a marker lands at a random position on a 1-metre strip. Let \(X\) be its distance from the left end, in metres. Its density is \[f(x)=\begin{cases}2x,&0\leq x\leq1,\\0,&\text{otherwise}.\end{cases}\] The player receives €12 per metre of distance, so the payout is \(R=12X\) euros and the net gain is \(G=12X-c\) euros. Find the expected payout, the fair entry fee and the expected net gain when entry costs €9.

Solution

1. First find the mean distance: \[\E(X)=\int_0^1 x(2x)\dd x=\left[\frac{2x^3}{3}\right]_0^1=\frac23\text{ metres}.\] Thus \(\E(R)=12\left(\frac23\right)=8\) euros.

2. The fair entry price equals the expected payout: \(c=8\) euros.

3. At a price of €9, \(\E(G)=\E(R)-9=8-9=-1\) euro.

Worked example 16 — Fair does not mean a certain result

For the same marker game, \(c=8\). The payout is still \(R=12X\) euros, and the net gain is now \(G=12X-8\) euros per play. Find the probability of making a profit and the variance of the net gain.

Solution

1. Find the probability of making a profit. \[G>0\iff12X-8>0\iff X>\frac23.\] \[\Prob(G>0)=\int_{2/3}^1 2x\dd x=\left[x^2\right]_{2/3}^1=\frac59.\]

2. Find the variance of the net gain. \[\E(X^2)=\int_0^1x^2(2x)\dd x=\frac12,\qquad \Var(X)=\frac12-\left(\frac23\right)^2=\frac1{18}.\] Hence \(\Var(G)=12^2\Var(X)=8\) euros\(^2\), and \(\sigma_G=2\sqrt2\) euros.