Hard Further Maths The continuous uniform distribution Questions

Challenging, exam-style Further Maths The continuous uniform distribution questions with worked solutions. Stretch yourself on the hardest continuous-uniform-distribution, second-moment, derivation, integration problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The continuous random variable XX is uniformly distributed over the interval [0,3]\left[0,3\right]. Which of the following is the exact value of E((X+1)2)E\left(\left(X+1\right)^{2}\right)?
Show worked solution

Worked solution

  1. Write down the probability density function

    f(x)=13−0=13,0≤x≤3f(x)=\frac{1}{3-0}=\frac{1}{3},\quad 0\le x\le 3

    For a rectangular distribution the height of the density is one over the width of the interval.

  2. Write down the integral defining E((X+1)2)E\left(\left(X+1\right)^{2}\right)

    E((X+1)2)=∫03(x+1)2×13 dxE\left(\left(X+1\right)^{2}\right)=\int_{0}^{3}\left(x+1\right)^{2}\times\frac{1}{3}\,dx

    The expectation of a function of XX weights g(x)g(x) by the density.

  3. Integrate

    E((X+1)2)=[x(x2+3x+3)9]03E\left(\left(X+1\right)^{2}\right)=\left[\frac{x \left(x^{2} + 3 x + 3\right)}{9}\right]_{0}^{3}

    The constant density 13\frac{1}{3} can be taken outside the integral.

  4. Substitute the limits

    E((X+1)2)=7−0E\left(\left(X+1\right)^{2}\right)=7-0

    Upper limit minus lower limit.

  5. Note that this is not the same as substituting the mean

    E((X+1)2)≠g(E(X))E\left(\left(X+1\right)^{2}\right)\neq g\left(E(X)\right)

    The expectation of a function is not the function of the expectation unless gg is linear.

  6. Compare with the value of gg at the mean

    g(E(X))=g(32)=254≠7g\left(E(X)\right)=g\left(\frac{3}{2}\right)=\frac{25}{4}\neq7

    Substituting the mean into gg is a common error; the two values differ.

  7. Identify the antiderivative used

    ∫(x+1)2×13 dx=x(x2+3x+3)9+c\int \left(x+1\right)^{2}\times\frac{1}{3}\,dx=\frac{x \left(x^{2} + 3 x + 3\right)}{9}+c

    Only a single power rule is needed once the density is taken outside.

  8. Interpret the answer as a weighted average

    E((X+1)2)=13∫03(x+1)2 dxE\left(\left(X+1\right)^{2}\right)=\frac{1}{3}\int_{0}^{3}\left(x+1\right)^{2}\,dx

    Because the density is constant, E((X+1)2)E\left(\left(X+1\right)^{2}\right) is just the average value of g(x)g(x) across the interval.

  9. Check the answer numerically

    E((X+1)2)≈7.0E\left(\left(X+1\right)^{2}\right)\approx 7.0

    A decimal check confirms the exact value is of a sensible size.

  10. Find the width of the interval

    b−a=3−0=3b-a=3-0=3

    Every result for a rectangular distribution is built from this width.

  11. Note the danger of leaving out the density

    ∫03(x+1)2 dx=21≠7\int_{0}^{3}\left(x+1\right)^{2}\,dx=21\neq7

    Integrating g(x)g(x) without the factor f(x)f(x) gives the wrong answer.

  12. Note the mean and the variance of XX

    E(X)=32,Var(X)=34E(X)=\frac{3}{2},\qquad \text{Var}(X)=\frac{3}{4}

    These are the standard results for the interval in the question.

  13. Integrate the density to obtain the cumulative distribution function

    F(x)=∫0x13 dt=x3F(x)=\int_{0}^{x}\frac{1}{3}\,dt=\frac{x}{3}

    The cdf is the accumulated area under the density up to xx.

  14. Recall the probability density function of a rectangular distribution

    f(x)=1b−a,a≤x≤b,f(x)=0 otherwisef(x)=\frac{1}{b-a},\quad a\le x\le b,\qquad f(x)=0\ \text{otherwise}

    The density is constant on the interval, which is what makes the distribution rectangular.

  15. Recall that the total area under the density must be 11

    ∫ab1b−a dx=b−ab−a=1\int_{a}^{b}\frac{1}{b-a}\,dx=\frac{b-a}{b-a}=1

    The graph is a rectangle of width b−ab-a and height 1b−a\frac{1}{b-a}, so its area is 11.

  16. State the exact value

    E((X+1)2)=7E\left(\left(X+1\right)^{2}\right)=7

    This is the exact value of E((X+1)2)E\left(\left(X+1\right)^{2}\right).

Answer
77
Question 2
9 markschallenging
The independent continuous random variables XX and YY are uniformly distributed over the intervals [0,4]\left[0,4\right] and [1,3]\left[1,3\right] respectively. Which of the following is the value of P(X>Y)P(X>Y)?
Show worked solution

Worked solution

  1. Write down the two densities

    fX(x)=14,fY(y)=12f_{X}(x)=\frac{1}{4},\quad f_{Y}(y)=\frac{1}{2}

    Each variable has a constant density equal to one over its own width.

  2. Condition on the value of YY

    P(X>Y)=∫13P(X>y)fY(y) dyP(X>Y)=\int_{1}^{3}P\left(X>y\right)f_{Y}(y)\,dy

    Fix Y=yY=y, find the probability that XX beats it, then average over yy.

  3. Write down P(X>y)P(X>y) for yy inside the range of XX

    P(X>y)=4−y4,0≤y≤4P\left(X>y\right)=\frac{4-y}{4},\quad 0\le y\le 4

    This is the tail probability of the rectangular distribution of XX.

  4. Identify the range of yy that contributes

    1≤y≤31\le y\le 3

    Outside this overlap the tail probability is 00 or 11, which is dealt with separately.

  5. Integrate over the overlap

    ∫1312×4−y4 dy=12\int_{1}^{3}\frac{1}{2}\times\frac{4-y}{4}\,dy=\frac{1}{2}

    The integrand is linear in yy, so this is a routine integration.

  6. Add the contribution from y<0y<0

    12×0=0\frac{1}{2}\times0=0

    There are no such values of yy, so this contributes nothing.

  7. Write down the joint density

    f(x,y)=fX(x)fY(y)=18f(x,y)=f_{X}(x)f_{Y}(y)=\frac{1}{8}

    Independence means the joint density is the product of the two densities.

  8. Write the probability as a double integral

    P(X>Y)=∬x>y18 dx dyP(X>Y)=\iint_{x>y}\frac{1}{8}\,dx\,dy

    The region x>yx>y is the part of the rectangle below the line y=xy=x.

  9. Interpret the answer geometrically

    P(X>Y)=area of the region x>yarea of the rectangleP(X>Y)=\frac{\text{area of the region}\ x>y}{\text{area of the rectangle}}

    The joint density is constant, so the probability is a ratio of areas.

  10. Check the complementary probability

    P(X<Y)=1−12=12P(X<Y)=1-\frac{1}{2}=\frac{1}{2}

    The event X=YX=Y has probability zero, so the two probabilities sum to 11.

  11. Compare the two means as a sanity check

    E(X)=2,E(Y)=2E(X)=2,\qquad E(Y)=2

    The larger mean belongs to the variable more likely to come out on top.

  12. Note that the answer is a probability, not an area of overlap

    0≤12≤10\le \frac{1}{2}\le 1

    The two intervals overlapping is not the same as one variable exceeding the other.

  13. Recall the probability density function of a rectangular distribution

    f(x)=1b−a,a≤x≤b,f(x)=0 otherwisef(x)=\frac{1}{b-a},\quad a\le x\le b,\qquad f(x)=0\ \text{otherwise}

    The density is constant on the interval, which is what makes the distribution rectangular.

  14. Recall that the total area under the density must be 11

    ∫ab1b−a dx=b−ab−a=1\int_{a}^{b}\frac{1}{b-a}\,dx=\frac{b-a}{b-a}=1

    The graph is a rectangle of width b−ab-a and height 1b−a\frac{1}{b-a}, so its area is 11.

  15. State the probability

    P(X>Y)=12+0=12P(X>Y)=\frac{1}{2}+0=\frac{1}{2}

    This is the exact probability that XX exceeds YY.

Answer
12\frac{1}{2}
Question 3
9 markschallenging
The independent continuous random variables XX and YY are uniformly distributed over the intervals [0,6]\left[0,6\right] and [1,5]\left[1,5\right] respectively. Which of the following is the value of Var(X+Y)\text{Var}\left(X+Y\right)?
Show worked solution

Worked solution

  1. Find the mean and variance of XX

    E(X)=3,Var(X)=3E(X)=3,\qquad \text{Var}(X)=3

    Use a+b2\frac{a+b}{2} and (b−a)212\frac{(b-a)^{2}}{12} for the first interval.

  2. Find the mean and variance of YY

    E(Y)=3,Var(Y)=43E(Y)=3,\qquad \text{Var}(Y)=\frac{4}{3}

    The same two results applied to the second interval.

  3. Recall the variance of a linear combination of INDEPENDENT variables

    Var(cX+dY)=c2Var(X)+d2Var(Y)\text{Var}\left(cX+dY\right)=c^{2}\text{Var}(X)+d^{2}\text{Var}(Y)

    Each coefficient is squared, so a minus sign makes no difference.

  4. Substitute the two variances

    Var(X+Y)=(1)2×3+(1)2×43\text{Var}\left(X+Y\right)=\left(1\right)^{2}\times3+\left(1\right)^{2}\times\frac{4}{3}

    Here c2=1c^{2}=1 and d2=1d^{2}=1.

  5. Note that the variances ADD

    Var(X−Y)=Var(X)+Var(Y)\text{Var}(X-Y)=\text{Var}(X)+\text{Var}(Y)

    Subtracting a variable still adds its variability.

  6. Note that independence is essential

    Cov(X,Y)=0 because X and Y are independent\text{Cov}(X,Y)=0\ \text{because}\ X\ \text{and}\ Y\ \text{are independent}

    Without independence a covariance term would be needed.

  7. State the two distributions

    X∼U[0,6],Y∼U[1,5]X\sim U\left[0,6\right],\qquad Y\sim U\left[1,5\right]

    Each variable is rectangular on its own interval.

  8. Find the standard deviation of the combination

    σ=133=393\sigma=\sqrt{\frac{13}{3}}=\frac{\sqrt{39}}{3}

    The standard deviation is the square root of the variance just found.

  9. Note the mean of the combination

    E(X+Y)=6E\left(X+Y\right)=6

    The mean is not needed for the variance, but it locates the distribution.

  10. Note that the combination is NOT uniform

    the sum of two independent uniforms is triangular, not rectangular\text{the sum of two independent uniforms is triangular, not rectangular}

    The rules used here need only the mean and the variance, not the shape.

  11. Recall the probability density function of a rectangular distribution

    f(x)=1b−a,a≤x≤b,f(x)=0 otherwisef(x)=\frac{1}{b-a},\quad a\le x\le b,\qquad f(x)=0\ \text{otherwise}

    The density is constant on the interval, which is what makes the distribution rectangular.

  12. Recall that the total area under the density must be 11

    ∫ab1b−a dx=b−ab−a=1\int_{a}^{b}\frac{1}{b-a}\,dx=\frac{b-a}{b-a}=1

    The graph is a rectangle of width b−ab-a and height 1b−a\frac{1}{b-a}, so its area is 11.

  13. Recall the cumulative distribution function

    F(x)=∫ax1b−a dt=x−ab−a,a≤x≤bF(x)=\int_{a}^{x}\frac{1}{b-a}\,dt=\frac{x-a}{b-a},\quad a\le x\le b

    The cdf grows linearly from 00 at x=ax=a to 11 at x=bx=b.

  14. Recall the values of FF at the ends of the interval

    F(a)=0,F(b)=1F(a)=0,\qquad F(b)=1

    Below aa nothing has happened yet; above bb everything has.

  15. Recall how a probability is read off the rectangle

    P(c<X<d)=d−cb−a,a≤c<d≤bP(c<X<d)=\frac{d-c}{b-a},\quad a\le c<d\le b

    A probability is the fraction of the interval that the event occupies.

  16. Recall that a single value has probability zero

    P(X=c)=0P(X=c)=0

    For a continuous variable the strict and weak inequalities give the same probability.

  17. State the value of Var(X+Y)\text{Var}\left(X+Y\right)

    Var(X+Y)=133\text{Var}\left(X+Y\right)=\frac{13}{3}

    This is the exact value for the combination X+YX+Y.

Answer
133\frac{13}{3}
Question 4
9 markschallenging
The continuous random variable XX has a rectangular distribution on the interval [a,b]\left[a,b\right], with probability density function f(x)=1b−af(x)=\frac{1}{b-a} for a≤x≤ba\le x\le b and f(x)=0f(x)=0 otherwise. Which of the following is E(X2)E(X^{2})?
Show worked solution

Worked solution

  1. Write down the integral defining E(X2)E(X^{2})

    E(X2)=∫abx2×1b−a dxE(X^{2})=\int_{a}^{b}x^{2}\times\frac{1}{b-a}\,dx

    For a function of XX the rule is E(g(X))=∫g(x)f(x) dxE\left(g(X)\right)=\int g(x)f(x)\,dx.

  2. Take the constant density outside the integral

    E(X2)=1b−a∫abx2 dxE(X^{2})=\frac{1}{b-a}\int_{a}^{b}x^{2}\,dx

    The density is constant across the whole interval.

  3. Integrate

    E(X2)=1b−a[x33]abE(X^{2})=\frac{1}{b-a}\left[\frac{x^{3}}{3}\right]_{a}^{b}

    Raise the power by one and divide by the new power.

  4. Substitute the limits

    E(X2)=b3−a33(b−a)E(X^{2})=\frac{b^{3}-a^{3}}{3\left(b-a\right)}

    Upper limit minus lower limit.

  5. Factorise the difference of two cubes

    b3−a3=(b−a)(b2+ab+a2)b^{3}-a^{3}=\left(b-a\right)\left(b^{2}+ab+a^{2}\right)

    This produces the factor b−ab-a that cancels with the denominator.

  6. Cancel the factor b−ab-a

    E(X2)=(b−a)(a2+ab+b2)3(b−a)=a2+ab+b23E(X^{2})=\frac{\left(b-a\right)\left(a^{2}+ab+b^{2}\right)}{3\left(b-a\right)}=\frac{a^{2}+ab+b^{2}}{3}

    The width cancels, leaving a symmetric expression in aa and bb.

  7. Check the result on a simple case

    a=0, b=1 ⇒ E(X2)=13a=0,\ b=1\ \Rightarrow\ E(X^{2})=\frac{1}{3}

    Integrating x2x^{2} over [0,1][0,1] does indeed give 13\frac{1}{3}.

  8. Note that this is not the square of the mean

    a2+ab+b23≠(a+b2)2\frac{a^{2}+ab+b^{2}}{3}\neq\left(\frac{a+b}{2}\right)^{2}

    Their difference is exactly the variance.

  9. Use the result to write down the variance

    Var(X)=a2+ab+b23−(a+b)24=(b−a)212\text{Var}(X)=\frac{a^{2}+ab+b^{2}}{3}-\frac{\left(a+b\right)^{2}}{4}=\frac{\left(b-a\right)^{2}}{12}

    This second moment is exactly what is needed for the variance.

  10. Check the result on a second case

    a=1, b=3 ⇒ E(X2)=1+3+93=133a=1,\ b=3\ \Rightarrow\ E(X^{2})=\frac{1+3+9}{3}=\frac{13}{3}

    Direct integration over [1,3][1,3] gives the same value.

  11. Note the symmetry of the expression

    a2+ab+b2 is unchanged when a and b are swappeda^{2}+ab+b^{2}\ \text{is unchanged when}\ a\ \text{and}\ b\ \text{are swapped}

    Swapping the endpoints cannot change any moment of the distribution.

  12. Recall the probability density function of a rectangular distribution

    f(x)=1b−a,a≤x≤b,f(x)=0 otherwisef(x)=\frac{1}{b-a},\quad a\le x\le b,\qquad f(x)=0\ \text{otherwise}

    The density is constant on the interval, which is what makes the distribution rectangular.

  13. Recall that the total area under the density must be 11

    ∫ab1b−a dx=b−ab−a=1\int_{a}^{b}\frac{1}{b-a}\,dx=\frac{b-a}{b-a}=1

    The graph is a rectangle of width b−ab-a and height 1b−a\frac{1}{b-a}, so its area is 11.

  14. Recall the cumulative distribution function

    F(x)=∫ax1b−a dt=x−ab−a,a≤x≤bF(x)=\int_{a}^{x}\frac{1}{b-a}\,dt=\frac{x-a}{b-a},\quad a\le x\le b

    The cdf grows linearly from 00 at x=ax=a to 11 at x=bx=b.

  15. Recall the values of FF at the ends of the interval

    F(a)=0,F(b)=1F(a)=0,\qquad F(b)=1

    Below aa nothing has happened yet; above bb everything has.

  16. State the result

    E(X2)=a2+ab+b23E(X^{2})=\frac{a^{2}+ab+b^{2}}{3}

    This is the standard result quoted in the formula book, now derived.

Answer
a2+ab+b23\frac{a^{2}+ab+b^{2}}{3}
Question 5
9 markschallenging
A thin rod of length 9090 cm is cut at a point chosen at random along its length. The distance XX cm from the left-hand end of the rod to the cut is modelled by a continuous uniform distribution over the interval [0,90]\left[0,90\right]. Find the probability that the shorter of the two pieces is less than 2020 cm long.
Show worked solution

Worked solution

  1. Write down the probability density function

    f(x)=190−0=190,0≤x≤90f(x)=\frac{1}{90-0}=\frac{1}{90},\quad 0\le x\le 90

    For a rectangular distribution the height of the density is one over the width of the interval.

  2. Express the length of the shorter piece in terms of XX

    shorter piece=min⁡(X,90−X)\text{shorter piece}=\min\left(X,90-X\right)

    The cut divides the rod into pieces of length XX and 90−X90-X.

  3. Translate the condition into a condition on XX

    min⁡(X,90−X)<20  ⟺  X<20 or X>70\min\left(X,90-X\right)<20\iff X<20\ \text{or}\ X>70

    The shorter piece is short exactly when the cut is near one of the ends.

  4. Add the probabilities of the two disjoint regions

    P=∫020190 dx+∫7090190 dxP=\int_{0}^{20}\frac{1}{90}\,dx+\int_{70}^{90}\frac{1}{90}\,dx

    The two regions cannot both happen, so the probabilities add.

  5. Evaluate the two integrals

    P=190×20+190×20=49P=\frac{1}{90}\times20+\frac{1}{90}\times20=\frac{4}{9}

    Each region has length 2020 cm.

  6. Interpret the answer as a fraction of the rod

    2×2090=49\frac{2\times20}{90}=\frac{4}{9}

    The cut must land in one of two end-sections, each of length 2020 cm.

  7. Note the symmetry of the situation

    P(X<20)=P(X>70)=29P\left(X<20\right)=P\left(X>70\right)=\frac{2}{9}

    The two end-sections are the same length, so they are equally likely.

  8. Write down the complementary probability

    P(shorter piece≥20)=1−49=59P\left(\text{shorter piece}\ge 20\right)=1-\frac{4}{9}=\frac{5}{9}

    The cut then lands in the middle section of the rod.

  9. State the mean position of the cut

    E(X)=0+902=45E(X)=\frac{0+90}{2}=45

    On average the cut is made at the midpoint, though any point is equally likely.

  10. Note the range of the shorter piece

    0≤min⁡(X,90−X)≤450\le\min\left(X,90-X\right)\le 45

    The shorter piece can never be longer than half the rod.

  11. Note that the shorter piece is NOT uniform on the whole rod

    min⁡(X,90−X)∼U[0,45]\min\left(X,90-X\right)\sim U\left[0,45\right]

    Folding the rod in half maps the cut position onto the shorter piece, so it is uniform on the HALF interval.

  12. Check the answer with the cumulative distribution function

    F(20)+1−F(70)=49F\left(20\right)+1-F\left(70\right)=\frac{4}{9}

    The same answer comes straight from the cdf of XX.

  13. Recall the probability density function of a rectangular distribution

    f(x)=1b−a,a≤x≤b,f(x)=0 otherwisef(x)=\frac{1}{b-a},\quad a\le x\le b,\qquad f(x)=0\ \text{otherwise}

    The density is constant on the interval, which is what makes the distribution rectangular.

  14. Recall that the total area under the density must be 11

    ∫ab1b−a dx=b−ab−a=1\int_{a}^{b}\frac{1}{b-a}\,dx=\frac{b-a}{b-a}=1

    The graph is a rectangle of width b−ab-a and height 1b−a\frac{1}{b-a}, so its area is 11.

  15. State the probability

    P(shorter piece<20)=49P\left(\text{shorter piece}<20\right)=\frac{4}{9}

    This is the exact probability that the shorter piece is under 2020 cm.

Answer
49\frac{4}{9}

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