Hard Further Maths Continuous random variables Questions

Challenging, exam-style Further Maths Continuous random variables questions with worked solutions. Stretch yourself on the hardest continuous-random-variables, expectation, variance, cumulative-distribution-function problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The continuous random variable XX has probability density function f(x)={6x250≤x≤1125−6x51≤x≤20otherwisef(x)=\begin{cases}\frac{6 x^{2}}{5}&0\le x\le 1\\\frac{12}{5} - \frac{6 x}{5}&1\le x\le 2\\0&\text{otherwise}\end{cases}. Which of the following is the cumulative distribution function of XX?
Show worked solution

Worked solution

  1. Build FF as the accumulated area under the density

    F(x)=∫0xf(u) duF(x)=\int_{0}^{x}f(u)\,du

    Integrate the density from the left-hand end of the support up to xx.

  2. Integrate the density on 0≤x≤10\le x\le 1

    F(x)=∫0x6t25 dt=2x35F(x)=\int_{0}^{x}\frac{6 t^{2}}{5}\,dt=\frac{2 x^{3}}{5}

    This is the area under the density from 00 up to xx.

  3. Integrate the next piece and add the area already accumulated

    F(x)=25+∫1x125−6t5 dt=−3x25+12x5−75F(x)=\frac{2}{5}+\int_{1}^{x}\frac{12}{5} - \frac{6 t}{5}\,dt=- \frac{3 x^{2}}{5} + \frac{12 x}{5} - \frac{7}{5}

    The area 25\frac{2}{5} under the earlier piece must be carried forward.

  4. Check the value at the top of the support

    F(2)=1F\left(2\right)=1

    A cumulative distribution function must finish at 11.

  5. Reject any option that does not start at 00

    F(0)=0F\left(0\right)=0

    No area has accumulated at the left-hand end of the support.

  6. Recall the defining property of a probability density function

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty}f(x)\,dx=1

    The total area under the density curve is always exactly 11.

  7. Recall the second condition for a valid density

    f(x)≥0 for all xf(x)\ge0\ \text{for all}\ x

    A density can never be negative, because areas under it are probabilities.

  8. Recall that a probability is an area

    P(a<X<b)=∫abf(x) dxP(a<X<b)=\int_{a}^{b}f(x)\,dx

    For a continuous variable, probability is the area under the density.

  9. Note that single values carry no probability

    P(X=a)=0P(X=a)=0

    This is why P(a<X<b)P(a<X<b) and P(a≤X≤b)P(a\le X\le b) are the same number.

  10. Recall the definition of the cumulative distribution function

    F(x)=P(X≤x)=∫−∞xf(u) duF(x)=P(X\le x)=\int_{-\infty}^{x}f(u)\,du

    FF accumulates the area under the density from the left.

  11. Recall how to recover the density from FF

    f(x)=dFdxf(x)=\frac{dF}{dx}

    Differentiation undoes the integration that produced FF.

  12. Recall the end conditions on FF

    F(−∞)=0,F(∞)=1F(-\infty)=0,\qquad F(\infty)=1

    FF increases from 00 to 11 across the support.

  13. Recall the definition of the mean

    E(X)=∫−∞∞xf(x) dxE(X)=\int_{-\infty}^{\infty}xf(x)\,dx

    The mean weights each value of xx by the density at that point.

  14. Recall the expectation of a function of XX

    E(g(X))=∫−∞∞g(x)f(x) dxE\left(g(X)\right)=\int_{-\infty}^{\infty}g(x)f(x)\,dx

    Apply gg to xx inside the integral; never apply it to E(X)E(X).

  15. Recall the working formula for the variance

    Var(X)=E(X2)−[E(X)]2\text{Var}(X)=E\left(X^{2}\right)-\left[E(X)\right]^{2}

    This is the form quoted in the formula book.

  16. Select the matching option

    F(x)={0x<02x350≤x≤1−3x25+12x5−751≤x≤21x>2F(x)=\begin{cases}0&x<0\\\frac{2 x^{3}}{5}&0\le x\le 1\\- \frac{3 x^{2}}{5} + \frac{12 x}{5} - \frac{7}{5}&1\le x\le 2\\1&x>2\end{cases}

    This is the only option that starts at 00, ends at 11 and differentiates back to ff.

Answer
F(x)={0x<02x350≤x≤1−3x25+12x5−751≤x≤21x>2F(x)=\begin{cases}0&x<0\\\frac{2 x^{3}}{5}&0\le x\le 1\\- \frac{3 x^{2}}{5} + \frac{12 x}{5} - \frac{7}{5}&1\le x\le 2\\1&x>2\end{cases}
Question 2
9 markschallenging
The continuous random variable XX has probability density function f(x)={x80≤x≤40otherwisef(x)=\begin{cases}\frac{x}{8}&0\le x\le 4\\0&\text{otherwise}\end{cases}. The random variable YY is defined by Y=X3Y=X^{3}. Which of the following is the probability density function of YY?
Show worked solution

Worked solution

  1. Work through the cumulative distribution function of YY

    FY(y)=P(X≤y3)=y2316F_{Y}(y)=P\left(X\le \sqrt[3]{y}\right)=\frac{y^{\frac{2}{3}}}{16}

    A change of variable must go through FF, never through ff directly.

  2. Differentiate with respect to yy

    fY(y)=124y3f_{Y}(y)=\frac{1}{24 \sqrt[3]{y}}

    Differentiating the cumulative function of YY gives its density.

  3. Find the interval on which fYf_{Y} is non-zero

    0≤y≤640\le y\le 64

    The support of XX maps onto this interval under the transformation.

  4. Reject the option that simply rewrites fXf_{X} with yy in place of xx

    fY(y)≠fX(y)f_{Y}(y)\neq f_{X}(y)

    The chain rule contributes an extra factor that this option ignores.

  5. Check that the chosen density integrates to 11

    ∫064124y3 dy=1\int_{0}^{64}\frac{1}{24 \sqrt[3]{y}}\,dy=1

    Only a genuine density encloses unit area.

  6. Recall the defining property of a probability density function

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty}f(x)\,dx=1

    The total area under the density curve is always exactly 11.

  7. Recall the second condition for a valid density

    f(x)≥0 for all xf(x)\ge0\ \text{for all}\ x

    A density can never be negative, because areas under it are probabilities.

  8. Recall that a probability is an area

    P(a<X<b)=∫abf(x) dxP(a<X<b)=\int_{a}^{b}f(x)\,dx

    For a continuous variable, probability is the area under the density.

  9. Note that single values carry no probability

    P(X=a)=0P(X=a)=0

    This is why P(a<X<b)P(a<X<b) and P(a≤X≤b)P(a\le X\le b) are the same number.

  10. Recall the definition of the cumulative distribution function

    F(x)=P(X≤x)=∫−∞xf(u) duF(x)=P(X\le x)=\int_{-\infty}^{x}f(u)\,du

    FF accumulates the area under the density from the left.

  11. Recall how to recover the density from FF

    f(x)=dFdxf(x)=\frac{dF}{dx}

    Differentiation undoes the integration that produced FF.

  12. Recall the end conditions on FF

    F(−∞)=0,F(∞)=1F(-\infty)=0,\qquad F(\infty)=1

    FF increases from 00 to 11 across the support.

  13. Recall the definition of the mean

    E(X)=∫−∞∞xf(x) dxE(X)=\int_{-\infty}^{\infty}xf(x)\,dx

    The mean weights each value of xx by the density at that point.

  14. Recall the expectation of a function of XX

    E(g(X))=∫−∞∞g(x)f(x) dxE\left(g(X)\right)=\int_{-\infty}^{\infty}g(x)f(x)\,dx

    Apply gg to xx inside the integral; never apply it to E(X)E(X).

  15. Select the matching option

    124y3, 0≤y≤64\frac{1}{24 \sqrt[3]{y}},\ 0\le y\le 64

    This is the density of YY together with the correct interval.

Answer
124y3, 0≤y≤64\frac{1}{24 \sqrt[3]{y}},\ 0\le y\le 64
Question 3
9 markschallenging
The continuous random variable XX has probability density function f(x)={x290≤x≤30otherwisef(x)=\begin{cases}\frac{x^{2}}{9}&0\le x\le 3\\0&\text{otherwise}\end{cases}. The random variable YY is defined by Y=X2Y=X^{2}. Find the probability density function of YY, and hence find the exact value of E(Y)E(Y).
Show worked solution

Worked solution

  1. Start from the cumulative distribution function of YY

    FY(y)=P(X2≤y)=P(X≤y)F_{Y}(y)=P\left(X^{2}\le y\right)=P\left(X\le \sqrt{y}\right)

    The transformation is increasing, so the inequality is unchanged.

  2. Write down the cumulative distribution function of XX

    FX(x)=x327F_{X}(x)=\frac{x^{3}}{27}

    This comes from integrating the density of XX.

  3. Substitute to obtain FYF_{Y}

    FY(y)=y3227F_{Y}(y)=\frac{y^{\frac{3}{2}}}{27}

    Replace xx by y\sqrt{y}.

  4. Differentiate to obtain fYf_{Y}

    fY(y)=y18f_{Y}(y)=\frac{\sqrt{y}}{18}

    The density of YY is the derivative of its cumulative function.

  5. State the range of YY

    0≤y≤90\le y\le 9

    The support of XX maps onto this interval.

  6. Set up the integral for E(Y)E(Y)

    E(Y)=∫09y(y18) dyE(Y)=\int_{0}^{9}y\left(\frac{\sqrt{y}}{18}\right)\,dy

    The mean of YY uses the density of YY in the usual way.

  7. Integrate and evaluate the limits

    E(Y)=275E(Y)=\frac{27}{5}

    This is the exact mean of YY.

  8. Check the answer against E(g(X))E\left(g(X)\right)

    ∫03(x2)f(x) dx=275\int_{0}^{3}\left(x^{2}\right)f(x)\,dx=\frac{27}{5}

    Working directly with the density of XX gives the same value, as it must.

  9. Check that fYf_{Y} integrates to 11

    ∫09y18 dy=1\int_{0}^{9}\frac{\sqrt{y}}{18}\,dy=1

    A correct density always encloses unit area.

  10. Note that E(Y)E(Y) is not g(E(X))g\left(E(X)\right)

    275≠8116\frac{27}{5}\neq \frac{81}{16}

    Transforming the mean is not the same as taking the mean of the transform.

  11. Recall the defining property of a probability density function

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty}f(x)\,dx=1

    The total area under the density curve is always exactly 11.

  12. Recall the second condition for a valid density

    f(x)≥0 for all xf(x)\ge0\ \text{for all}\ x

    A density can never be negative, because areas under it are probabilities.

  13. Recall that a probability is an area

    P(a<X<b)=∫abf(x) dxP(a<X<b)=\int_{a}^{b}f(x)\,dx

    For a continuous variable, probability is the area under the density.

  14. Note that single values carry no probability

    P(X=a)=0P(X=a)=0

    This is why P(a<X<b)P(a<X<b) and P(a≤X≤b)P(a\le X\le b) are the same number.

  15. Recall the definition of the cumulative distribution function

    F(x)=P(X≤x)=∫−∞xf(u) duF(x)=P(X\le x)=\int_{-\infty}^{x}f(u)\,du

    FF accumulates the area under the density from the left.

  16. Recall how to recover the density from FF

    f(x)=dFdxf(x)=\frac{dF}{dx}

    Differentiation undoes the integration that produced FF.

  17. State the density of YY and its mean

    fY(y)=y18 (0≤y≤9),E(Y)=275f_{Y}(y)=\frac{\sqrt{y}}{18}\ \left(0\le y\le 9\right),\qquad E(Y)=\frac{27}{5}

    Both results follow from the cumulative function of YY.

Answer
fY(y)=y18 for 0≤y≤9, E(Y)=275f_{Y}(y)=\frac{\sqrt{y}}{18}\ \text{for}\ 0\le y\le 9,\ E(Y)=\frac{27}{5}
Question 4
9 markschallenging
The continuous random variable XX has probability density function f(x)={x80≤x≤40otherwisef(x)=\begin{cases}\frac{x}{8}&0\le x\le 4\\0&\text{otherwise}\end{cases}. The random variable YY is defined by Y=X3Y=X^{3}. Find the probability density function of YY, and hence find the exact value of E(Y)E(Y).
Show worked solution

Worked solution

  1. Start from the cumulative distribution function of YY

    FY(y)=P(X3≤y)=P(X≤y3)F_{Y}(y)=P\left(X^{3}\le y\right)=P\left(X\le \sqrt[3]{y}\right)

    The transformation is increasing, so the inequality is unchanged.

  2. Write down the cumulative distribution function of XX

    FX(x)=x216F_{X}(x)=\frac{x^{2}}{16}

    This comes from integrating the density of XX.

  3. Substitute to obtain FYF_{Y}

    FY(y)=y2316F_{Y}(y)=\frac{y^{\frac{2}{3}}}{16}

    Replace xx by y3\sqrt[3]{y}.

  4. Differentiate to obtain fYf_{Y}

    fY(y)=124y3f_{Y}(y)=\frac{1}{24 \sqrt[3]{y}}

    The density of YY is the derivative of its cumulative function.

  5. State the range of YY

    0≤y≤640\le y\le 64

    The support of XX maps onto this interval.

  6. Set up the integral for E(Y)E(Y)

    E(Y)=∫064y(124y3) dyE(Y)=\int_{0}^{64}y\left(\frac{1}{24 \sqrt[3]{y}}\right)\,dy

    The mean of YY uses the density of YY in the usual way.

  7. Integrate and evaluate the limits

    E(Y)=1285E(Y)=\frac{128}{5}

    This is the exact mean of YY.

  8. Check the answer against E(g(X))E\left(g(X)\right)

    ∫04(x3)f(x) dx=1285\int_{0}^{4}\left(x^{3}\right)f(x)\,dx=\frac{128}{5}

    Working directly with the density of XX gives the same value, as it must.

  9. Check that fYf_{Y} integrates to 11

    ∫064124y3 dy=1\int_{0}^{64}\frac{1}{24 \sqrt[3]{y}}\,dy=1

    A correct density always encloses unit area.

  10. Note that E(Y)E(Y) is not g(E(X))g\left(E(X)\right)

    1285≠51227\frac{128}{5}\neq \frac{512}{27}

    Transforming the mean is not the same as taking the mean of the transform.

  11. Recall the defining property of a probability density function

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty}f(x)\,dx=1

    The total area under the density curve is always exactly 11.

  12. Recall the second condition for a valid density

    f(x)≥0 for all xf(x)\ge0\ \text{for all}\ x

    A density can never be negative, because areas under it are probabilities.

  13. Recall that a probability is an area

    P(a<X<b)=∫abf(x) dxP(a<X<b)=\int_{a}^{b}f(x)\,dx

    For a continuous variable, probability is the area under the density.

  14. Note that single values carry no probability

    P(X=a)=0P(X=a)=0

    This is why P(a<X<b)P(a<X<b) and P(a≤X≤b)P(a\le X\le b) are the same number.

  15. Recall the definition of the cumulative distribution function

    F(x)=P(X≤x)=∫−∞xf(u) duF(x)=P(X\le x)=\int_{-\infty}^{x}f(u)\,du

    FF accumulates the area under the density from the left.

  16. State the density of YY and its mean

    fY(y)=124y3 (0≤y≤64),E(Y)=1285f_{Y}(y)=\frac{1}{24 \sqrt[3]{y}}\ \left(0\le y\le 64\right),\qquad E(Y)=\frac{128}{5}

    Both results follow from the cumulative function of YY.

Answer
fY(y)=124y3 for 0≤y≤64, E(Y)=1285f_{Y}(y)=\frac{1}{24 \sqrt[3]{y}}\ \text{for}\ 0\le y\le 64,\ E(Y)=\frac{128}{5}
Question 5
9 markschallenging
The continuous random variable XX has probability density function f(x)={kx(2−x)20≤x≤20otherwisef(x)=\begin{cases}k x \left(2 - x\right)^{2}&0\le x\le 2\\0&\text{otherwise}\end{cases}, where kk is a constant. Find the value of kk, the mode of XX and the exact value of Var(X)\text{Var}(X).
Show worked solution

Worked solution

  1. Use the fact that the total area under a density is 11

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty}f(x)\,dx=1

    This is the condition that fixes the unknown constant kk.

  2. Write the total area as an integral over the support

    ∫02kx(2−x)2 dx=1\int_{0}^{2}k x \left(2 - x\right)^{2}\,dx=1

    The density is zero outside the support, so only these pieces contribute.

  3. Integrate each piece

    [kx44−4kx33+2kx2]02\left[\frac{k x^{4}}{4} - \frac{4 k x^{3}}{3} + 2 k x^{2}\right]_{0}^{2}

    Each piece is a standard integral, linear in kk.

  4. Collect the terms and solve for kk

    4k3=1 ⇒ k=34\frac{4 k}{3}=1\ \Rightarrow\ k=\frac{3}{4}

    The total area is a multiple of kk, so kk follows at once.

  5. Write down the integral for the mean

    E(X)=∫−∞∞xf(x) dx=∫023x44−3x3+3x2 dxE(X)=\int_{-\infty}^{\infty}xf(x)\,dx=\int_{0}^{2}\frac{3 x^{4}}{4} - 3 x^{3} + 3 x^{2}\,dx

    Multiply the density by xx and integrate over the support.

  6. Integrate

    =[3x520−3x44+x3]02=\left[\frac{3 x^{5}}{20} - \frac{3 x^{4}}{4} + x^{3}\right]_{0}^{2}

    Antidifferentiate each piece of xf(x)xf(x).

  7. Evaluate the limits

    E(X)=45E(X)=\frac{4}{5}

    Substituting the limits gives the exact mean.

  8. Write down the integral for E(X2)E\left(X^{2}\right)

    E(X2)=∫−∞∞x2f(x) dx=∫023x54−3x4+3x3 dxE\left(X^{2}\right)=\int_{-\infty}^{\infty}x^{2}f(x)\,dx=\int_{0}^{2}\frac{3 x^{5}}{4} - 3 x^{4} + 3 x^{3}\,dx

    The same integral as for the mean, but with x2x^{2} in place of xx.

  9. Integrate and evaluate the limits

    E(X2)=45E\left(X^{2}\right)=\frac{4}{5}

    This is the second moment of XX about the origin.

  10. Apply the variance formula

    Var(X)=E(X2)−[E(X)]2=45−(45)2=425\text{Var}(X)=E\left(X^{2}\right)-\left[E(X)\right]^{2}=\frac{4}{5}-\left(\frac{4}{5}\right)^{2}=\frac{4}{25}

    Subtracting the square of the mean gives the variance.

  11. Differentiate the density

    f′(x)=9x24−6x+3f'(x)=\frac{9 x^{2}}{4} - 6 x + 3

    A stationary point of ff is the candidate for the maximum.

  12. Solve f′(x)=0f'(x)=0

    9x24−6x+3=0 ⇒ x=23\frac{9 x^{2}}{4} - 6 x + 3=0\ \Rightarrow\ x=\frac{2}{3}

    Only the root inside the support can be the mode.

  13. Confirm that this stationary point is a maximum

    f′′(23)=−3<0f''\left(\frac{2}{3}\right)=-3<0

    A negative second derivative confirms a maximum.

  14. Check that the density is non-negative throughout

    f(x)≥0 on 0≤x≤2f(x)\ge0\ \text{on}\ 0\le x\le 2

    A valid density never dips below the axis.

  15. State all the required quantities

    k=34, mode=23, Var(X)=425k=\frac{3}{4},\ \text{mode}=\frac{2}{3},\ \text{Var}(X)=\frac{4}{25}

    These are the exact values asked for.

Answer
k=34, mode=23, Var(X)=425k=\frac{3}{4},\ \text{mode}=\frac{2}{3},\ \text{Var}(X)=\frac{4}{25}

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