Hard Further Maths Combinations of random variables Questions

Challenging, exam-style Further Maths Combinations of random variables questions with worked solutions. Stretch yourself on the hardest modelling, linear-combination-of-normals, probability, normal-distribution problems.

modellinglinear-combination-of-normalsprobabilitynormal-distributionsum-of-poisson-variablespoisson-distribution
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The independent random variables XX and YY have X∼Po⁡(4)X\sim\operatorname{Po}\left(4\right) and Y∼Po⁡(3)Y\sim\operatorname{Po}\left(3\right). Which of the following statements about X−YX-Y is correct?
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Worked solution

  1. Write down the mean and the variance of each Poisson variable

    E(X)=Var⁡(X)=4,E(Y)=Var⁡(Y)=3E(X)=\operatorname{Var}(X)=4,\qquad E(Y)=\operatorname{Var}(Y)=3

    For a Poisson variable the mean and the variance are both λ\lambda.

  2. Find the mean of the difference

    E(X−Y)=4−3=1E(X-Y)=4-3=1

    Means subtract, exactly as the linearity of expectation says they must.

  3. Find the variance of the difference

    Var⁡(X−Y)=12×4+(−1)2×3=7\operatorname{Var}(X-Y)=1^{2}\times 4+\left(-1\right)^{2}\times 3=7

    The variances ADD, so the difference is more variable than either count.

  4. Compare the mean with the variance

    1≠71\neq 7

    A Poisson variable must have equal mean and variance, so X−YX-Y cannot be Poisson.

  5. Note the other obstruction

    P(X−Y<0)>0P\left(X-Y<0\right)>0

    A Poisson variable cannot take negative values, but a difference certainly can.

  6. Contrast with the SUM, which is Poisson

    X+Y∼Po⁡(7)X+Y\sim\operatorname{Po}\left(7\right)

    Adding independent Poisson variables preserves the family; subtracting does not.

  7. Check the mean of the sum against its variance

    E(X+Y)=Var⁡(X+Y)=7E\left(X+Y\right)=\operatorname{Var}\left(X+Y\right)=7

    For the sum the two agree, which is exactly what a Poisson requires.

  8. Write the difference as a linear combination

    X−Y=1×X+(−1)×YX-Y=1\times X+\left(-1\right)\times Y

    Seeing the coefficient −1-1 explicitly is what fixes the variance.

  9. Square the coefficients

    12=1,(−1)2=11^{2}=1,\qquad\left(-1\right)^{2}=1

    Both contributions therefore enter with a plus sign.

  10. Reject the claim that the variances subtract

    4−3=1≠74-3=1\neq 7

    That would make the difference LESS variable than XX alone, which is absurd.

  11. Give the standard deviation of the difference

    sd⁡(X−Y)=7=2.6458\operatorname{sd}\left(X-Y\right)=\sqrt{7}=2.6458

    The difference is more spread out than either count on its own.

  12. Note that the means DO subtract

    E(X−Y)=4−3=1E\left(X-Y\right)=4-3=1

    Linearity of expectation handles the minus sign exactly as written.

  13. Note what the difference is used for

    X−Y measures the excess of one count over the otherX-Y\ \text{measures the excess of one count over the other}

    Its mean and variance are still available, even though it has no standard name.

  14. Check that every contribution is positive

    Var⁡=7>0\operatorname{Var}=7>0

    A squared coefficient can never be negative, so a variance never shrinks when another independent variable is brought in.

  15. Note what independence has bought

    Var⁡(aX+bY)=a2Var⁡(X)+b2Var⁡(Y)+2abCov⁡(X,Y)\operatorname{Var}\left(aX+bY\right)=a^{2}\operatorname{Var}(X)+b^{2}\operatorname{Var}(Y)+2ab\operatorname{Cov}(X,Y)

    Independence makes the covariance zero; without it the last term survives.

  16. Select the correct statement

    E(X−Y)=1,Var⁡(X−Y)=7E(X-Y)=1,\qquad\operatorname{Var}(X-Y)=7

    Poisson variables add, but they do not subtract.

Answer
E(X−Y)=1, Var⁡(X−Y)=7E(X-Y)=1,\ \operatorname{Var}(X-Y)=7
Question 2
9 markschallenging
The independent random variables XX and YY have X∼N(20,9)X\sim N\left(20,9\right) and Y∼N(14,16)Y\sim N\left(14,16\right). Which of the following is the distribution of 3X−2Y+13X-2Y+1?
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Worked solution

  1. State the two independent normal distributions

    X∼N(20,9),Y∼N(14,16)X\sim N\left(20,9\right),\qquad Y\sim N\left(14,16\right)

    The second parameter of NN is the variance, not the standard deviation.

  2. Find the mean of the combination

    E(3X−2Y+1)=3×20+−2×14+1=33E\left(3X-2Y+1\right)=3\times 20+-2\times 14+1=33

    Expectation is linear; a constant is simply added on.

  3. Find the variance of the combination

    Var⁡(3X−2Y+1)=(3)2×9+(−2)2×16=145\operatorname{Var}\left(3X-2Y+1\right)=\left(3\right)^{2}\times 9+\left(-2\right)^{2}\times 16=145

    The coefficients are squared and the constant contributes nothing.

  4. Use the closure property of the normal distribution

    aX+bY+c∼N(aμX+bμY+c, a2σX2+b2σY2)aX+bY+c\sim N\left(a\mu_{X}+b\mu_{Y}+c,\ a^{2}\sigma_{X}^{2}+b^{2}\sigma_{Y}^{2}\right)

    A linear combination of INDEPENDENT normal variables is normal.

  5. Find the contribution of XX to the mean

    3×E(X)=3×20=603\times E(X)=3\times 20=60

    Each coefficient multiplies its own mean, and only its own mean.

  6. Find the contribution of YY to the mean

    −2×E(Y)=−2×14=−28-2\times E(Y)=-2\times 14=-28

    Each coefficient multiplies its own mean, and only its own mean.

  7. Add the constant to the mean

    E(…+1): add 1E\left(\ldots+1\right):\ \text{add }1

    A constant is added to the mean exactly as it stands.

  8. Find the contribution of XX to the variance

    (3)2×Var⁡(X)=9×9=81\left(3\right)^{2}\times\operatorname{Var}(X)=9\times 9=81

    The coefficient is squared, so this contribution is positive whether 33 is positive or negative.

  9. Find the contribution of YY to the variance

    (−2)2×Var⁡(Y)=4×16=64\left(-2\right)^{2}\times\operatorname{Var}(Y)=4\times 16=64

    The coefficient is squared, so this contribution is positive whether −2-2 is positive or negative.

  10. Confirm that the constant contributes nothing

    Var⁡(1)=0\operatorname{Var}\left(1\right)=0

    A constant never varies, so it adds no spread.

  11. Add the separate contributions

    81+64=14581+64=145

    Every contribution is added; none of them can cancel another.

  12. Reject the option that subtracts the variances

    81−64=17 is wrong81-64=17\ \text{is wrong}

    Squaring the coefficient −2-2 removes the minus sign entirely.

  13. Reject the option that quotes a standard deviation

    145=12.0416 is a spread, not a variance\sqrt{145}=12.0416\ \text{is a spread, not a variance}

    The second parameter of NN must be the variance.

  14. Check the effect of the constant

    it shifts the mean by 1 and leaves the variance alone\text{it shifts the mean by }1\text{ and leaves the variance alone}

    A constant is not random, so it can add no spread.

  15. Select the correct distribution

    3X−2Y+1∼N(33,145)3X-2Y+1\sim N\left(33,145\right)

    The mean and the variance determine the normal distribution completely.

Answer
3X−2Y+1∼N(33,145)3X-2Y+1\sim N\left(33,145\right)
Question 3
9 markschallenging
A chocolatier fills a box with truffles. The mass of a randomly chosen item, in grams, is X∼N(30,4)X\sim N\left(30,4\right), and X1X_{1} and X2X_{2} are the masses of two independently chosen items. Which of the following correctly gives P(2X>63)P\left(2X>63\right) and P(X1+X2>63)P\left(X_{1}+X_{2}>63\right)?
Show worked solution

Worked solution

  1. Write down the distribution of a single observation

    X∼N(30,4)X\sim N\left(30,4\right)

    Both combinations are built from this one distribution.

  2. Find the distribution of 2X2X

    2X∼N(60,16)2X\sim N\left(60,16\right)

    One observation, doubled: the variance is multiplied by 22=42^{2}=4.

  3. Find the distribution of X1+X2X_{1}+X_{2}

    X1+X2∼N(60,8)X_{1}+X_{2}\sim N\left(60,8\right)

    Two independent observations: the two variances add.

  4. Note that the means are identical

    E(2X)=E(X1+X2)=60E(2X)=E\left(X_{1}+X_{2}\right)=60

    The distributions differ only in their spread.

  5. Standardise the boundary for 2X2X

    z=63−6016=0.7500z=\frac{63-60}{\sqrt{16}}=0.7500

    Divide by the standard deviation of 2X2X.

  6. Standardise the boundary for X1+X2X_{1}+X_{2}

    z=63−608=1.0607z=\frac{63-60}{\sqrt{8}}=1.0607

    The same boundary, but a smaller standard deviation, so a larger zz.

  7. Evaluate the two probabilities

    P(2X>63)=0.2266,P(X1+X2>63)=0.1444P\left(2X>63\right)=0.2266,\qquad P\left(X_{1}+X_{2}>63\right)=0.1444

    The wider distribution puts more probability beyond the boundary.

  8. Compare the two standard deviations

    sd⁡(2X)=4.0000,sd⁡(X1+X2)=2.8284\operatorname{sd}\left(2X\right)=4.0000,\qquad\operatorname{sd}\left(X_{1}+X_{2}\right)=2.8284

    Doubling doubles the spread; sampling twice multiplies it only by 2\sqrt{2}.

  9. Explain why 2X2X has the larger tail probability

    same mean, larger σ ⇒ fatter tail\text{same mean, larger }\sigma\ \Rightarrow\ \text{fatter tail}

    The boundary lies above the common mean, so the more variable variable exceeds it more often.

  10. Reject the claim that the two probabilities are equal

    0.2266≠0.14440.2266\neq 0.1444

    Equal means do not imply equal probabilities: the spreads differ.

  11. Reject the two lower-tail values

    1−0.2266=0.7734,1−0.1444=0.85561-0.2266=0.7734,\qquad 1-0.1444=0.8556

    These are P(⋅<63)P(\cdot<63), the complements, not the required probabilities.

  12. Explain the modelling difference

    one item weighed twice ≠ two items weighed once\text{one item weighed twice}\ \neq\ \text{two items weighed once}

    Only X1+X2X_{1}+X_{2} describes a random sample of two items.

  13. See what would happen without independence

    X2=X1 ⇒ X1+X2=2X1X_{2}=X_{1}\ \Rightarrow\ X_{1}+X_{2}=2X_{1}

    The two probabilities would then coincide; independence is what separates them.

  14. Check both answers are valid probabilities

    0<0.2266<1,0<0.1444<10<0.2266<1,\qquad 0<0.1444<1

    Every probability obtained must lie between 00 and 11.

  15. Note that a mean comparison could not separate them

    E(2X)−E(X1+X2)=0E\left(2X\right)-E\left(X_{1}+X_{2}\right)=0

    Only the variance distinguishes the two constructions.

  16. Note what independence has bought

    Var⁡(aX+bY)=a2Var⁡(X)+b2Var⁡(Y)+2abCov⁡(X,Y)\operatorname{Var}\left(aX+bY\right)=a^{2}\operatorname{Var}(X)+b^{2}\operatorname{Var}(Y)+2ab\operatorname{Cov}(X,Y)

    Independence makes the covariance zero; without it the last term survives.

  17. Select the correct pair of probabilities

    P(2X>63)=0.2266,P(X1+X2>63)=0.1444P\left(2X>63\right)=0.2266,\qquad P\left(X_{1}+X_{2}>63\right)=0.1444

    Equal means, unequal variances, and therefore unequal tail probabilities.

Answer
P(2X>63)=0.2266, P(X1+X2>63)=0.1444P\left(2X>63\right)=0.2266,\ P\left(X_{1}+X_{2}>63\right)=0.1444
Question 4
9 markschallenging
The random variable XX has E(X)=9E(X)=9 and Var⁡(X)=7\operatorname{Var}(X)=7. The random variables X1X_{1}, X2X_{2} and X3X_{3} are independent, each having the same distribution as XX. Find Var⁡(X1+X2)−Var⁡(X1−X2)\operatorname{Var}\left(X_{1}+X_{2}\right)-\operatorname{Var}\left(X_{1}-X_{2}\right).
Show worked solution

Worked solution

  1. Find the variance of the sum

    Var⁡(X1+X2)=7+7=14\operatorname{Var}\left(X_{1}+X_{2}\right)=7+7=14

    Both coefficients are +1+1, so both variances are added.

  2. Find the variance of the difference

    Var⁡(X1−X2)=12×7+(−1)2×7=14\operatorname{Var}\left(X_{1}-X_{2}\right)=1^{2}\times 7+\left(-1\right)^{2}\times 7=14

    Squaring −1-1 gives +1+1, so the variances are added here too.

  3. Subtract the two results

    14−14=014-14=0

    The sum and the difference of two independent copies are equally variable.

  4. Write down the variance of one observation

    Var⁡(X)=Var⁡(X1)=Var⁡(X2)=7\operatorname{Var}(X)=\operatorname{Var}(X_{1})=\operatorname{Var}(X_{2})=7

    Every copy of XX carries exactly the same variance.

  5. Derive the variance of the sum from the general rule

    Var⁡(X1+X2)=12Var⁡(X)+12Var⁡(X)=2Var⁡(X)=14\operatorname{Var}\left(X_{1}+X_{2}\right)=1^{2}\operatorname{Var}(X)+1^{2}\operatorname{Var}(X)=2\operatorname{Var}(X)=14

    Two independent variables, each with coefficient 11.

  6. Derive the variance of the doubled observation

    Var⁡(2X)=22Var⁡(X)=4Var⁡(X)=28\operatorname{Var}\left(2X\right)=2^{2}\operatorname{Var}(X)=4\operatorname{Var}(X)=28

    One variable with coefficient 22: the coefficient is squared.

  7. Compare the means of the two constructions

    E(2X)=E(X1+X2)=18E\left(2X\right)=E\left(X_{1}+X_{2}\right)=18

    The means agree exactly, which is why the two are so easily confused.

  8. See what the independence is doing

    X2=X1 ⇒ Var⁡(X1+X2)=Var⁡(2X1)=28X_{2}=X_{1}\ \Rightarrow\ \operatorname{Var}\left(X_{1}+X_{2}\right)=\operatorname{Var}\left(2X_{1}\right)=28

    If the second observation merely copied the first, the sum would be as variable as 2X2X; independence is what reduces the spread.

  9. State the general results

    Var⁡(nX)=n2Var⁡(X),Var⁡(X1+⋯+Xn)=nVar⁡(X)\operatorname{Var}\left(nX\right)=n^{2}\operatorname{Var}(X),\qquad\operatorname{Var}\left(X_{1}+\cdots+X_{n}\right)=n\operatorname{Var}(X)

    Scaling squares the multiplier; sampling merely adds the variances.

  10. Compare the two standard deviations

    sd⁡(2X)=5.2915,sd⁡(X1+X2)=3.7417\operatorname{sd}\left(2X\right)=5.2915,\qquad\operatorname{sd}\left(X_{1}+X_{2}\right)=3.7417

    Doubling doubles the standard deviation; sampling multiplies it by 2\sqrt{2}.

  11. Note the ratio of the two variances

    Var⁡(2X)Var⁡(X1+X2)=4Var⁡(X)2Var⁡(X)=2\frac{\operatorname{Var}\left(2X\right)}{\operatorname{Var}\left(X_{1}+X_{2}\right)}=\frac{4\operatorname{Var}(X)}{2\operatorname{Var}(X)}=2

    The ratio is always 22, whatever the variance of XX happens to be.

  12. Explain why the sample is the less variable of the two

    independent fluctuations partly cancel\text{independent fluctuations partly cancel}

    A high X1X_{1} can be offset by a low X2X_{2}; doubling one value offers no such cancellation.

  13. Note that the difference is as variable as the sum

    Var⁡(X1−X2)=Var⁡(X1+X2)=14\operatorname{Var}\left(X_{1}-X_{2}\right)=\operatorname{Var}\left(X_{1}+X_{2}\right)=14

    Squaring the −1-1 leaves the two variances still being added.

  14. Contrast with the mean of the two observations

    Var⁡(X1+X22)=14×14=3.5\operatorname{Var}\left(\frac{X_{1}+X_{2}}{2}\right)=\frac{1}{4}\times 14=3.5

    Averaging halves the variance; doubling quadruples it.

  15. Note the three-observation version

    Var⁡(3X)=63,Var⁡(X1+X2+X3)=21\operatorname{Var}\left(3X\right)=63,\qquad\operatorname{Var}\left(X_{1}+X_{2}+X_{3}\right)=21

    The gap widens as more observations are involved.

  16. State the value

    Var⁡(X1+X2)−Var⁡(X1−X2)=0\operatorname{Var}\left(X_{1}+X_{2}\right)-\operatorname{Var}\left(X_{1}-X_{2}\right)=0

    This is the required value.

Answer
Var⁡(X1+X2)−Var⁡(X1−X2)=0\operatorname{Var}\left(X_{1}+X_{2}\right)-\operatorname{Var}\left(X_{1}-X_{2}\right)=0
Question 5
9 markschallenging
The independent random variables XX, YY and WW have E(X)=7E(X)=7, Var⁡(X)=3\operatorname{Var}(X)=3, E(Y)=11E(Y)=11, Var⁡(Y)=8\operatorname{Var}(Y)=8, E(W)=2E(W)=2 and Var⁡(W)=5\operatorname{Var}(W)=5. Find Var⁡(2X−3Y+4W−5)\operatorname{Var}\left(2X-3Y+4W-5\right).
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Worked solution

  1. Quote the rule for the variance of this combination

    Var⁡(2X−3Y+4W−5)=(2)2Var⁡(X)+(−3)2Var⁡(Y)+(4)2Var⁡(W)\operatorname{Var}\left(2X-3Y+4W-5\right)=\left(2\right)^{2}\operatorname{Var}(X)+\left(-3\right)^{2}\operatorname{Var}(Y)+\left(4\right)^{2}\operatorname{Var}(W)

    Each coefficient is SQUARED, and the constant term contributes nothing.

  2. Substitute the given variances

    Var⁡(2X−3Y+4W−5)=(2)2×3+(−3)2×8+(4)2×5\operatorname{Var}\left(2X-3Y+4W-5\right)=\left(2\right)^{2}\times 3+\left(-3\right)^{2}\times 8+\left(4\right)^{2}\times 5

    Replace each variance by the number supplied in the question.

  3. Evaluate the squared coefficients

    Var⁡(2X−3Y+4W−5)=4×3+9×8+16×5\operatorname{Var}\left(2X-3Y+4W-5\right)=4\times 3+9\times 8+16\times 5

    A negative coefficient becomes positive once it is squared.

  4. Find the contribution of XX to the variance

    (2)2×Var⁡(X)=4×3=12\left(2\right)^{2}\times\operatorname{Var}(X)=4\times 3=12

    The coefficient is squared, so this contribution is positive whether 22 is positive or negative.

  5. Find the contribution of YY to the variance

    (−3)2×Var⁡(Y)=9×8=72\left(-3\right)^{2}\times\operatorname{Var}(Y)=9\times 8=72

    The coefficient is squared, so this contribution is positive whether −3-3 is positive or negative.

  6. Find the contribution of WW to the variance

    (4)2×Var⁡(W)=16×5=80\left(4\right)^{2}\times\operatorname{Var}(W)=16\times 5=80

    The coefficient is squared, so this contribution is positive whether 44 is positive or negative.

  7. Confirm that the constant contributes nothing

    Var⁡(−5)=0\operatorname{Var}\left(-5\right)=0

    A constant never varies, so it adds no spread.

  8. Add the separate contributions

    12+72+80=16412+72+80=164

    Every contribution is added; none of them can cancel another.

  9. Check the sign trap

    (−1)2=+1\left(-1\right)^{2}=+1

    Subtracting a variable still ADDS its variance; variances never cancel.

  10. Compare with the all-positive combination

    Var⁡(2X−3Y+4W−5)=Var⁡(2X+3Y+4W−5)=164\operatorname{Var}\left(2X-3Y+4W-5\right)=\operatorname{Var}\left(2X+3Y+4W-5\right)=164

    Changing the sign of a coefficient cannot change the variance.

  11. Write down the information given

    E(X)=7,E(Y)=11,E(W)=2,Var⁡(X)=3,Var⁡(Y)=8,Var⁡(W)=5E(X)=7,\qquad E(Y)=11,\qquad E(W)=2,\qquad \operatorname{Var}(X)=3,\qquad \operatorname{Var}(Y)=8,\qquad \operatorname{Var}(W)=5

    Everything needed for the combination is in this list.

  12. Check that every contribution is positive

    Var⁡=164>0\operatorname{Var}=164>0

    A squared coefficient can never be negative, so a variance never shrinks when another independent variable is brought in.

  13. Confirm the variance from its definition

    E((2X−3Y+4W−5)2)−[E(2X−3Y+4W−5)]2=420−256=164E\left(\left(2X-3Y+4W-5\right)^{2}\right)-\left[E\left(2X-3Y+4W-5\right)\right]^{2}=420-256=164

    The rule used above is only a shortcut for Var⁡=E(L2)−[E(L)]2\operatorname{Var}=E(L^{2})-\left[E(L)\right]^{2}.

  14. Note what independence has bought

    Var⁡(aX+bY)=a2Var⁡(X)+b2Var⁡(Y)+2abCov⁡(X,Y)\operatorname{Var}\left(aX+bY\right)=a^{2}\operatorname{Var}(X)+b^{2}\operatorname{Var}(Y)+2ab\operatorname{Cov}(X,Y)

    Independence makes the covariance zero; without it the last term survives.

  15. State the variance

    Var⁡(2X−3Y+4W−5)=164\operatorname{Var}\left(2X-3Y+4W-5\right)=164

    This is the variance of the combination.

Answer
Var⁡(2X−3Y+4W−5)=164\operatorname{Var}\left(2X-3Y+4W-5\right)=164

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