Further Maths Combinations of random variables Practice Questions

Free Further Maths Combinations of random variables practice questions with full step-by-step worked solutions. Covers linear-transformation, expectation, variance, scaling-versus-sampling. Practise exam-style problems and check your method.

linear-transformationexpectationvariancescaling-versus-samplingindependencesum-of-poisson-variables
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The random variable XX has E(X)=6E(X)=6 and Var⁡(X)=4\operatorname{Var}(X)=4. Find E(3X+5)E\left(3X+5\right).
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Worked solution

  1. Quote the rule for the expectation of this combination

    E(3X+5)=3E(X)+5E\left(3X+5\right)=3E(X)+5

    Expectation is linear, so each coefficient simply multiplies its own mean.

  2. Substitute the given means

    E(3X+5)=3×6+5E\left(3X+5\right)=3\times 6+5

    Replace each EE by the number supplied in the question.

  3. State the expectation

    E(3X+5)=23E\left(3X+5\right)=23

    This is the mean of the combination.

Answer
E(3X+5)=23E\left(3X+5\right)=23
Question 2
2 markseasy
The independent random variables XX and YY have Var⁡(X)=9\operatorname{Var}(X)=9 and Var⁡(Y)=4\operatorname{Var}(Y)=4. Which of the following statements about Var⁡(X−Y)\operatorname{Var}\left(X-Y\right) is correct?
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Worked solution

  1. Write down the given variances

    Var⁡(X)=9,Var⁡(Y)=4\operatorname{Var}(X)=9,\qquad\operatorname{Var}(Y)=4

    Only the variances matter here; the means are irrelevant.

  2. Write the difference as a linear combination

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

    Seeing the −1-1 explicitly is the whole point.

  3. Square each coefficient

    Var⁡(X−Y)=12×9+(−1)2×4\operatorname{Var}(X-Y)=1^{2}\times 9+\left(-1\right)^{2}\times 4

    Squaring the −1-1 turns it into +1+1.

  4. Select the correct statement

    Var⁡(X−Y)=Var⁡(X)+Var⁡(Y)=13\operatorname{Var}(X-Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)=13

    Variances of independent variables always add, whatever the signs.

Answer
Var⁡(X−Y)=13\operatorname{Var}(X-Y)=13
Question 3
4 marksintermediate
The independent random variables XX and YY have Var⁡(X)=9\operatorname{Var}(X)=9 and Var⁡(Y)=16\operatorname{Var}(Y)=16. Which of the following is the standard deviation of X+YX+Y?
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Worked solution

  1. Add the variances, not the standard deviations

    Var⁡(X+Y)=9+16=25\operatorname{Var}\left(X+Y\right)=9+16=25

    This is the only quantity that is additive for independent variables.

  2. Take the square root of the total variance

    sd⁡(X+Y)=25=5\operatorname{sd}\left(X+Y\right)=\sqrt{25}=5

    The root is taken once, after the variances have been combined.

  3. Reject the sum of the standard deviations

    9+16=7≠5\sqrt{9}+\sqrt{16}=7\neq 5

    Standard deviations are never added: this is the classic error.

  4. Check with the triangle picture

    sd⁡(X+Y)2=sd⁡(X)2+sd⁡(Y)2\operatorname{sd}\left(X+Y\right)^{2}=\operatorname{sd}(X)^{2}+\operatorname{sd}(Y)^{2}

    Standard deviations combine like the sides of a right-angled triangle.

  5. Reject the variance itself as an answer

    25 is a variance, not a standard deviation25\ \text{is a variance, not a standard deviation}

    A variance is in squared units, so it cannot be the spread.

  6. Confirm by squaring the answer back

    (5)2=25\left(5\right)^{2}=25

    Squaring must return the combined variance.

  7. Select the correct standard deviation

    sd⁡(X+Y)=5\operatorname{sd}\left(X+Y\right)=5

    Variances add, and only then is the square root taken.

Answer
sd⁡(X+Y)=5\operatorname{sd}\left(X+Y\right)=5
Question 4
6 markshard
The independent random variables XX and YY have X∼Po⁡(3)X\sim\operatorname{Po}\left(3\right) and Y∼Po⁡(4)Y\sim\operatorname{Po}\left(4\right). Which of the following is the distribution of X+YX+Y?
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Worked solution

  1. Write down the parameters given

    X∼Po⁡(3),Y∼Po⁡(4)X\sim\operatorname{Po}\left(3\right),\qquad Y\sim\operatorname{Po}\left(4\right)

    Each count is Poisson with its own rate.

  2. Recall the additive property of the Poisson distribution

    Po⁡(λ1)+Po⁡(λ2)=Po⁡(λ1+λ2)(independent)\operatorname{Po}\left(\lambda_{1}\right)+\operatorname{Po}\left(\lambda_{2}\right)=\operatorname{Po}\left(\lambda_{1}+\lambda_{2}\right)\quad\text{(independent)}

    The parameters ADD; they are not multiplied or averaged.

  3. Add the parameters

    λ=3+4=7\lambda=3+4=7

    The combined count occurs at the combined rate.

  4. Check with the mean and the variance

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

    Both the means and the variances add, and for a Poisson they must be equal.

  5. Justify the additivity by convolution

    P(X+Y=s)=∑j=0sP(X=j)P(Y=s−j)P\left(X+Y=s\right)=\sum_{j=0}^{s}P(X=j)P(Y=s-j)

    Carrying out this sum gives exactly e−λλss!e^{-\lambda}\frac{\lambda^{s}}{s!} with λ=7\lambda=7: that is the proof, not merely a rule.

  6. Reject the product of the parameters

    λ≠3×4=12\lambda\neq 3\times 4=12

    Rates add over independent streams; they are not multiplied.

  7. Reject the average of the parameters

    λ≠72=3.5\lambda\neq\frac{7}{2}=3.5

    Averaging would describe one typical stream, not the two combined.

  8. Reject the normal option

    X+Y takes only the values 0,1,2,…X+Y\ \text{takes only the values }0,1,2,\dots

    The combined count is discrete, so it cannot have a normal distribution.

  9. Record the modelling assumption

    X and Y independentX\ \text{and}\ Y\ \text{independent}

    The additivity of the Poisson distribution fails without independence.

  10. Note that the difference is NOT Poisson

    Var⁡(X−Y)=7≠E(X−Y)\operatorname{Var}\left(X-Y\right)=7\neq E\left(X-Y\right)

    Only the SUM keeps the Poisson form; the difference cannot.

  11. Select the correct distribution

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

    A sum of independent Poisson variables is Poisson with the summed parameter.

Answer
X+Y∼Po⁡(7)X+Y\sim \operatorname{Po}\left(7\right)
Question 5
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

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