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Worked solution

Normalise the parameterised distribution before transforming variance

The probability function of X is

P(X=x)={(2x−1)θ,x=1,2,3,k,x=4,0,otherwise,P(X=x)=\begin{cases}(2x-1)\theta,&x=1,2,3,\\k,&x=4,\\0,&\text{otherwise},\end{cases}

where 0<θ<1/90<\theta<1/9.

(i) Show that k=1−9θk=1-9\theta. Find the distribution of X in terms of θ.

(ii) Find E(X)E(X) in terms of θ and hence show that Var⁡(X)=26θ−196θ2\operatorname{Var}(X)=26\theta-196\theta^2.

(iii) Y=a+bXY=a+bX, with a,b non-zero constants. Given Var⁡(Y)=b2/3\operatorname{Var}(Y)=b^2/3, find θ.

[2017/MJC/II/6]

CJC · Discussion · Question 5

Answer

k=1−9θk=1-9\theta; distribution above; E(X)=4−14θE(X)=4-14\theta; variance 26θ−196θ226\theta-196\theta^2; θ=(39−933)/588≈0.0144\theta=(39-\sqrt{933})/588\approx0.0144.

(i) Retain the source’s initial table and normalisation

xP(X=x)P(X=x) before substitution
1θ\theta
23θ3\theta
35θ5\theta
4kk

Summing all probabilities gives θ+3θ+5θ+k=1\theta+3\theta+5\theta+k=1, so k=1−9θk=1-9\theta.

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(i) Complete distribution after substituting k

xP(X=x)P(X=x)
1θ\theta
23θ3\theta
35θ5\theta
41−9θ1-9\theta

The stated interval ensures all four probabilities are positive; other values of X have probability 0.

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(ii) Mean with every source expansion term

E(X)=1(θ)+2(3θ)+3(5θ)+4(1−9θ)=θ+6θ+15θ+4−36θ=4−14θE(X)=1(\theta)+2(3\theta)+3(5\theta)+4(1-9\theta)=\theta+6\theta+15\theta+4-36\theta=4-14\theta.

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(ii) Second moment and variance

E(X2)=12(θ)+22(3θ)+32(5θ)+42(1−9θ)=θ+12θ+45θ+16−144θ=16−86θE(X^2)=1^2(\theta)+2^2(3\theta)+3^2(5\theta)+4^2(1-9\theta)=\theta+12\theta+45\theta+16-144\theta=16-86\theta. Hence Var⁡(X)=16−86θ−(4−14θ)2=16−86θ−(16−112θ+196θ2)=26θ−196θ2\operatorname{Var}(X)=16-86\theta-(4-14\theta)^2=16-86\theta-(16-112\theta+196\theta^2)=26\theta-196\theta^2.

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(iii) Cancel the non-zero scale factor

Var⁡(Y)=Var⁡(a+bX)=b2Var⁡(X)\operatorname{Var}(Y)=\operatorname{Var}(a+bX)=b^2\operatorname{Var}(X). Thus b2/3=b2(26θ−196θ2)b^2/3=b^2(26\theta-196\theta^2). Since b≠0, divide by b2b^2 to obtain 196θ2−26θ+1/3=0196\theta^2-26\theta+1/3=0.

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(iii) Apply the parameter restriction

The source calculator roots are θ≈0.0144 and θ≈0.118. The latter exceeds 1/9 and is rejected. Hence θ≈0.0144. An exact expression for the retained root is (39−933)/588(39-\sqrt{933})/588; the source’s rounded answer is preserved.

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Common mistakes

The second quadratic root is inadmissible under 0<θ<1/9. Do not retain both numerical roots.

Solution notes

All three asks, piecewise support/zero-elsewhere condition, initial and final source tables, every mean/second-moment/variance expansion, quadratic and both source roots/rejection retained. Exact root supplements the printed calculator approximation. No marks printed.

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