Porisma

Worked solution

Use a weighted difference grid for two independent totals

An unbiased disc has one dot on one side and two dots on the other. The disc and an unbiased ordinary die are thrown; X is the sum of the disc dots and the dots on top of the die. Tabulate the distribution of X.

Show that P(X≥4∣X≤7)=8/11P(X\ge4\mid X\le7)=8/11. Write down E(X)E(X) and show that Var⁡(X)=19/6\operatorname{Var}(X)=19/6.

Two independent observations X1,X2X_1,X_2 are taken.

(i) Find Var⁡(X1−X2)\operatorname{Var}(X_1-X_2).

(ii) Find P(X1−X2≥5)P(X_1-X_2\ge5).

[J1995/II/6]

CJC · Review problems · Question 3

Answer

P(X≥4∣X≤7)=8/11P(X\ge4\mid X\le7)=8/11; E(X)=5E(X)=5; variance 19/6; (i)19/3; (ii)5/144.

Retain the complete disc-and-die sum grid

The source uses independent devices, giving twelve equiprobable ordered outcomes.

Disc / die123456
1234567
2345678
How did this part go?

Full distribution

xP(X=x)P(X=x)
21/121/12
32/122/12
42/122/12
52/122/12
62/122/12
72/122/12
81/121/12
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Conditional probability

P(X≥4∣X≤7)=P(X≥4∩X≤7)/P(X≤7)=[P(X=4)+P(X=5)+P(X=6)+P(X=7)]/P(X≤7)=(8/12)/(11/12)=8/11P(X\ge4\mid X\le7)=P(X\ge4\cap X\le7)/P(X\le7)=[P(X=4)+P(X=5)+P(X=6)+P(X=7)]/P(X\le7)=(8/12)/(11/12)=8/11.

How did this part go?

Symmetry and every second-moment term

By symmetry E(X)=5E(X)=5. Also E(X2)=22(1/12)+32(2/12)+42(2/12)+52(2/12)+62(2/12)+72(2/12)+82(1/12)=169/6E(X^2)=2^2(1/12)+3^2(2/12)+4^2(2/12)+5^2(2/12)+6^2(2/12)+7^2(2/12)+8^2(1/12)=169/6. Hence Var⁡(X)=169/6−52=19/6\operatorname{Var}(X)=169/6-5^2=19/6.

How did this part go?

(i) Add independent variances despite the subtraction

Var⁡(X1−X2)=Var⁡(X1)+Var⁡(X2)=2Var⁡(X)=2(19/6)=19/3\operatorname{Var}(X_1-X_2)=\operatorname{Var}(X_1)+\operatorname{Var}(X_2)=2\operatorname{Var}(X)=2(19/6)=19/3.

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(ii) Retain all 49 difference-grid cells

Rows give X1X_1, columns X2X_2; each cell gives X1−X2X_1-X_2.

X1X_1 / X2X_22345678
20−1−2−3−4−5−6
310−1−2−3−4−5
4210−1−2−3−4
53210−1−2−3
643210−1−2
7543210−1
86543210

Cells are not equiprobable: multiply the marginal probabilities for each ordered pair.

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(ii) Retain all three contributing pair products

P(X1−X2≥5)=P(X1−X2=5)+P(X1−X2=6)=P(X1=7)P(X2=2)+P(X1=8)P(X2=3)+P(X1=8)P(X2=2)=(1/6)(1/12)+(1/12)(1/6)+(1/12)(1/12)=5/144P(X_1-X_2\ge5)=P(X_1-X_2=5)+P(X_1-X_2=6)=P(X_1=7)P(X_2=2)+P(X_1=8)P(X_2=3)+P(X_1=8)P(X_2=2)=(1/6)(1/12)+(1/12)(1/6)+(1/12)(1/12)=5/144.

How did this part go?

Common mistakes

Uniform elementary disc/die outcomes do not make the seven possible totals equally likely.

Solution notes

All initial asks and both numbered asks retained. Full 12 sum cells, seven distribution entries, conditional calculation, all moment terms, full 49 difference cells and three final pair products retained. No printed marks.

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