Porisma

Worked solution

Average absolute values and build the sum distribution

X takes values −2,0,2 with probabilities 1/4,1/2,1/4 respectively. Find Var⁡(X)\operatorname{Var}(X) and E(∣X∣)E(|X|).

Y=X1+X2Y=X_1+X_2, where X1,X2X_1,X_2 are independent observations of X. Find the distribution of Y, Var⁡(Y)\operatorname{Var}(Y) and E(Y+3)E(Y+3).

[J1989/II/7]

CJC · Review problems · Question 4

Answer

Var⁡(X)=2\operatorname{Var}(X)=2; E(∣X∣)=1E(|X|)=1; Y distribution above; Var⁡(Y)=4\operatorname{Var}(Y)=4; E(Y+3)=3E(Y+3)=3.

Retain the X distribution and moments

xP(X=x)P(X=x)
−21/41/4
01/21/2
21/41/4

By symmetry E(X)=0E(X)=0. Then E(X2)=(−2)2(1/4)+02(1/2)+22(1/4)=2E(X^2)=(-2)^2(1/4)+0^2(1/2)+2^2(1/4)=2 and Var⁡(X)=E(X2)−[E(X)]2=2−0=2\operatorname{Var}(X)=E(X^2)-[E(X)]^2=2-0=2.

E(∣X∣)=∣−2∣(1/4)+∣0∣(1/2)+∣2∣(1/4)=1E(|X|)=|-2|(1/4)+|0|(1/2)+|2|(1/4)=1.

How did this part go?

Retain every source sum-grid cell

X2X_2 / X1X_1−202
−2−4−20
0−202
2024
How did this part go?

Retain each source probability derivation

P(Y=−4)=P(X1=−2)P(X2=−2)=(1/4)(1/4)=1/16P(Y=-4)=P(X_1=-2)P(X_2=-2)=(1/4)(1/4)=1/16.

P(Y=−2)=P(X1=−2)P(X2=0)+P(X1=0)P(X2=−2)=(1/4)(1/2)+(1/2)(1/4)=1/4P(Y=-2)=P(X_1=-2)P(X_2=0)+P(X_1=0)P(X_2=-2)=(1/4)(1/2)+(1/2)(1/4)=1/4.

P(Y=0)=P(X1=0)P(X2=0)+P(X1=−2)P(X2=2)+P(X1=2)P(X2=−2)=(1/2)(1/2)+(1/4)(1/4)+(1/4)(1/4)=3/8P(Y=0)=P(X_1=0)P(X_2=0)+P(X_1=-2)P(X_2=2)+P(X_1=2)P(X_2=-2)=(1/2)(1/2)+(1/4)(1/4)+(1/4)(1/4)=3/8.

P(Y=2)=P(X1=2)P(X2=0)+P(X1=0)P(X2=2)=(1/4)(1/2)+(1/2)(1/4)=1/4P(Y=2)=P(X_1=2)P(X_2=0)+P(X_1=0)P(X_2=2)=(1/4)(1/2)+(1/2)(1/4)=1/4.

P(Y=4)=P(X1=2)P(X2=2)=(1/4)(1/4)=1/16P(Y=4)=P(X_1=2)P(X_2=2)=(1/4)(1/4)=1/16.

How did this part go?

Complete Y distribution

yP(Y=y)P(Y=y)
−41/161/16
−21/41/4
03/83/8
21/41/4
41/161/16
How did this part go?

Retain the source second-moment method and shifted mean

By symmetry E(Y)=0E(Y)=0. E(Y2)=(−4)2(1/16)+(−2)2(1/4)+02(3/8)+22(1/4)+42(1/16)=4E(Y^2)=(-4)^2(1/16)+(-2)^2(1/4)+0^2(3/8)+2^2(1/4)+4^2(1/16)=4. Thus Var⁡(Y)=E(Y2)−[E(Y)]2=4−0=4\operatorname{Var}(Y)=E(Y^2)-[E(Y)]^2=4-0=4. Finally E(Y+3)=E(Y)+E(3)=0+3=3E(Y+3)=E(Y)+E(3)=0+3=3.

How did this part go?

Common mistakes

E(|X|) is not |E(X)|. Sum-grid cells have weights from the two marginal distributions.

Solution notes

Every ask retained, including absolute-value bars. Full X table, nine sum-grid cells, all five pair-product derivations, complete Y table and every moment term retained. No printed marks.

Read the hint