Porisma

Worked solution

Normalize a symmetric law and average an absolute distance

The discrete random variable XX has probability function

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

where k>0k>0. (i) Show that k=1/6k=1/6 and state E(X)E(X). (ii) Find E(∣X−2∣)E(|X-2|). (iii) Find Var⁡(X)\operatorname{Var}(X).

[2006/IJC/II/23]

CJC · Lecture · Example 12

(i) Normalize and use symmetry

The completed Teacher table is

xx01234
P(X=x)P(X=x)2k2kkk0kk2k2k

2k+k+0+k+2k=12k+k+0+k+2k=1, so k=1/6k=1/6, as required. The masses are symmetric about x=2x=2, hence E(X)=2E(X)=2.

How did this part go?

(ii) Transform each value into an absolute distance

Retain the full Teacher-completed mapping table:

xx01234
P(X=x)P(X=x)2/62/61/61/601/61/62/62/6
∣x−2∣\lvert x-2\rvert∣0−2∣=2\lvert 0-2\rvert=2∣1−2∣=1\lvert 1-2\rvert=1∣2−2∣=0\lvert 2-2\rvert=0∣3−2∣=1\lvert 3-2\rvert=1∣4−2∣=2\lvert 4-2\rvert=2

E(∣X−2∣)=2(2/6)+1(1/6)+0+1(1/6)+2(2/6)=5/3E(|X-2|)=2(2/6)+1(1/6)+0+1(1/6)+2(2/6)=5/3.

How did this part go?

(iii) Calculate the second moment

The probabilities at x=0,1,2,3,4x=0,1,2,3,4 are 2/6,1/6,0,1/6,2/62/6,1/6,0,1/6,2/6. Thus E(X2)=02(2/6)+12(1/6)+22(0)+32(1/6)+42(2/6)=7E(X^2)=0^2(2/6)+1^2(1/6)+2^2(0)+3^2(1/6)+4^2(2/6)=7. Therefore Var⁡(X)=E(X2)−[E(X)]2=7−22=3\operatorname{Var}(X)=E(X^2)-[E(X)]^2=7-2^2=3. This working is completed in the Teacher copy.

How did this part go?

Common mistakes

The printed ask uses absolute value: E(|X−2|), which is 5/3. It is not E(X−2)=0. The listed centre x=2 has zero probability, but symmetry still gives mean 2.

Read the hint