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Worked solution

Normalize a piecewise mass function

Find cc such that the following is a probability distribution, then find P(∣X−12∣>14)P(|X-\frac12|>\frac14).

P(X=x)={cx2,x=−1,−12,x,x=0,cx,x=34,1,0,otherwise.P(X=x)=\begin{cases}cx^2,&x=-1,-\frac12,\\x,&x=0,\\cx,&x=\frac34,1,\\0,&\text{otherwise}.\end{cases}

CJC · Lecture · Example 4

Find c using all listed values

P(X=−1)+P(X=−12)+P(X=0)+P(X=34)+P(X=1)=1P(X=-1)+P(X=-\frac12)+P(X=0)+P(X=\frac34)+P(X=1)=1. Substitute the correct piece in each case: c(−1)2+c(−12)2+0+c(34)+c(1)=1c(-1)^2+c(-\frac12)^2+0+c(\frac34)+c(1)=1. Hence c+14c+34c+c=1c+\frac14c+\frac34c+c=1, giving c=1/3c=1/3. These masses are nonnegative and sum to one; the listed value zero has zero mass.

How did this part go?

Method 1: split the absolute-value event

P(∣X−12∣>14)=P(X−12>14)+P(X−12<−14)=P(X>34)+P(X<14)P(|X-\frac12|>\frac14)=P(X-\frac12>\frac14)+P(X-\frac12<-\frac14)=P(X>\frac34)+P(X<\frac14). From the listed values this is P(X=1)+P(X=−1)+P(X=−12)+P(X=0)=13(1)+13(−1)2+13(−12)2+0=3/4P(X=1)+P(X=-1)+P(X=-\frac12)+P(X=0)=\frac13(1)+\frac13(-1)^2+\frac13(-\frac12)^2+0=3/4.

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Method 2: take the closed-interval complement

P(∣X−12∣>14)=1−P(∣X−12∣≤14)=1−P(−14≤X−12≤14)=1−P(14≤X≤34)P(|X-\frac12|>\frac14)=1-P(|X-\frac12|\le\frac14)=1-P(-\frac14\le X-\frac12\le\frac14)=1-P(\frac14\le X\le\frac34). The only listed value in this interval is 3/43/4. Thus 1−P(X=34)=1−13(34)=3/41-P(X=\frac34)=1-\frac13(\frac34)=3/4. This is the completion supplied in the Teacher copy.

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Common mistakes

The strict absolute-value inequality excludes X=3/4. Its complement includes both endpoints. Evaluate negative x with the cx² branch; using cx there would create invalid negative probabilities.

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