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Worked solution

Square a discrete random variable

The random variable YY has distribution

yy−3-3−2-2−1-101
P(Y=y)P(Y=y)0.10.30.340.250.01

If V=Y2V=Y^2, find the probability distribution of VV.

CJC · Lecture · Example 5

Map each original value to its square

Keep the probability attached to each original value while completing the source mapping table:

yy−3-3−2-2−1-101
P(Y=y)P(Y=y)0.10.30.340.250.01
v=y2v=y^294101
How did this part go?

Merge original values that have the same image

Both Y=−1Y=-1 and Y=1Y=1 produce V=1V=1. Add their probabilities: P(V=1)=0.34+0.01=0.35P(V=1)=0.34+0.01=0.35. The completed Teacher table is

vv0149
P(V=v)P(V=v)0.250.01+0.34=0.350.01+0.34=0.350.30.1

The other masses are P(V=0)=P(Y=0)=0.25P(V=0)=P(Y=0)=0.25, P(V=4)=P(Y=−2)=0.3P(V=4)=P(Y=-2)=0.3 and P(V=9)=P(Y=−3)=0.1P(V=9)=P(Y=-3)=0.1. Their sum is 0.25+0.35+0.3+0.1=10.25+0.35+0.3+0.1=1 (a supplementary check).

How did this part go?

Common mistakes

Transform values, not probabilities. A many-to-one transformation combines the masses of distinct original values; do not list V=1 twice.

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