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In an old homework I found from my university while preparing for the incoming year, I found this question:

Find PDF of Y=g(X) while:

$g(x)=x-c$ for $x>c$

$g(x)=0$ for $-c<x<c$

$g(x)=x+c$ for $x<-c$

Now the school’s answer is that since we have delta on 0, then the rest of PDF is zero. BUT THE REST IS WELL DEFINED! For x>c we clearly don’t get PDF zero. We get the PDF of X=0.5, X=2, X=3 etc. THAT makes the CDF of g(X) explode and that makes this transformation of random variable to be undefined.

Am I wrong? Could someone correct me and explain me?

I claim that we didn’t get the PDF from 100% mathematical calculations and that we got it from assumptions and conclusion we own assumed and concluded without math and hence by the CDF which is greater than 1, we have a contradiction => hence an undefined random variable = impossible transformation.

Vitali Pom
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1 Answers1

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You are partly correct. The school's answer is wrong, the PDF is not zero, it doesn't exist. However the distribution is a perfectly good one. It is a mixture of a discrete random variable which is always zero, and a continuous random variable. A graph of the cumulative probabilities is continuous except for a jump at zero which corresponds to the probability that X=0.

Terry Moore
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