1
$\begingroup$

Suppose you have a random variable $X$ with a probability density function $f(x)$. Now, we define a new random variable $Y$ as follows:

\begin{equation} Y = \max(0,X-q), \end{equation} where $q$ is a constant.

What is the functional form for the probability density function $f(y)$? Specifically, is it possible to write the PDF of $y$ as a function $f(x)$ and $q$ somehow?

This page shows proofs for some similar but simpler cases: https://www.chem.purdue.edu/courses/chm621/text/stat/combined/constant/constant.htm

However, because my function is more complicated I have trouble applying the methodology given on that page to my case. Could somebody help me?

$\endgroup$
1
$\begingroup$

Functionally, the operation that transforms $X$ to $Y$ works as follows:

  • Shift the pdf of $X$ to the left by $q$ units: $X\to X-q$
  • Take all the probability of this new random variable being negative and crush it into a single point at 0: $X-q\to\max(0,X-q)$

Therefore $Y$ has the following pdf: $$g(x)=\begin{cases}0&x<0\\ c=\int_{-\infty}^qf(t)\,dt&x=0\\ f(x+q)&x>0\end{cases}$$ Assuming $f$ was continuous, this is an example of a mixed random variable – neither continuous, neither discrete, but combining the characteristics of both. The discrete part is the spike at $x=0$ of $Y$, a Dirac delta function, and the continuous part is $x>0$, which has not been affected by anything other than the shift.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.