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I'm trying to derive the distribution of a random variable $Y$ given that I know the distribution of a random variable $X$ and the relationship they share.

The $pdf$ of $X$ is expressed as:

$ f_{X} = \begin{cases} 1/3 & \text{if -2 $<$ $x$ < 1}\\ 0 & \text{otherwise} \end{cases} $

I also know that $Y = g(X) = |{X}|$

In problems like this, I learned one must first calculate the $CDF$ of $Y$ and then derived in relation to $y$.

$F_{Y}(y) = \mathbb{P}( Y \le y) = \mathbb{P}(|X| \leq y) = \mathbb{P}( X < y) + \mathbb{P}( X \geq -y) $

Given that $Y$ is the absolute value of $X$, the inequality can be seen as the area of triangles (as far as I understand it).

But considering these are continuous random variables, I can't see how I'll integrate to find $\mathbb{P}(.)$

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Here is a figure based on a simulation in R that suggests the answer. The simulation uses a million observations of $X \sim \mathsf{Unif}(-2,1).$ Then we show histograms of the samples of $X$ and of $Y = |X|.$

You have made a reasonable start on the formal derivation. Now, you need to express the right-hand side of the CDF of $Y$ as a function of $y$ and then take its derivative to get the PDF of $Y.$

 set.seed(2020)
 x = runif(10^6, -2, 1)                  # simulate sample of X's
 y = abs(x)                              # transform to get Y's
 par(mfrow=c(1,2))                       # make histogram of X's and of Y's
  hist(x, prob=T, br=12, col="skyblue2")
  hist(y, prob=T, br=12, col="skyblue2")
 par(mfrow=c(1,1))

enter image description here

The plots below (based on Empirical CDFs of the two samples) suggest the linear functions that make up the CDFs of $X$ and $Y.$

enter image description here

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  • $\begingroup$ This is an inventive solution, BruceET. Thank you. I was hoping for a more clear mathematical derivation. At least, the beginning of it so I could follow from there. =) $\endgroup$ – Mason Beau Mar 27 at 23:32

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