If X ∼ Uniform(−1, 1) find the pdf of Y = |X|

[BobblyHats97]

This question is killing me.

I know the graph is non-monotonic so i have to split up finding F(Y) for -1<Y and Y<1 but then what do I do with the modulus? >.<

Any help would be greatly appreciated! Thank you so much x

 

[andrewkirk]

You know the pdf of X, which is as simple as a pdf can be.
You have correctly assessed that the next step is to work out the cdf $F_Y$ of Y. THat is, for a given $y\in[0,1]$, we need to work out what is $F_Y(y)$?

$F_Y(y)$ is the probability that $Y\leq y$. What values of $X$ produce a value of $Y$ that satisfies that? Can you work out the probability that $X$ has such a value?

 

[mathman]

By inspection, Y is uniform between 0 and 1.

 

[WWGD]

YOu can use the change-of-variable formula for a random variable: https://onlinecourses.science.psu.edu/stat414/node/157

 

[andrewkirk]

YOu can use the change-of-variable formula for a random variable: https://onlinecourses.science.psu.edu/stat414/node/157

Unfortunately that only works for increasing or decreasing functions, so it can't be applied to the absolute value function.

 

[WWGD]

Unfortunately that only works for increasing or decreasing functions, so it can't be applied to the absolute value function.

Why not apply it piecewise, from -1 to 0 then from 0 to 1?

 

[BobblyHats97]

You know the pdf of X, which is as simple as a pdf can be.
You have correctly assessed that the next step is to work out the cdf $F_Y$ of Y. THat is, for a given $y\in[0,1]$, we need to work out what is $F_Y(y)$?

$F_Y(y)$ is the probability that $Y\leq y$. What values of $X$ produce a value of $Y$ that satisfies that? Can you work out the probability that $X$ has such a value?

I'm sorry, can you explain what you just said? I'm tired and feeling slightly brain dead but this homework is due in tomorrow :(

I've calculated that the pdf is 1/2 for -1<x<1 and 0 otherwise.

I understand that to find the probability that Y≤y i have to fine the probability that |x|≤y for -1<Y<0 and 0<Y<1 but how do I do that? Am I missing something obvious? Can I just get rid of the modulus because I'm considering all possible values on Y?

 

[andrewkirk]

I understand that to find the probability that Y≤y i have to fine the probability that |x|≤y for -1<Y<0 and 0<Y<1 but how do I do that?

You need to be more careful with upper and lower case, in order to avoid confusing yourself. The standard convention, which works nicely, is to use upper case for a random variable and lower case for a plain old number. So when we write $Prob(Y\leq y)$ we mean the probability that the random variable $Y$ is less than or equal to the number $y$.
So don't write things like $|x|≤y$ as you did above, because you'll just confuse yourself. If you instead write $|X|\leq y$ then it's easy to see how to proceed. You are trying to find the probability that the random variable $X$ has an absolute value in the range $[0,y]$. So ask yourself:
1. What range does $X$ have to be in for that to be the case?
2. What is the probability of $X$ being in that range?