# intuitive reason absolute values are used for transformations in statistics?

by phiiota
Tags: absolute, intuitive, statistics, transformations, values
 P: 29 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution this isn't really homework, but I was just wondering if someone could offer an intuitive reason as to why when random variables are transformed, we use absolute values of derivative of those functions, as opposed to the functions themselves?
Mentor
P: 19,758
 Quote by phiiota 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution this isn't really homework, but I was just wondering if someone could offer an intuitive reason as to why when random variables are transformed, we use absolute values of derivative of those functions, as opposed to the functions themselves?
Can you give an example where this is happening?
 P: 29 well, if say X distributed as f(X), and say i have Y=g(X), and I want to know the distribution of Y, in a simple case I can say that Y~f(g-1(y))|d/dy g-1(y)|. I don't know why we'd always want the absolute value, as opposed to the derivative in general.
P: 29

## intuitive reason absolute values are used for transformations in statistics?

or did you want a specific case? say X~U(0,1), and I want to know say Y=X-1.
then I can say g-1(y)=y-1., and d/dy g-1(y)=-y2, so then my distribution of Y would be

f(g-1(y))|d/dy g-1(y)|=y-2.

I get that if we didn't take the absolute value, then this function would be negative... But aside from that, I'm not seeing an intuitive idea as to why we'd always take the absolute value.
HW Helper
P: 4,129
 Quote by phiiota well, if say X distributed as f(X), and say i have Y=g(X), and I want to know the distribution of Y, in a simple case I can say that Y~f(g-1(y))|d/dy g-1(y)|. I don't know why we'd always want the absolute value, as opposed to the derivative in general.
$$P\{y < Y < y + dy\} = P\{y < g(X) < y + dy\}.$$ If g is an increasing function we have
$$P\{y < g(X) < y+dy\} = P\{g^{-1}(y) < X < g^{-1}(y+dy) \} = P\{ g^{-1}(y) < X < g^{-1}(y) + {g^{-1}}^{\prime} (y) dy\} \doteq f[g^{-1}(y)] {g^{-1}}^{\prime}(y) \, dy.$$
If g is a decreasing function we have
$$P\{y < g(X) < y+dy\} = P \{ g^{-1}(y+dy) < X < g^{-1}(y) \} = P\{ g^{-1}(y) - |{g^{-1}}^{\prime}(y)| \, dy < X < g^{-1}(y) \} \doteq f[g^{-1}(y)]\, | {g^{-1}}^{\prime}(y) | \, dy.$$ The point is that the *length* of the x-interval corresponding to dy is $| {g^{-1}}^{\prime}(y) | \, dy$, and you need the absolute value so that the length cannot be negative.
 P: 29 Okay, that makes sense. Thank you.

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