# Relationship between dist. of x and dist. of 1/x?

by Mapes
Tags: 1 or x, dist, relationship
 Sci Advisor HW Helper PF Gold P: 2,532 I've been working recently in the area of tissue cell mechanics; specifically, I'm measuring mechanical stiffness (or compliance, the reciprocal of stiffness) and considering its possible underlying distribution. I was wondering about the following: If the distribution of stiffness measurements is approximately Gaussian (or lognormal, or gamma distributed, etc.), then what can we say about the distribution of the corresponding compliance (= 1/stiffness) values? More generally, if $x$ is distributed in a certain way, what about $1/x$? Is there a simple relationship?
 HW Helper P: 1,334 No - there is no universal statement that can be made. Each case needs to be considered on its own. (The mathematical ideas behind studying the distributions is the same in each case, but unless I'm totally off that wasn't the point of your inquiry.)
Mentor
P: 11,966
 Quote by Mapes More generally, if $x$ is distributed in a certain way, what about $1/x$? Is there a simple relationship?
There is a relationship, how simple depends on the details of your example.

Given a probability distribution f(x), we seek the distribution g(y) where y is a function of x. A simple probability conservation argument tells us that
f(x) |dx| = g(y) |dy|
so that
g(y) = f(x) / |dy/dx|
Take y = 1/x, and f(x) is whatever you think, you can get g(y).

EDIT:
Continuing the example for y = 1/x

Since |dy/dx| = 1/x2 = y2, we have
g(y) = f(x) / y2
And, of course, you would substitute 1/y for x in the expression for f(x).

HW Helper
P: 1,334

## Relationship between dist. of x and dist. of 1/x?

The transformation approach is correct (modulo being careful around x = 0); my intention was to say there is nothing simple to say about the type of distribution for X and the type for 1/X (normal to normal, t to t, etc).
 Mentor P: 11,966 True

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