# Computing density and cdf

## Homework Statement

Density of f_x (x) = 4x^4 for 0<x<1

Y=(x-1/4)^2 Z= X^-2

Determine density of Y
and Distribution of Z

## Homework Equations

The cdf of f_x (x) is invalid since F_x (x) = (4/5)x^5 so the limit to infinity does not equal 1 as a cdf should have. Am I missing something?

## The Attempt at a Solution

density of Y P ((x-1/4)^2 =< x) = P(x =< sqrt(x) +1/4) = f (sqrt(x) +1/4) *(1/2)x^(-1/2) = (2/sqrt(2)) (sqrt(x) +1/4)^4 what are the bounds?

cdf of Z P(X^-2 =< x) = P(x <=-1/x) +P (X>= 1/x) = F(-1/x)+1-F(1/x) = 1-8/(5x^5) what are the bounds?

Related Calculus and Beyond Homework Help News on Phys.org
Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Density of f_x (x) = 4x^4 for 0<x<1

Y=(x-1/4)^2 Z= X^-2

Determine density of Y
and Distribution of Z

## Homework Equations

The cdf of f_x (x) is invalid since F_x (x) = (4/5)x^5 so the limit to infinity does not equal 1 as a cdf should have. Am I missing something?

## The Attempt at a Solution

density of Y P ((x-1/4)^2 =< x) = P(x =< sqrt(x) +1/4) = f (sqrt(x) +1/4) *(1/2)x^(-1/2) = (2/sqrt(2)) (sqrt(x) +1/4)^4 what are the bounds?

cdf of Z P(X^-2 =< x) = P(x <=-1/x) +P (X>= 1/x) = F(-1/x)+1-F(1/x) = 1-8/(5x^5) what are the bounds?
You are right about f(x): it should probably be 5x^4.

Never, never, never write something like P ((x-1/4)^2 =< x): you are using the same letter x to stand for two totally different things, and that is a sure invitation to error (or being marked wrong). You want P{Y ≤ y} = P{(X - 1/2)^2 ≤ y}. Look at the graph of the function (x - 1/2)^2; for which values of x on [0,1] is it ≤ y? For what values of y will the probability be < 1?

Note: in the above you attempted to find the CDF of Y, not the density! How would you get the density if you know the cdf?

In the second solution, can you see how wrong it is to write P(x <= -1/2), etc? Again, you are inviting a poor mark and are increasing your chances of making a serious error.

RGV