What is the Density Function of Y=√X?

[Bertrandkis]

Homework Statement

Suppose a random variable X has probability density function(pdf)
f(x) { 1/3 for 1 \leq x \leq 4

find the density function of Y= \sqrt{X}

The Attempt at a Solution

y=g(x)=\sqrt{x}
so g^-1(y)=x=y^2

A= \{ x: 1 \leq x \leq 4 \}
is monotonic onto
B= \{y: 1 \leq y \leq 2 \}

(g^-1(y))^'=2y

f(y)=fx (g^-1(y). |(g^-1(y))^'|
which gives me
f(x)=\{ 2y/3 for 1 \leq y \leq 2 \}

This seems to be an invalid pdf as for y=2 f(y)=1.3333 which is >1
Can anyone tell me where I went wrong?
thanks

 

[Dick]

It's not invalid. You did it right. A probability DENSITY can be greater than 1. A probability can't be greater than 1. The integral of f(y)dy from 1 to 2 is one.

 

[Bertrandkis]

Thanks, You are the man Dick.