# Function of a random variable

## 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
Homework Helper
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.

Last edited:
Thanks, You are the man Dick.