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Function of a random variable

  • #1

Homework Statement


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

find the density function of [tex] Y= \sqrt{X}[/tex]

The Attempt at a Solution


[tex] y=g(x)=\sqrt{x}[/tex]
so [tex]g^-1(y)=x=y^2[/tex]

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

[tex](g^-1(y))^'[/tex][tex]=2y[/tex]

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

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
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
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:
  • #3
Thanks, You are the man Dick.
 

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