Square a random variable

  • #1

Homework Statement


Let the random variable X represent the length of the side of a square. It has a uniform distribution over the interval (0, 5).

What is the cumulative distribution function for the area of the square, Y?


Homework Equations


F(x) = 0.2x (the cdf of the side).


The Attempt at a Solution


So I tried simply squaring F(x), giving 0.04x^2, which is incorrect since F(25) = 25 instead of 1. Also, it wouldn't make sense for the probability of the largest areas to have the highest probability, since:
P(4.9 < X < 5.0) = 0.02
Therefore P(4.9^2 < X < 5.0^2) = 0.02
But going by 0.04x^2 we get 0.08 or something.

Also, I couldn't find anything like this in my textbook (I'm a highschool student), is there any good website to line this stuff from?

Thanks in advance.
 

Answers and Replies

  • #2
VietDao29
Homework Helper
1,424
3

Homework Statement


Let the random variable X represent the length of the side of a square. It has a uniform distribution over the interval (0, 5).

What is the cumulative distribution function for the area of the square, Y?


Homework Equations


F(x) = 0.2x (the cdf of the side).


The Attempt at a Solution


So I tried simply squaring F(x), giving 0.04x^2, which is incorrect since F(25) = 25 instead of 1. Also, it wouldn't make sense for the probability of the largest areas to have the highest probability, since:
P(4.9 < X < 5.0) = 0.02
Therefore P(4.9^2 < X < 5.0^2) = 0.02
But going by 0.04x^2 we get 0.08 or something.

Also, I couldn't find anything like this in my textbook (I'm a highschool student), is there any good website to line this stuff from?

Thanks in advance.

No, why are you squaring F(x)?

To handle this type of problem, we often try to find the cdf of S first, then (if the problem asks further for pdf function) we can obtain it by differentiating the cdf of S.

So, it goes like this:
[tex]\begin{align*}F_S(x) = P(S \le x) = P(X^2 \le x) &= \left\{ \begin{array}{ll} 0 & , \mbox{ if } x \le 0 \\ P(-\sqrt{x} \le X \le \sqrt{x}) & , \mbox{ if } x > 0 \end{array} \right. \\
&= \left\{ \begin{array}{ll} 0 & , \mbox{ if } x \le 0 \\ P(-\sqrt{x} \le X \le 0) + P(0 < X \le \sqrt{x}) & , \mbox{ if } x > 0 \end{array} \right. \\
&= ...
\end{align}[/tex]

Can you go from here, :)
 
Last edited:
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966
You are going the wrong way. In order that [itex]x^2< a[/itex], we must have [itex]x<\sqrt{a}[/itex]. Take the square root of 0.2x, not the square!
 
  • #4
Ah I see where I went wrong.

But it should be [itex]0.2\sqrt{x}[/itex] and not [itex]\sqrt{0.2x}[/itex] right?

Thanks for the speedy responses!
 
  • #5
VietDao29
Homework Helper
1,424
3
Ah I see where I went wrong.

But it should be [itex]0.2\sqrt{x}[/itex] and not [itex]\sqrt{0.2x}[/itex] right?

Thanks for the speedy responses!

Yup, that's correct. Congratulations. :)

But, remember that [itex]F_S(x) = 0.2\sqrt{x}[/itex] on the interval (0; 25), it takes other values elsewhere.
 
  • #6
yup. it's 0 for x < 0 and 1 for x > 25. Thanks for the help :)
 

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