Find the probability distribution

[relinquished™]

Given the probability distribution function

$$f(x) = \frac{2}{9}(x-1), -1<x<2$$

find the pdf of $$Y = X^2$$

My Solution:

When x = -1, y = 1 and when x = 2, y = 4, so the range of y is

$$1 \leq y \leq 4$$

So to find the pdf of Y = X^2, we need to find the cdf of Y. Since Y is nontrivial over the interval $$1 \leq y \leq 4$$,

$$F(y) = P(1 \leq Y \leq y) =P(1 \leq x^2 \leq y)$$

I'm at a loss here. I do not know how I should deal with square rooting the x, since $$x^2 \leq y$$ is rewritten as $$-\sqrt{y} \leq x \leq \sqrt{y}$$

Any suggestions?

 

[HallsofIvy]

What you have is correct: $-\sqrt{y}\leqx\leq\sqrt{y}$. Now integrate your pdf for x, (2/9)(x-1) from -y to y (be careful about y> 1!).

 

[relinquished™]

err... when you mean about be careful of y>1, do you mean that I have to separate the integral like this:

$$\int^{1}_{-\sqrt{y}} f(x)dx + \int^{\sqrt{y}}_{1} f(x)dx$$

Because I know that y=1 should not be within the range of y (since y is just greater than 1)

Edit:

I got a cdf of $$-4\sqrt{y}$$. I am sure I did something wrong and it must have something to do with that y>1 precaustion...

 

[irony of truth]

Hello relinquished! Isn't it that it should be 2/9 (x + 1) as the given?

 

[relinquished™]

well... if the pdf really is $$\frac{2}{9}(x+1)$$ then I should look back at the book ^^;

 

[irony of truth]

I don't think it is found in the book.