1
$\begingroup$

The problem is :

Let $f(x) = {1\over 3} , -1<x<2 $, zero elsewhere, be the pdf of $X$. Find the cdf and pdf of $Y=X^2$.

The answer is:

Consider case 1: When $ 0<y<1 $

$P(X^2\le y) =P(-\sqrt{y} \le x \le \sqrt{y}) = \int_{-\sqrt{y}}^{\sqrt{y}}\,{1\over 3}dx = {2\sqrt{y}\over 3}$

Consider case 2: When $ 1\le y \le4 $

$P(X^2\le y) =P( X^2 \le 1) + P( 1 \le X^2 \le y) = P( x \le 1) + P( 1 < x \le \sqrt{y}) = {2\over 3} + \int_{1}^{\sqrt{y}}\, {1\over 3} dx = {1\over 3}+{\sqrt{y}\over3} $

Therefore, $f_{Y}(y) = \begin{cases} {1\over 3\sqrt{y}}, & \text{$0<y<1$} \\ {1\over 6\sqrt{y}}, & \text{$1\le y \le4$} \\ {0}, & \text{otherwise} \end{cases} $

My questions:

(1)When encountering this kind of not one-to-one function, how to determine the intervals for $y$ ?

(2)How to determine the number of partition of the interval? Like this case, 2 intervals.

(3)Is there any way I can draw a graph to figure out the answers?

(4)If possible, can someone please give me some related practice questions? Since my text book contains only one not one-to-one transformation function problem.

$\endgroup$
  • $\begingroup$ (1) & (2): count carefully. $\endgroup$ – BruceET Sep 24 '16 at 21:52
1
$\begingroup$

Comment: It seems you have done everything correctly. Perhaps the following simulation will help with part (3). In R statistical software, I simulated 100,000 iterations of $X \sim Unif(-1, 2)$ and found $Y = X^2$ for each. Then made a histogram of the simulated distribution of $Y$, plotted your density curve through the histogram as a check; and plotted values of $Y$ against values of $X,$ to make a 'curve' consisting of 100,000 points.

x = runif(10^5, -1, 2)
y = x^2
mean(y);  sd(y)
## 0.9916676  # aprx E(Y), about 2 place accuracy
## 1.09006    # aprx SD(Y)

enter image description here

$\endgroup$
  • $\begingroup$ Thank you for the graph in right hand side, it helps a lot ! I couldn't interpret the meaning of transformation at all, but simulated by the graph I am getting more understanding on it now, thank you ! $\endgroup$ – Pak Long Sep 25 '16 at 4:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.