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.

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

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

  • $\begingroup$ 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

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