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I have a PDF of X defined as $f(x) = e^{-x}\text{ for } x \geq 0,$ 0 otherwise, and a RV $Y$ defined as X if X $\leq 1$, and $\frac{1}{X}$ if X>1. I need to find a pdf of Y. I graphed Y versus X, and can see that Y varies from 0 to 1, the curve goes as y=x for x between 0 and 1, and $\frac{1}{x}$ from 1 onwards. Could someone show how to get to the pdfs? I tried doing pdf of y = $F_x(t)' + F_x(\frac{1}{t})'$ at t between 0 and 1, but I don't know -- should I be adding them or subtracting them? This is in preparation for the test tomorrow morning, many thanks.

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For every $y$ in $(0,1)$, write the event $[Y\leqslant y]$ as the union of two disjoint events involving $X$ only. Deduce from this the value of $F_Y(y)=\mathrm P(Y\leqslant y)$. –  Did Nov 2 '11 at 7:54
    
@DidierPiau: I think I got to that step in my thought process -- every y in 0, 1 can come from either x=y or 1/x = y. But I am not sure how to deduce the pdf described above. I do not want to deduce the distribution function because our teacher emphasizes finding PDFs by differentiating dist functions of the original variable, and not doing the integration.. –  fasthelp Nov 2 '11 at 8:00
    
You go too fast and hit a wall without realizing it... Note that I did not mention $x=y$ or $x=1/y$ for $y$ in $(0,1)$. Sticking to what I wrote, did you write $[Y\leqslant y]$ as a union? (Not $[Y=y]$, mind you, but $[Y\leqslant y]$.) If you do that, you will realize your $F_X(t)+F_X(1/t)$ is not relevant. –  Did Nov 2 '11 at 8:12
    
The question said $f(x)=e^{-x} x \ge 0$. I changed it to $f(x)=e^{-x} \text{ for }x \ge 0$. It seems that was probably intended since otherwise "$\ge 0$" wouldn't have been there. –  Michael Hardy Nov 2 '11 at 11:06

2 Answers 2

up vote 3 down vote accepted

Here's one way. For $0<y\le1$, $$ \begin{align} f_Y(y) & = \frac{d}{dy} F_Y(y) = \frac{d}{dy}\Pr(Y\le y) = \frac{d}{dy}\Pr\left( X \le y \text{ or }X\ge\frac1y \right) = \frac{d}{dy}\left( (1-e^{-y}) + e^{-1/y} \right) \\ \\ & = e^{-y} + \frac{e^{-1/y}}{y^2}. \end{align} $$

(And of course $f_Y(y) = 0$ if $y>1$ or $y<0$.)

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I think that You should use following formula:

$f_Y(y)=|\frac{1}{g'(x)}|f_X(x)$ , where $Y=g(X)$

$a)$ $x<0 , f_X(x)=0 , g(x)=x$

$f_Y(y)=1 \cdot 0=0$

$b)$ $x\in [0,1] , f_X(x)=e^{-x} , g(x)=x$

$f_Y(y)=1 \cdot e^{-x}=e^{-x}$

$c)$ $x>1 , f_X(x)=e^{-x} , g(x)=\frac{1}{x}$

$f_Y(y)=\frac{1}{x^2} \cdot e^{-x}$

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For $y>1$, the density $f_Y(y)$ is $0$. You're running into complications resulting from the fact that $Y$ is not a one-to-one function of $X$. When $X>1$, the other variable $Y$ is still $<1$. The bottom line seems to be that the two functions you come up with, $e^{-x}$ and $e^{-x}/x^2$ should be added together when $0<x<1$, and you should get $0$ when $x>1$ or $x<0$. –  Michael Hardy Nov 2 '11 at 11:40
    
But $\frac1{x^2}e^{-x}$ does not enter the picture. –  Did Nov 2 '11 at 13:00
    
@Didier You're right: $\frac{1}{x^2}e^{-\text{something}}$ does enter the picture. My examination of this posting didn't get as far as that exponent. –  Michael Hardy Nov 2 '11 at 13:13

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