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.
Finding a PDF of Y given a PDF of X
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probability
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0The 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. – 2011-11-02
2 Answers
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Here's one way. For $0
(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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0@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. – 2011-11-02