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If the pdf of a continuous random variable, $X$, is defined such that:

$$ f_{X}(x)= \frac{3}{(1+x)^{4}}, x\geqslant0 $$

And a random variable, $Y$, is defined as:

$$ Y=\frac{1}{X} $$

What is the method of finding the pdf of $Y$?

From what I've cobbled together (and I could be wrong), I need to get the cdf of X and also calculate $P(Y\geqslant y)$ to get the pdf of Y?

I'm not looking for the solution but rather the procedure. Happy to be directed towards any resources!

edit 1

With some confidence thanks to Michael, I went ahead to get a cdf of X, evaluated over the limits from $0$ to x:

$$ F_X(x) = 1-\frac{1}{(1+x)^3} $$

And I have an expression for $P(Y\geqslant y) = P(X\leqslant y^{-1})$.

Am I on the right track?

edit 2

Continuing:

$$ F_X(y^{-1}) = 1-\frac{1}{(1+\frac{1}{y})^3} = 1-\frac{y^3}{(y+1)^3} $$

And since $F_Y(y) = \int f_Y(y)$, we take the derivative of $F_Y(y)$ to get the pdf.

$$ f_Y(y) = -\frac{3y^2}{(y+1)^4} $$

Defined over $1 \geqslant y > 0$.

Does that seem correct?

PizzAzzra
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  • That is a good procedure. Can you give an expression for $P[Y>y]$ in terms of something about $X$? – Michael Sep 20 '16 at 16:59
  • So if I've got it correctly, $P(Y>y) \rightarrow P(\frac{1}{X} >y) \rightarrow P(X< \frac{1}{y})$ would be the progression of my expression? – PizzAzzra Sep 20 '16 at 17:07
  • Would the limits of the integral of the cdf be from $0$ to $x$? – PizzAzzra Sep 20 '16 at 17:09
  • I would prefer equal signs, rather than arrows, for your second-to-last comment [and mention something about positivity to avoid flipping inequalities and/or dividing by zero]. How would you evaluate $P[X < 1/y]$, given the PDF of $X$? [also, I do not understand your last comment about $x$. What is $x$?] – Michael Sep 20 '16 at 17:24
  • @Michael: I guess I'd mention that $y \neq 0$. As to how I'd evaluate $P(X<y^{-1})$ given the pdf of X, that step eludes me. Sorry, I used x as the upper bound (and $0$ as the lower) of the integral of $f_X(x)$ to get $F_X(x)$. Is that the right procedure? – PizzAzzra Sep 20 '16 at 17:32
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    Given a PDF $f_W(w)$ and an interval $w \in [a,b]$, can you give an expression for $P[W \in [a,b]]$? – Michael Sep 20 '16 at 17:33
  • Let us continue this discussion in chat. Ignore this. – PizzAzzra Sep 20 '16 at 17:43
  • @Michael That's the bit I'm struggling with. The expressions for $P$ w.r.t. pdfs. (Among other things). – PizzAzzra Sep 20 '16 at 17:45
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    You integrate a PDF over the desired interval. $P[W\in [a,b]] = \int_a^b f_W(w)dw$. – Michael Sep 20 '16 at 18:06
  • Ahhh, got it! Thanks! – PizzAzzra Sep 20 '16 at 18:14
  • @Michael I wrote out my solution in the question: does it seem right? – PizzAzzra Sep 20 '16 at 18:38
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    $\frac{d}{dy}P[Y>y] = \frac{d}{dy} (1-F_Y(y))= -f_Y(y)$. And you need your $y$ to satisfy $y>0$ for your previous manipulations to be valid. – Michael Sep 20 '16 at 21:37
  • Is it possible for a pdf to be negative? Or is the answer basically written as: $-f_Y(y) = \frac{3y^2}{(y+1)^4}$ defined when $1 \geqslant y > 0$? – PizzAzzra Sep 20 '16 at 22:23
  • No a pdf cannot be negative. – Michael Sep 20 '16 at 22:49

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