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I have a problem where $X$ is uniform on the interval $[4,7]$ and $Y = 20/X$. I am asked to find $F_Y(y)$ and $f_Y(y)$ using the CDF and PDF.

This is a uniform distribution, so it's easy enough finding the CDF and PDF $F_X(x)$ and $f_X(x)$.

I am assuming I found $F_Y(y)$ and $f_Y(y)$ correctly from the CDF of $X$. I am unsure of how to find them from the PDF, though. The PDF in this case is just $1/3$ for $4<x<7$, correct? How to go from there, I'm not quite sure. Any pointers would be greatly appreciated!

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2 Answers 2

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Note

$y = 20/x$ is one-to-one while $x$ is in the range of $4$ to $7$.

However, when transforming a continuous variable you need to also consider what is known as the Jacobian of the transformation. This shows up when transforming continuous variables.

In general, for one-to-one transformations, the PDF of $Y = g(X)$ will be $$f_Y(y) = f_X(g^{-1}(Y) ) \left| \frac{d}{dy} g^{-1} (Y)\right|$$

For this specific example, $g^{-1}(Y) = 20/y$, so it's derivative would be equal to $-20/y^2$. Plugging in $g^{-1}(Y)$ into the PDF of $X$ doesn't change it because the PDF of $X$ is constant.

Plugging it all in gives $$f_Y(y) = \frac13 *\frac{20}{y^2}$$ while $Y$ is in the range of $20/7$ to $5$

From this the CDF of $Y$ can be found easily from here

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Well, you know how to find it through $F_X(x)$, and should know how to find that from $f_X(x)$.

$$\begin{align}F_Y(y)&=1-F_X(20/y)\\[1ex]&=\int_{20/y}^\infty f_X(x)~\mathrm d x\\[4ex]f_Y(y)&=\dfrac{\mathrm d ~~}{\mathrm d y}F_Y(y)\\&=\dfrac{\mathrm d ~~}{\mathrm d y}\int_{20/y}^\infty f_X(x)~\mathrm d x\end{align}$$

So...by using the fundamental principle of calculus...

$$\begin{align}f_Y(y)&=-\dfrac{\mathrm d (20/y)}{\mathrm dy}\cdot f_X(20/y)\\[3ex] F_Y(y)&= \int_{-\infty}^y f_Y(y)\mathrm d y\end{align}$$

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