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Suppose the probability density function of $X$ is given by $f(x)=\frac{k x^3}{(1+2x)^6}$ if $x \in D=(0,\infty)$.

Let $Y=u(X)=\frac{2X}{1+2X}$. Then $u(D)=(0,1)$. Since $u(x)$ is increasing on $D$, $u(x)$ is one-to-one on $D$ and has an inverse $w(y)$.

From there, I tried to apply this formula: $f_Y(y)=f_X\left[w(y)\right]\left|w'(y)\right|$ if $y \in u(D)$.

$w(y)=\frac{y}{2(1-y)}$.

$f[w(y)]=\frac{k}{8} \frac{y^3}{(1-y)^3\left(1+\frac{y}{1-y}\right)^6}$

$|w'(y)|=\frac{1}{2(1-y)^2}$

$f(w(y))|w'(y)|=\frac{k}{16} \frac{y^3}{(1-y)^5\left(1+\frac{y}{1-y}\right)^6}=\frac{k}{16}y^3(1-y)$

However, something doesn't seem right here. I am asked to "identify" the distribution and from there, deduce the value of $k$. From my result, the only way I can think to find $k$ is to integrate it over the whole domain and use the properties of probability density functions, when the implication seems to be that I should have gotten a more familiar distribution.

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    It's just a beta distribution2012-09-19

1 Answers 1

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Thank you, Cocopuffs.

$f_Y(y)$ is Beta with $\alpha = 4$ and $\beta = 2$.

$B[4,2]=\frac{1}{20} $

$\frac{k}{16}=\frac{1}{B[4,2]} \Longrightarrow k=320$