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Let Y be a random variable which has pdf $$f_Y(y) = \begin{cases}4y^3, & 0 < y < 1, \\ 0, &\text{elsewhere}.\end{cases}$$

Show that $-2 \ln (Y^4)$ ~ $X_{(2)}^2$.

Could anyone get me started on this? How to approach this question?

Thanks a lot

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    $\begingroup$ Is $X^2$ supposed to be chi-sqaured $\chi ^2$? $\endgroup$
    – homegrown
    Mar 1, 2014 at 18:06

2 Answers 2

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Hint: Start with the cdf of $W:=-2\ln(Y^4)$, for $0<y$. You have that $$\begin{align*}F_W(y)&=P(W\le y)=P(-2\ln(Y^4)\le y)=P(-8\ln Y \le y)=P(\ln Y \ge \frac{y}{-8})=\\&=P(Y\ge e^{-\frac{y}{8}})=1-P(Y\le e^{-\frac{y}{8}})=\\&=1-F_Y(e^{-\frac{y}{8}})\end{align*}$$ Taking the derivative: $$\begin{align*}f_W(y)&=\frac{d}{dy}F_W(y)=\frac{d}{dy}\left(1-F_Y(e^{-\frac{y}{8}})\right)=-f_y(e^{-\frac{y}{8}})\cdot(e^{-\frac{y}{8}})'=\frac{1}{8}\cdot4\left(e^{-\frac{y}{8}}\right)^3e^{-\frac{y}{8}}=\\&\\&=\frac{1}{2}\left(e^{-\frac{y}{8}}\right)^4=\frac{1}{2^{\frac{2}{2}}\Gamma(\frac{2}{2})}y^{\frac{2}{2}-1}e^{-\frac{y}{2}}\end{align*}$$ which is the pdf of $\chi^2_{(2)}$ for $y>0$.

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  • $\begingroup$ so there is no typo right? $\endgroup$ Mar 1, 2014 at 18:20
  • $\begingroup$ no, is ok, is just that $k=2$ and $x$ disappears $\endgroup$
    – Jimmy R.
    Mar 1, 2014 at 18:21
  • $\begingroup$ the thing after the last equal sign should be $\frac{1}{2^{\frac{2}{2}}\Gamma(\frac{2}{2})}y^{\frac{2}{2}-1}e^{-\frac{y}{2}}$? $\endgroup$ Mar 1, 2014 at 18:28
  • $\begingroup$ x should not be there since y is the variable for the function $\endgroup$ Mar 1, 2014 at 18:37
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    $\begingroup$ @afsdfdfsaf you are right, it is a typo, but it won't matter as $x^0 = y^0 =1$ anyway. $\endgroup$
    – ir7
    Mar 2, 2014 at 5:22
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More succinctly, you can use the change of variables formula

$$g(x) = \frac{f\bigl( h^{-1}(x) \bigr)}{\bigl|h'\bigl(h^{-1}(x)\bigr)\bigr|},$$

where $f$ is the density you started with, $g$ is the density of your transformed variable and $h$ is the transformation, here $h(y)=-2 \log y^4=-8\log y$. This works as long as $h$ is invertible, which it is in this instance. Note that $h^{-1}$ denotes the inverse function, so $\exp(-x/8)$.

Since this is homework I don't want to plug in all the details, but if you do things right then $g(x)$ will be the density of a $\chi_2^2$.

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