# Random var. Y with pdf $f_Y(y) = 4y^3$. Show that $-2\ln (Y^4)$ ~ $X_{(2)}^2$.

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

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

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 $y>0$.

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

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 $\chi_2^2$.