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I am trying to teach myself probability by solving problems and studying solutions. One of the solutions I have encountered had several points I did not understand. I would greatly appreciate if you could clarify where that is coming from, I am attaching a solution and the parts that I did not understand are circled with the red marker. My questions might be basic, but I kinda got stuck at that point and hope someone could provide an explanation.

  1. Where is that denominator coming from? I calculated the density function of V and am getting a different denominator.

  2. Why are we taking the absolute value of the derivatives of Z and U? What does that give us? What is it equal to? How are we getting the result that is there? Problem solution

Thanks! I am attaching the question first and the solution I did not understand

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If $V \sim \chi^2_k$ then its density is $f_V(x)=\frac{x^{\frac{k}{2}-1}e^{-\frac{x}{2}}}{\Gamma\big(\frac{k}{2}\big)2^{\frac{k}{2}}}$ for $x\geq0$ and $f_V(x)=0$ elsewhere. I think you might be confusing the independent variable "$v$" in the density $f_V$ with the number of degrees of freedom $\nu$. To avoid confusion you may want to replace $\nu$ with $k$ (like I did) to represent the number of degrees of freedom. Moreover, I think you're confusing the gamma function $\Gamma$ with a square root symbol. Now if $(Z,V)\sim f_{ZV}$ and $T$ is an invertible transformation, then the density of $(U,W)=T(Z,V)$ has formula $$f_{UW}(u,w)=f_{ZV}\Big(T^{-1}(u,w)\Big)\left\| \frac{\partial(z,v)}{\partial(u,w)} \right\|$$ The absolute values come with the formula. So if you define coordinates $(u,w)$ to be $$(u,w)=\Bigg(\frac{z}{\sqrt{v/k}},v\Bigg)=T(z,v)$$ then we have that $$(z,v)=T^{-1}(u,w)=\Bigg(u\sqrt{\frac{w}{k}},w\Bigg)$$ and so $$\frac{\partial(z,v)}{\partial(u,w)}=\text{det}\begin{bmatrix}z_u&z_w\\v_u&v_w\end{bmatrix}=\text{det}\begin{bmatrix}\sqrt{\frac{w}{k}}&\frac{u}{2\sqrt{kw}}\\0&1\end{bmatrix}=\sqrt{\frac{w}{k}}$$ Putting everything together, $$f_{UW}(u,w)=f_{ZV}\Bigg(u\sqrt{\frac{w}{k}},w\Bigg)\sqrt{\frac{w}{k}}=\sqrt{\frac{w}{k}}f_{Z}\Bigg(u\sqrt{\frac{w}{k}}\Bigg)f_{V}(w)$$

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