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I am trying to derive the probability distribution of $Z=Z_1^2 + Z_2^2$. I am doing this in the following way,

$$ \begin{align}p(Z=x) = &\int_0^x p(Z_1^2 = p)\times p(Z_2^2 = x-p) dp\\ \Leftrightarrow ~ &\int_0^x p(Z_1 = \pm\sqrt{p})\times p(Z_2 = \pm\sqrt{x-p})dp \\ \Leftrightarrow &\int_0^x [2p(Z_1 = \pm\sqrt{p})]\times [2p(Z_2 = \pm\sqrt{x-p})]dp \\ \Leftrightarrow 4&\int_0^x (\frac{1}{\sqrt{2\pi}} e^{-0.5p}) \times (\frac{1}{\sqrt{2\pi}} e^{-0.5(x-p)})dp\\ \Leftrightarrow \frac{4}{2\pi}&\int_0^x e^{-0.5p} \times e^{-0.5(x-p)}dp \\ \Leftrightarrow \frac{4}{2\pi}&\int_0^x e^{-0.5x}dp \\ \Leftrightarrow \frac{4}{2\pi} &[p\times e^{-0.5x}]_0^x \\ \Leftrightarrow \frac{4}{2\pi}&\int_0^x e^{-0.5x}dp \\ \Leftrightarrow \frac{2}{\pi} &x\times e^{-0.5x}\end{align}$$

However, this should be the same as the gamma distribution, $\operatorname{Gamma}(1,\frac{1}{2})$, which has pdf

$$ \frac{1}{2}e^{-0.5t} $$

With the value $x=0.3 = t$, we get two different answers, so the pdfs cannot be the same. What am I doing wrong? The limits should be fine as any values of $p$ outside of the interval $[0,x]$ would give a negative inside the square roots.

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  • $\begingroup$ In the second step you make a change of variables, can you do that? It is more foolproof when you compute $p[Z \leq x]$ instead of $p[Z = x]$ $\endgroup$ Dec 5, 2022 at 1:21
  • $\begingroup$ I did wonder if this step is wrong, but for a normal random variable, the probability of $-\sqrt{p}$ is the same as $\sqrt{p}$, which is where I get the factors of 2 from. Hmm... $\endgroup$
    – Cai
    Dec 5, 2022 at 1:27
  • $\begingroup$ Are you computing with probabilities or with probability densities? Also, It may help to make a drawing and picture the area of the region between $Z$ and $Z+dZ$. How would you compute the area? As $\int_0^x 1 dp$? $\endgroup$ Dec 5, 2022 at 1:31

1 Answer 1

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You are missing a term $\frac{1}{\sqrt{x^2-p^2}}$ and you should be ending up with using an integral like this

$$\int_0^x \frac{1}{\sqrt{x^2-p^2}} e^{-x} dp = \frac{\pi}{2} e^{-x}$$

instead of

$$\int_0^x e^{-x} dp = x e^{-x}$$

This error happens in the beginning when you change the variables from $Z_1^2$ and $Z_2^2$ to $Z_1$ and $Z_2$, your term $dp$ got changed here. The extra term is how the area changes (you integrate along a 1 dimensional line but this is a shortcut for integrating the area for an infinitesimal change of $dZ$ along this line).

It is more foolproof when you compute $p[Z≤x]$ instead of $p[Z=x]$.

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  • $\begingroup$ Also, of course, you can more easily solve this when you change to polar coordinates. $\endgroup$ Dec 5, 2022 at 6:54

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