Given two chi-squared random variables $X$ and $Y$ with $n$ respectively $m$ degrees of freedom. Denote by $f_X$ the density function of $X$. Moreover, $X$ is independent of $Y$. I am interested in the density function of $Z=X(1+Y)$. My approach is the following:

  1. Define $Q:=1+Y$. $Q$ is a random chi-squared distributed variable. Compute the density function $f_Q(x)$ of $Q$.
  2. $Z=XQ$ follows a product distribution (since $Q$ is still independent of $X$ right?). Therefore the density function is given by $f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Q(z/x)\frac{1}{|x|}dx$.

I have the following questions about this approach:

  1. Is it valid?
  2. How do I compute $f_Q(x)$. I know that by adding a constant the new random variable $Q$ is still chi-squared distributed but with a different scale and shift. But how do I get these?
  3. The density function of a chi-squared variable is only defined for positive values. Hence the integral should only go from $0$ to $\infty$ and is only defined for $z>0$. Right?

The overall approach is fine, however you state that $Q=Y+1$ follows a $\chi$-squared distribution, which is wrong. The $\chi$-squared distribution is concentrated on $[0,\infty)$ but $Q$ is concentrated on $[1,\infty)$. The density of $Q$ can however easily be calculated as $f_Q(q) = f_Y(q-1)$.

You are correct that we should only consider $x>0$, when integrating, but you also have to consider the fact that $f_Q(q)=0$ for $q<1$. This means that $f_Q(\frac{z}{x}) = 0$ for $0<z<x$ and thus $$f(z)=\int_{-\infty}^\infty f_X(x)f_Q(\frac{z}{x})\frac{1}{|x|} \: dx = \int_0^z f_X(x) f_Q(\frac{z}{x}) \frac{1}{|x|} \: dx = \int_0^z f_X(x) f_Y(\frac{z}{x} - 1) \frac{1}{|x|} \: dx$$ for any $z>0$.

  • $\begingroup$ Hello, thank you very much for our answer! One small question, since the integral goes from 0 to z we can get rid of the absolute value of x in the fraction and just write x right? So in conclusion we have $f(z)=\int_0^z f_X(x)f_Y(\frac{z}{x}-1)\frac{1}{x}dx$ $\endgroup$ – samabu Jun 1 at 6:58
  • $\begingroup$ Yes, you can replace $|x|$ by $x$. $\endgroup$ – Leander Tilsted Kristensen Jun 1 at 14:31
  • $\begingroup$ Thank you very much! $\endgroup$ – samabu Jun 2 at 7:09

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