# Overlapping $\chi^2$ random variables

I have 3 independent random variables that follow $$\chi^2$$ laws, with $$m$$ and $$n$$ the degrees of freedom: \begin{align}A&\sim\chi^2_m\\B&\sim\chi^2_n\\C&\sim\chi^2_m\end{align} I am interested to know the conditional probability distribution of $$B+C$$, knowing that $$A+B=x$$. In equation, this means: $$f_{B+C|A+B}(y|x)=\frac{f_{A+B,B+C}(x,y)}{f_{A+B}(x)}$$

I can compute the denominator as it's simply a $$\chi^2_{m+n}$$ distribution, given that $$A$$ and $$B$$ are independent. However, I don't know how to compute the term in the numerator. It is not two independent $$\chi^2$$ distributions as the $$B$$ overlaps.

• Welcome to Cross Validated! You're dealing with continuous random variables, so the probability of an "equals" is going to be zero.
– Dave
Jun 28 at 16:17
• @Dave Understood, I meant the density of probability here. I will amend accordingly.
– PC1
Jun 28 at 16:19
• Do you mean the "y-axis value of the probability density function?
– Dave
Jun 28 at 16:21
• Here we have a joint probability for independent $\chi^2$ random variables $A$, $B$, and $C$, with the constraint that $A+B$ is fixed. I am interested to get the pdf of $B+C$.
– PC1
Jun 28 at 16:28
• It sounds like you want to know $f(y)$, where $f$ is the PDF of $B+C$ conditioned on $A + B = x$. If this is the case, please edit your question to say so, as you can't just flip around PDFs like you do probabilities in Bayes' rule.
– Dave
Jun 28 at 16:36

Allowing for some slight abuse of notation, you can find the numerator directly by solving the following integral. \begin{align*} f(A+B=x, C+B=y) &= f(A= x-B, C+y-B) \\[1.5ex] &= \int_0^\infty f(A=x-b, C=y-b|B=b)f(B=b) db \\[1.5ex] &= \int_0^\infty f_A(x-b)f_C(y-b)f_B(b) db \\[1.5ex] &= \frac{1}{2^{m/2}\Gamma(m/2)}\frac{1}{2^{m/2}\Gamma(m/2)}\frac{1}{2^{n/2}\Gamma(n/2)} \times \\ &\quad\int_0^{\min\{x, y\}}(x-b)^{m/2-1}(y-b)^{m/2-1}b^{n/2-1}\exp\left(-\frac{1}{2}\left(x+y-b\right)\right) db \\[1.5ex] &= c\int_0^{\min\{x, y\}}\left[(x-b)(y-b)\right]^{m/2-1}b^{n/2}\exp\left(\frac{b}{2}\right)db \end{align*}

where $$c = \left(2^{m+2/n}\Gamma(m/2)^2\Gamma(n/2)\right)^{-1}\exp(-(x+y)/2)$$

I will revisit this later today if I have the time, but if @whuber cannot obtain an analytical solution then I doubt one exists.

• I think that you are correct. It's basically the joint distribution $f_{A,B,C}=f_Af_Bf_C$ as the 3 variables are independent, integrated on the constraint.
– PC1
Jun 28 at 17:01
• I cannot obtain an analytical solution to this, even when it is corrected to limit the integral to $\min(x,y)$ (it does not extend to $\infty$).
– whuber
Jun 28 at 17:02
• I used Mathematica and I don't get an analytic answer either, even with all the constraints on the range of the variables.
– PC1
Jun 28 at 17:06
• I have corrected the bounds of integration. @PC1, note still that for certain values of $m$ and $n$, you may be able to integrate this. For instance if the leading terms of the integrand forms a polynomial, you should be able to express this as a linear combination of truncated moments for some distribution. Jun 28 at 17:24
• I tried with $m=1$ but even there I don't get an analytic result. I will try to change my model slightly to ensure that I can get a result that I can integrate.
– PC1
Jun 28 at 17:32