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.

  • 1
    $\begingroup$ Welcome to Cross Validated! You're dealing with continuous random variables, so the probability of an "equals" is going to be zero. $\endgroup$
    – Dave
    Jun 28 at 16:17
  • $\begingroup$ @Dave Understood, I meant the density of probability here. I will amend accordingly. $\endgroup$
    – PC1
    Jun 28 at 16:19
  • $\begingroup$ Do you mean the "y-axis value of the probability density function? $\endgroup$
    – Dave
    Jun 28 at 16:21
  • $\begingroup$ 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 $B+C$. $\endgroup$
    – PC1
    Jun 28 at 16:28
  • $\begingroup$ 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. $\endgroup$
    – Dave
    Jun 28 at 16:36

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – PC1
    Jun 28 at 17:01
  • $\begingroup$ 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 $\infty$). $\endgroup$
    – whuber
    Jun 28 at 17:02
  • $\begingroup$ I used Mathematica and I don't get an analytic answer either, even with all the constraints on the range of the variables. $\endgroup$
    – PC1
    Jun 28 at 17:06
  • $\begingroup$ 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. $\endgroup$
    – knrumsey
    Jun 28 at 17:24
  • $\begingroup$ 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. $\endgroup$
    – PC1
    Jun 28 at 17:32

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