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Let X and Y be independent $\chi^2$ variables with $k$ degrees of freedom. What is the distribution of $\frac{X-E[X]}{Y-E[Y]}$?

I know that without subtracting the means it would be F distribution, but I can't find anything about the ratio when both variables are centered.

In particular, is it true that the median of this distribution is 1?

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  • 1
    $\begingroup$ It doesn't seem to be the case that the median is 1, no; for example a largish simulation with k=2 indicates that a median of 1 is untenable; it appears to be more like 0.133 or so. For k=1 it appears to be about 0.31. $\endgroup$ – Glen_b Aug 11 '17 at 6:39
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If I'm not mistaken, at least for even $k$, this is solvable, but it is very tedious. Following is the outline, in which, for simplicity, I'm omitting constant terms from any integral (i.e., those not involved in the integration).

The ratio distribution of two independent $\chi^2$ RVs is

$$ \sim \int_y |y| (zy)^{\frac{k}{2} - 1}e^{-\frac{zy}{2}} y^{\frac{k}{2} - 1}e^{-\frac{y}{2}} \text{d}y = \int_y |y| (zy^2)^{\frac{k}{2} - 1}e^{-\frac{y(z + 1)}{2}} \text{d}y \; (1) $$

(again, note that constants are omitted from the integral).

Similarly, if you shift the RVs by $a$ (which, in your case is the mean), then the ratio distribution is

$$ \sim \int_y |y| (zy - a)^{\frac{k}{2} - 1}e^{-\frac{zy}{2}} (y - a)^{\frac{k}{2} - 1}e^{-\frac{y}{2}} \text{d}y \\ = \int_y |y| (zy^2 - ay(z + 1) + a^2)^{\frac{k}{2} - 1}e^{-\frac{y(z + 1)}{2}} \text{d}y \; (2) $$

(once again, note that constants are omitted from the integral).

Note that

$$ \int_y |y| (zy^2 - ay(z + 1) + a^2)^{\frac{k}{2} - 1}e^{-\frac{y(z + 1)}{2}} \text{d}y \; (3) \\ = \int_{y > 0} y (zy^2 - ay(z + 1) + a^2)^{\frac{k}{2} - 1}e^{-\frac{y(z + 1)}{2}} \text{d}y - \int_{y < 0} y (zy^2 - ay(z + 1) + a^2)^{\frac{k}{2} - 1}e^{-\frac{y(z + 1)}{2}} \text{d}y . $$

If $k$ is even, then $q = \frac{k}{2} - 1$ is an integer, and, in (3),

$$ (zy^2 - ay(z + 1) + a^2)^{\frac{k}{2} - 1} = (zy^2 - ay(z + 1) + a^2)^q $$

so that trinomial expansion can be used. The integral corresponding to each term in the expansion has a closed form solution.

I wish you the best of luck!!!

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The behaviour of the distribution is quite 'wild' for small values of parameter $k$.

Here is a Monte Carlo simulation of the pdf of the ratio when $k = 1$:

enter image description here

... and when $k = 2$:

enter image description here

However, it rapidly stabilises as $k$ increases, and when $k$ is larger, is extremely well approximated by a Cauchy distribution (same as Student's $t$ with 1 df), i.e.

$$f(x) = \frac{1}{\pi \left(x^2+1\right)}$$

... which seems very pretty, considering how messy the symbolics are.

Here is a plot of the pdf of the ratio when $k = 25$:

enter image description here

  • the rough blue curve is the simulated Monte Carlo pdf of the ratio
  • the dashed red curve underneath is the exact Cauchy pdf

The fit appears amazing - too good to be arbitrary.

Touchy-feely explanation

To get a feel for why this is the case, first observe that the distribution of $X-k$ which has form:

$$f(x) = \frac{2^{-\frac{k}{2}} e^{-\frac{1}{2} (k+x)} (k+x)^{\frac{k}{2}-1}}{\Gamma \left(\frac{k}{2}\right)}$$

... tends to normality (in particular, it seems $N(0, 2k)$) as $k$ becomes large: see for instance: enter image description here

Then, note that the ratio of two Normals with zero mean is Cauchy, and we have the gist of it.

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