# Probability that one chi-square distributed variable is greater than another based on ratios

Supposing $\ X$ ~ $\chi_1^2$ and $\ Y$ ~ $\chi_{10}^2$ and $\ X$ and $\ Y$ are independent. How would I calculate the probability that $\ X$ is bigger than $\ Y$?

I know that the ratio of $\ X/1$ and $\ Y/10$ would be $\ F$ distributed, but I don't know if that's helpful or not.

Can anyone help?

Thanks

• Because $X$ and $Y$ are positive a.s., the statement $X\gt Y$ is algebraically equivalent to $(X/1)/(Y/10)\gt 10.$ Does that help?
– whuber
Commented Feb 16, 2014 at 18:01
• Thanks for helping, whuber. Could you explain where the >10 comes in? Commented Feb 16, 2014 at 18:43
• The left hand side is the ratio you mention. The right hand side comes from the multiplicative factor of $10$ you supplied. The inequality is equivalent to $X/Y\gt 1$ which in turn is equivalent to $X \gt Y$ when $Y \gt 0.$
– whuber
Commented Feb 16, 2014 at 18:45
• Tom: in case it's not already obvious, think about how to apply what whuber said to your idea about the F ratio. Commented Feb 17, 2014 at 0:03
• I think I have it now. So the probability would be equal to $\ P(F>10) = 1- P(F<=10) = 0.9899$? Commented Feb 17, 2014 at 6:12

To close this one: For $Y>0$, $$P(X > Y) = P\left(\frac XY > 1\right) = P\left(\frac {X/1}{Y/10}\cdot \frac 1{10} > 1\right)$$
$$= P(F > 10) = 1- P(F\leq 10) = 0.9899$$
since $F$ follows an $F$-distribution with $(1,10)$ degrees of freedom.
These are continuous r.v's, so "ignoring" the single-point case $Y=0$ causes no harm.