5
$\begingroup$

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

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ 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? $\endgroup$
    – whuber ♦
    Commented Feb 16, 2014 at 18:01
  • $\begingroup$ Thanks for helping, whuber. Could you explain where the >10 comes in? $\endgroup$
    – user123965
    Commented Feb 16, 2014 at 18:43
  • 1
    $\begingroup$ 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 $Y \gt 0.$ $\endgroup$
    – whuber ♦
    Commented Feb 16, 2014 at 18:45
  • $\begingroup$ Tom: in case it's not already obvious, think about how to apply what whuber said to your idea about the F ratio. $\endgroup$
    – Glen_b
    Commented Feb 17, 2014 at 0:03
  • $\begingroup$ I think I have it now. So the probability would be equal to $\ P(F>10) = 1- P(F<=10) = 0.9899 $? $\endgroup$
    – user123965
    Commented Feb 17, 2014 at 6:12

1 Answer 1

Reset to default
4
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.