1

I would like to specify radius r in a case where: $$\mathbb{P}(Z_1^2 + Z_2^2>r)=a\,,$$ $a$ is known, and $Z_1,Z_2$ are i.i.d $t_\nu$-distributed.

Question: Any ideas on how to do this?

Attempt: Since this can be really challenging to do analytically, I decided to do it by sampling.

My first thought was to simply draw N random values a from $t_\nu$ distributed variable and then count how many are there that fulfill the $|t_k|\geq r$ condition, then update $r$ until I cross $a$, but I'm not really convinced that this is the correct way. The $\nu$ value doesn't really matter.

I have checked these two posts: (1) (2).

Edit As was pointed out in the comments by whuber, it may be possible to numerically compute the probability through the integral.

I would also use this solution, but as I understood from the posts linked above, it will be really tricky.

Community
  • 1
Trelokoritso
  • 101
  • 5
  • 1
    Your question--which asks to compute the inverse distribution function of a sum of squared t variables--does not seem to agree with the title, which asks to draw a sample. Could you please edit one (or both) to make them consistent? – whuber Dec 02 '15 at 17:46
  • Thank you--the nature of your question is clear now. Would it still be worth pointing out that a numerical evaluation of the integral would be quicker and far more accurate than any amount of sampling? – whuber Dec 02 '15 at 18:05
  • I m not sure, i can write down the integral of the $f_{(Z_1^2,Z_2^2)}(z_1,z_2)$ (the pdf) correctly. If you could help me with that, i d appreciate it. – Trelokoritso Dec 02 '15 at 18:10
  • 1
    If that calculation would help you, then please edit your question to mention that. I'm confident that many community members would know of answers and I believe you're likely to get several different, but good, ones. – whuber Dec 02 '15 at 18:12
  • Ι don't really believe that it can be done through the integral though, since you are looking for a very tricky pdf there. So sampling is probably the best solution. – Trelokoritso Dec 02 '15 at 18:26
  • 1
    This PDF is benign! There isn't anything terribly tricky about it. There are many ways to perform the computation accurately and fairly quickly. – whuber Dec 02 '15 at 18:49
  • Please feel free to propose it! – Trelokoritso Dec 02 '15 at 18:54
  • 1
    Sure: for $\nu=3$ and $r=3.5$, for example, the complementary integral can be found by [Wolfram Alpha](http://www.wolframalpha.com/input/?i=NIntegrate[%28%283^%283%2F2%29+%28x+%2B+3%29^%281%2F2+%28-1+-+3%29%29%29+%283^%283%2F2%29+%283.5+-+x+%2B+3%29^%28+++++1%2F2+%28-1+-+3%29%29%29%29%2F%28%28Sqrt[x]+Beta[1%2F2%2C+3%2F2]%29+%28Sqrt[3.5+-+x]++++++Beta[1%2F2%2C+3%2F2]%29%29%2C+{x%2C+0%2C+3.5}]) – whuber Dec 02 '15 at 20:43
  • the pdf you must integrate should be the pdf of $Z_1^2 + Z_2^2$ and not the pdf of $Z_1,Z_2$. That's why it is hard. – Trelokoritso Dec 03 '15 at 10:41
  • 1
    The event $\{Z_1^2+Z_2^2\ge r\}$ is also an event for $(Z_1,Z_2)$ so all you need is a pdf on $(Z_1,Z_2)$. – Xi'an Dec 03 '15 at 13:32

0 Answers0