In the notes I'm working through, distributions are often "divided" by other distribution, and while I sort of understand what is meant, i would rather a rigorous explanation.

Let me provide an example:


$V$ ~$\chi_{9}^{2}$

U and V are independent random variables. Find $P(U-3<4.33\sqrt{V})$

$P(U-3<4.33\sqrt{V})$= $P(N(3,16)-3<4.33\sqrt{V})$= $P(N(0,16)<4.33\sqrt{V})$ =$P(4 \times N(0,1)<4.33\sqrt{V})$

=$P(\frac { N(0,1)}{\sqrt{V}} <\frac{4.33}{4})$

=$P(\frac {N(0,1)}{\sqrt{\chi_{9}}} <\frac{4.33}{4})$

=$P(3\times\frac {N(0,1)}{\sqrt{\chi_{9}}} <3\times\frac{4.33}{4})$

=$P(\frac {N(0,1)}{\sqrt{\frac{\chi_{9}}{9}}} <3\times\frac{4.33}{4})$ =$P(t_{9} <3\times\frac{4.33}{4})$

(where $t_{9}$ denotes the t distribution with 9 degrees of freedom)

I understand that "If $X_1,...,X_n$ are a random sample from $N(\mu,\sigma^2)$ then $\frac{\overline{X}-\mu}{\frac{s}{\sqrt{n}}}$ is distributed $t_{(n-1)}$ where $s=\sqrt{\frac{1}{n-1}\sum(X_i-\overline{X})^2}$.

But what does it mean to divide $N(0,1)$ by $\chi_{9}$?? why is $\frac {N(0,1)}{\sqrt{\frac{\chi_{9}}{9}}}$ equivalent to $t_{9}$??

  • 3
    $\begingroup$ The notation is truly awful ... (and not your fault, since it seems you're quoting somebody). $\endgroup$
    – whuber
    May 17 '19 at 14:26
  • 2
    $\begingroup$ @stochasticmrfox One further problem is that their notation seems to conflate the chi-square with the chi, a recipe for confusion. $\endgroup$
    – Glen_b
    May 17 '19 at 16:58

They're not dividing distributions. They're using a lot of shortcut notation. The piece that you're confused with appears to be this one: $$\frac { N(0,1)}{\sqrt{\frac {\chi_9} 9}} $$

What they mean here is the new variable: $$X=\frac{\xi}{u}$$ where $\xi$ is from standard normal distribution and $u$ is from $\chi$ distribution with 9 degrees of freedom scaled by the degrees of freedom. So, this variable $X$ will have its own distribution, which happens to be Student t distribution.

How do they know that it's Student t? They must know the history: it's how Gosset derived this distribution. In fact, I think this is a homework, and it was specifically designed to test whether you can spot the possibility to reduce the problem to the Student t distribution by noticing that it can be re arranged to have a ratio of normal and square root of chi-squared random variables.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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