What does it mean exactly to divide a distribution by another distribution?

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:

$$U$$~$$N(3,16)$$

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

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 $$t_{9}$$??

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