$U$~$N(3,16)$ $V$ ~$\chi_{9}^{2}$ U and V are independent random variables. Find $P(U-3<4.33\sqrt{V})$ $U$~$N(3,16)$
$V$ ~$\chi_{9}^{2}$
U and V are independent random variables.
Find $P(U-3<4.33\sqrt{V})$
(The notes I'm working through don't seem to approach this rigorously...)
The answer is $P(t_{9} <3\times\frac{4.33}{4})$ and I'd appreciate a rigorous solution. where $t_{9}$ is the t distribution with 9 degrees of freedom.
 A: For independent random variables $Z$ and $V$ it can be proven that $\frac{Z}{\sqrt{\frac{V(m)}{m}}}\sim t_m \quad (1)$ 
with $Z$ is standard normally distributed, $V$ is Chi-sqared distributed with $m$ the degrees of freedom and $t_m$ is t-distributed with $m$ degrees of freedom as well. For a proof see here.
First of all we have to standardize $U$: $Z=\frac{U-3}{4}\sim \mathcal N(0,1)$
We have $\frac{U-3}4<\frac{4.33\sqrt{V}}4$
We can divide the equation by $\sqrt{V}$
$\frac{U-3}{4\cdot \sqrt{V}}<\frac{4.33}4$
To obtain $(1)$ we have to divide the denominator of by $\sqrt{m}=\sqrt{9}$. This is the same as multiplying the equation by $\sqrt{9}=3$.
$\frac{U-3}{4\cdot \sqrt{\frac{V}{9}}}<3\cdot \frac{4.33}4$, where $\frac{U-3}{\sqrt{4\cdot \frac{V}{3}}}\sim t_9$
A: Note that
$$
\frac{U-3}{4}\sim N(0,1)
$$
whence
$$
W=\frac{U-3}{4}\bigg/\sqrt {V/9}\sim t_9
$$
since $U$ and $V$ are independent. Now to compute the probability we note that
$$
U-3<4.33\sqrt{V}\iff W=\frac{3(U-3)}{4\sqrt{V}}<\frac{3}{4}\times 4.33
$$
so
$$
P(U-3<4.33\sqrt{V})=P\left(W<\frac{3}{4}\times 4.33\right)
$$
as desired.
