I'm revising for an upcoming exam with old assignment questions, but I got this one wrong at the time and we aren't given model solutions. Looking for advice on whether or not my second attempt for A) is correct,and if not a tip in the right direction, and any hints on part B), thank you.
Let $X$ and $Y$ denote the respective outcomes when two fair dice are thrown. Let $U=\text{min}(X,Y)$, $V=\text{max}(X,Y)$, $S = U+V$, and $T=V-U$
A) Determine the conditional probability mass function for $U$ given $V=v$
B) Determine the joint mass function for $S$ and $T$
My Attempt:
A)
$\begin{align}
P(U=u|V=v)&=P(\text{min}(X,Y)=u|\text{Max}(X,Y)=v)\\
&=[P(X=u,Y=v)+P(X=v,Y=u)]/P(\text{max}(x,y)=v)\\
&=\frac{1}{18 \times P(\text{max}(x,y)=v)}
\end{align}$
$\begin{align}
P(\text{max}(x,y)=v)&=P(X=v,Y\leq v)+P(Y=v,X\leq v)\\
&=2 \times P(X=v,Y\leq v)\quad \text{(By symmetry)} \\
&=2\times(1/6)\times (v/6)\\
&=v/18
\end{align}$
Substituting back into the above gives the conditional distribution for $U$ given $V=v$ as $1/v$.
Edit: After revision the PMF for the maximum came out as (2v-1)/36 which means the above conditional pmf is definitely wrong
$B)$
$\begin{align}
P(S=s,T=t)&=P(U+V=s,V-U=t)\\
&=P(X+Y=s,|X-Y|=t)\\
&=P(X+Y=s,X-Y=t)+P(X+Y=s,X-Y=-t)\\
\vdots \\
&=P(Y=(s-t)/2,X=(s+t)/2)+P(X=(s-t)/2,Y=(s+t)/2)
\end{align}$
I'm stuck here. A hint in the right direction would be greatly appreciated.
Best Answer
Draw a $6$" $\times$ $6$" square and divide it into a $6\times 6$ array of $36$ one-inch squares. Label the rows and columns with numbers $1$-$6$ and in each square, write down the values of $(X,Y)$, $U$, $V$, $S$ and $T$ in each square . Then, count!