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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.

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1 Answer 1

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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!

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  • $\begingroup$ That's what I answered initially, and it got marked wrong. I just can't see a way of expressing it in an actual pmf though, unless I do it in some sort of case by case basis. Any ideas? $\endgroup$
    – Patrick
    Jun 6, 2015 at 21:11
  • $\begingroup$ @Patrick Did you forget to count? If $\max(X,Y) = V$, its probability mass function is very definitely not $\frac{v}{18}$ for $1 \leq v \leq 6$. Since the numerators of $\frac{v}{18}$ sum to $1+2+3+4+5+6 = 21$, the pmf that you found does not satisfy the requirement $\sum_{v=1}^6 p_V(v) = 1$. $\endgroup$
    – Dilip Sarwate
    Jun 6, 2015 at 21:53
  • $\begingroup$ You're right, thanks for pointing that out. It just agreed with what I would expect the conditional pmf to be so I didn't bother checking. Will work on the problem again later and update it when I get something that works. $\endgroup$
    – Patrick
    Jun 6, 2015 at 23:03
  • $\begingroup$ I guess a rest helped me think because just after reading your comment I got the following: $P(max(X,Y)=v) = P(max(X,Y)<=v)-P(max(X,Y)<=v-1)$ =$P(X<=v,Y<=v)-P(X<=v-1,Y<=v-1)$ =$v^2/36 - (v-1)^2/36$ =$(2*v-1)/36$ Which does indeed sum to 1 over all possible values of v. Unless you can spot some flawed leap of logic in my reasoning when I changed the max into terms of X,Y. $\endgroup$
    – Patrick
    Jun 6, 2015 at 23:11
  • $\begingroup$ Consequently this must mean the joint PMF I derived in A) to find the conditional must be wrong, as 2/(2v-1) definitely does not sum to 1 as it should for some fixed v. It occurs to me that if you had $max(X,Y)=min(X,Y)$ then clearly the probability cannot be 1/18 as there is only one instance that could happen. Again now I'm confused how to express this without doing a case by case pmf. $\endgroup$
    – Patrick
    Jun 6, 2015 at 23:24

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