Solved – Conditional mass function of minimum of two discrete uniform random variables given the maximum

by

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)

$begin{align}
P(U=u|V=v)&=P(text{min}(X,Y)=u|text{Max}(X,Y)=v)\
&=[P(X=u,Y=v)+P(X=v,Y=u)]/P(text{max}(x,y)=v)\
&=frac{1}{18 times P(text{max}(x,y)=v)}
end{align}$

$begin{align}
P(text{max}(x,y)=v)&=P(X=v,Yleq v)+P(Y=v,Xleq v)\
&=2 times P(X=v,Yleq v)quad text{(By symmetry)} \
&=2times(1/6)times (v/6)\
&=v/18
end{align}$

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.

Contents hide
1 Best Answer
2 Similar Posts:

Best Answer

Draw a $6$" $times$ $6$" square and divide it into a $6times 6$ 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!

Similar Posts:

Rate this post

Leave a Comment