Find conditional probability mass function

[psie]

Homework Statement

Consider a fair die thrown twice. Let $U_1$ be the dots on the first throw and $U_2$ on the second (intuitively, they are independent). Let $X=U_1+U_2$ and $Y=\min(U_1,U_2)$. Find the conditional probability (mass) function $p_{Y|X=x}(y)$ and the conditional distribution $F_{Y|X=x}(y)$.

Relevant Equations

We have $$p_{Y|X=x}(y)=\frac{p_{X,Y}(x,y)}{p_X(x)}=\frac{p_{X,Y}(x,y)}{\sum_z p_{X,Y}(x,z)},$$and $$F_{Y|X=x}(y)=\frac{\sum_{z\leq y}p_{X,Y}(x,z)}{p_X(x)}=\frac{\sum_{z\leq y} p_{X,Y}(x,z)}{\sum_z p_{X,Y}(x,z)}.$$

I don't really know how to approach this problem, but my plan is to find $p_{X,Y}(x,y)=P(X=x,Y=y)$. The two conditions $X=x$ and $Y=y$ in terms of $U_1$ and $U_2$ read (I think) $$U_1=y,U_2 = x-y \text{ or }U_2 = y, U_1 = x-y,\qquad x\geq 2y.$$ So $$P(X = x, Y = y) = \begin{cases} P(U_1 = y, U_2 = x-y\text{ or }U_2 = y, U_1 = x-y) & x\geq 2y \\ 0 & x < 2y\end{cases}.$$ I don't know how to proceed and find $p_{Y|X=x}(y)$.

 

[Orodruin]

I think you are severely overcomplicating things. Both $X$ and $Y$ are functions of $U_1$ and $U_2$ whose probability distributions are very simple. For a given $X = x$, what are the possible values of $U_1$ and $U_2$? What are the probabilities of each combination? What does this mean for $Y$?

 

[psie]

I think you are severely overcomplicating things. Both $X$ and $Y$ are functions of $U_1$ and $U_2$ whose probability distributions are very simple. For a given $X = x$, what are the possible values of $U_1$ and $U_2$? What are the probabilities of each combination? What does this mean for $Y$?

I appreciate the reply, but I feel very hopeless about this. If we fix an $x$, then the possible values of $U_1,U_2$ are $U_1=y,U_2=x-y$ or $U_2=y,U_2=x-y$, and we want $x\geq 2y$. But then I'm just getting back into what I did above. Could you elaborate on your approach a bit more? Maybe show how you would obtain the conditional pmf, and from there I can see if I can either modify my approach or go with yours.

 

[Orodruin]

Could you elaborate on your approach a bit more?

Please try to answer the questions I posed in the previous post. You only answered the first:

For a given $X = x$, what are the possible values of $U_1$ and $U_2$?

The second and third are very significant:

What are the probabilities of each combination? What does this mean for $Y$?

Edit: Note: I am talking about the probabilities conditioned on $X = x$ here. It is the only thing you need to look at.

 

[Orodruin]

Here is an illustration to help your thought process:

If the general case is too hard to start with, focus on this special case of x = 9. What are the conditional distributions of $U_1$ and $U_2$. Can you fill out the values of $Y$ in this square? What is the conditional probability to find $Y = 4$?

 

[Orodruin]

If we fix an $x$, then the possible values of $U_1,U_2$ are $U_1=y,U_2=x-y$ or $U_2=y,U_2=x-y$, and we want $x\geq 2y$.

Ok, i just reread this and realized it actually did not really answer my first question either. The first question asks what is the possible combinations of $U_1$ and $U_2$ given $X = x$. The value of $Y$ does not enter into the question, just the conditioning on $X$.

 

[psie]

Here is an illustration to help your thought process:
View attachment 347103
If the general case is too hard to start with, focus on this special case of x = 9. What are the conditional distributions of $U_1$ and $U_2$. Can you fill out the values of $Y$ in this square? What is the conditional probability to find $Y = 4$?

Ok, thank you for elaborating. I don't see the general case right now. In this special case, the conditional pmf of $U_1$ and $U_2$ are, I believe, $\frac14$ (since $3,4,5$ and $6$ have equal probability of appearing). The values of $Y$ are $3$ or $4$, also with equal probability, so the conditional probability must be $\frac12$ to find $Y=4$ given $X=9$. Is this correct?

 

[haruspex]

the conditional probability must be 1/2 to find Y=4 given X=9. Is this correct?

Yes. Even values of X are a little trickier.

 

[Orodruin]

Yes. Even values of X are a little trickier.

Although still not very tricky …