joint pmf of Y1=X1-X2 and Y2=X1+X2 This is not homework. 
I am just bothered about question 2.2.1 of Introduction to Mathematical Statistics (Sixth or seventh edition) of Hogg,McKean and Craig. Question for ready reference is: 
If the joint pmf of $X_1$ and $X_2$ is...
 $$p(x_1,x_2)=(2/3)^{x_1+x_2} (1/3)^{2-x_1-x_2}$$ 
for $(x1,x2)\in(0,0), (1,0), (1,0), (1,1)$ and zero elsewhere, find the joint pmf of $Y_1=X_1-X_2$ and $Y_2=X_1+X_2$.
First I checked that $p(x_1,x_2)$ is a proper pmf i.e.
$$ \sum_{X_1,X_2} p(X_1,X_2) = 1$$
Which for this pdf is true:
$$ p(0,0)+p(0,1)+p(1,0)+p(1,1)=1/9+2/9+2/9+4/9=1$$ 
Next, starting with $y_1=X_1-X_2$ and $Y_2=X_1+X_2$ will give us $x_1=(y_1+y_2)/2$ and $x_2=(y_2-y_1)/2$. By putting these values of $x_1$ and $x_2$ in the given equation above i.e. $p(x_1,x_2)$ this gives us
$$p(y_1,y_2)=(2/3)^{(y_1+y_2)/2+(y_2-y_1)/2} (1/3)^{2-(y1+y2)/2-(y2-y1)/2}$$.
Simplifying we obtain:
$$p(y_1,y_2)=(2/3)^{y_2} (1/3)^{2-y_2}$$
What is bothering me that last expression that is joint pmf of $Y_1$ and $Y_2$ does not have $Y_1$. So how it can be proved to joint pmf?
 A: Ok, it is fairly easy to show that it is a legitimate pmf. There are only four cases that you have to look at:
If $y_2 = 0$ then $y_1 = 0$ because $x_1$ and $x_2$ must have both been $0$.
If $y_2 = 1$ then $y_1= 1$ or $y_2 = -1$ because one of the x's was 1 and the other 0.
If $y_2 = 2$ then $y_1=0$ because both of the x's were 1.
So, $P_{Y_1, Y_2}(0, 0) = \frac{1}{9}$, $P_{Y_1, Y_2}(-1, 1) = \frac{2}{9}$, $P_{Y_1, Y_2}(1, 1) = \frac{2}{9}$ and $P_{Y_1, Y_2}(0, 2) = \frac{4}{9}$.
That's it, it is a pmf which depends only on $Y_1$ through the possible values admitted by $Y_2$.
A: $y_1$ does not be in the pdf because $y_2$ contains all the information about how the events (or values) of $x_1$ and $x_2$ affect the original distribution $p(x_1,x_2)$. So we don't need $y_1$ at all to know what probability occurred in $p(x_1,x_2)$. Remember that:
$$Y_2 = X_1 + X_2$$
Which has support $(Y_2 \in 0,1,2 \ )$. Consider the following scenarios:


*

*$Y_2 = 0 \Rightarrow p(x_1,x_2) = p(0,0) = \frac{1}{9} $

*$Y_2 = 1 \Rightarrow p(x_1,x_2) = p(1,0) = p(0,1) = \frac{2}{9} $

*$Y_2 = 2 \Rightarrow p(x_1,x_2) = p(0,0) = \frac{4}{9} $


As we can see, $y_2$ and thus $p(y_1,y_2)$, tells us which value the function $p(x_1,x_2)$ took on. $y_2$ contains all the information about $p(x_1,x_2)$ and we thus we don't need $y_1$ in our joint pmf.
Secondly, your answer is also a valid pmf. @Jim Baldwin is correct the two conditions to prove that an equation is a proper pdf. These can be found in (Casella & Berger, Statistical Inference, 2002 p.36) which states:
Theorem 1.6.5 A function  $\; p_X (x) $ is a pdf (or pmf) of a random variable $X$ if and only if:


*

*$p_{X}(x) \geq 0$ for all $x$ 

*$\sum_{x} p_X (x) =1$


The pdf you solved for satisfies both conditions.
A: That all values are non-negative, sum to 1, and cover all of the possibilities of the values of y1 and y2 (along with one-to-one correspondence with the x1,x2 pairs) should be enough to satisfy that this is a legitimate joint probability mass function.
If you didn't see the formula for the density but only the 4 positive probabilities, I suspect there wouldn't be any doubt.  (And note that a uniform distribution probability density function has a constant value which does not include the symbol for the random variable and it is a legitimate probability density.)
