# A multi-part PMF/joint PMF question

#### nacho

##### Active member
Please refer to the attached image.

for Question 1,
Referring to the solutions, I know how to derive the values of
$z$ in the table, but not of $p_Z(z)$. What have they done there?

#### chisigma

##### Well-known member
Please refer to the attached image.

for Question 1,
Referring to the solutions, I know how to derive the values of
$z$ in the table, but not of $p_Z(z)$. What have they done there?

View attachment 1610
First step it to express Z = F (X,Y) = 3 X - 2 Y as function of X and Y...

F(1,1) = 1

F(1,2) = -1

F(1,3) = - 3

F(2,1) = 4

F(2,2) = 2

F(2,3) = 0

F(3,1) = 7

F(3,2) = 5

F(3,3) = 3

Setting P (Z) the PMF of Z we have...

P(-3) = 1/84 (1 + 9) = 5/42

P (-1) = 1/84 (1 + 4) = 5/84

Are You able to proceed?...

Kind regards

$\chi$ $\sigma$

#### nacho

##### Active member
First step it to express Z = F (X,Y) = 3 X - 2 Y as function of X and Y...

Setting P (Z) the PMF of Z we have...

P(-3) = 1/84 (1 + 9) = 5/42

P (-1) = 1/84 (1 + 4) = 5/84

$\chi$ $\sigma$
where do you get the bracketed terms from, such as (1 + 9) and (1 + 4)

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#### nacho

##### Active member
Sorry, I know this is bumping and at mhb this is against the rules, but I am really quite desperate (and promise never to do this again).
I was able to do this previously so i know it can't be difficult, i've just forgotten something basic.

but how exactly is $p_Z(z)$ obtained?
I would be so grateful if someone could tell me. I have an exam in two days and don't want to lose easy marks like this !

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
where do you get the bracketed terms from, such as (1 + 9) and (1 + 4)
This is recalculating the values in the first table in the image in post #1. Thus,
$P(Z=-3)=P(X=1,Y=3)=p_{X,Y}(1,3)=c(1^2+3^2) =(1+9)/84=10/84$
and
$P(Z=-1)=P(X=1,Y=2)=p_{X,Y}(1,2)=c(1^2+2^2)= (1+4)/84=5/84$
Here it turns out that each value of $Z$ can be produced by one and only one pair of values of $X$ and $Y$, i.e., the function $F$ from post #2 is injective. Therefore, $P(Z=z)=P(X=x,Y=y)$ where $(x,y)$ is the unique pair such that $F(x,y)=z$. If $F$ were not unique, then
$P(Z=z)=\sum_{F(x,y)=z}P(X=x,Y=y)$
(the sum is over all $(x,y)$ such that $F(x,y)=z$).

#### nacho

##### Active member
Thank you so much, I really appreciate that!