A multi-part PMF/joint PMF question

  • MHB
  • Thread starter nacho-man
  • Start date
In summary, the conversation discusses how to derive the values of $z$ in a table and the associated probability function $p_Z(z)$. The first step is to express $z$ as a function of $x$ and $y$, and then use the probabilities from the table to calculate the values of $p_Z(z)$. It is important to note that $z$ can be produced by one and only one pair of values of $x$ and $y$, making the function $F$ injective. If $F$ were not unique, the probability of $z$ would be calculated differently.
  • #1
nacho-man
171
0
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
 

Attachments

  • JOINT PMF Q.jpg
    JOINT PMF Q.jpg
    36.9 KB · Views: 75
Physics news on Phys.org
  • #2
nacho said:
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) = 5/84Are You able to proceed?... Kind regards $\chi$ $\sigma$
 
  • #3
chisigma said:
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)
 
Last edited:
  • #4
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 !
 
  • #5
nacho said:
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 $F(x,y)=z$).
 
  • #6
Thank you so much, I really appreciate that!
 

FAQ: A multi-part PMF/joint PMF question

What is a PMF and how is it different from a joint PMF?

A PMF, or probability mass function, is a function that assigns probabilities to discrete random variables. It shows the probability of each possible outcome of a random variable. A joint PMF, on the other hand, is a function that assigns probabilities to multiple discrete random variables. It shows the probability of each combination of outcomes of the random variables.

How do you calculate a multi-part PMF?

To calculate a multi-part PMF, you first need to identify all the possible outcomes of the random variables. Then, you can assign probabilities to each outcome based on the given information or data. Finally, you can add up the probabilities of all the outcomes to get the total probability of the multi-part PMF.

What is the purpose of using a multi-part PMF?

A multi-part PMF is useful for analyzing the probabilities of multiple events occurring simultaneously. It can also be used to calculate the joint probabilities of different combinations of outcomes, which can be helpful in decision-making and risk assessment.

Can a multi-part PMF be used for continuous random variables?

No, a multi-part PMF is only applicable to discrete random variables. Continuous random variables have an infinite number of possible outcomes, making it impossible to assign probabilities to each individual outcome.

How can a multi-part PMF be visualized?

A multi-part PMF can be visualized using a joint probability table or a joint probability plot. The table shows the probabilities of each combination of outcomes, while the plot displays the probabilities as points on a graph.

Similar threads

Replies
2
Views
2K
Replies
1
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
14
Views
6K
Replies
2
Views
2K
Replies
8
Views
848
Back
Top