Probability & Expectation Value of X + Y

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Homework Help Overview

The discussion revolves around calculating the expectation value of the sum of two random variables, X and Y, given their joint probability distribution function f(m,n). The original poster expresses uncertainty about how to begin solving the problem, particularly regarding the definition of expected value.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants question the accuracy of the function notation and the values assigned to f(m,n), seeking clarification on the possible values for X and Y. There is a suggestion to create a probability table based on the provided probabilities, and two methods for calculating the expectation value are introduced. Additionally, some participants express confusion about the table's structure and commands.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and attempting to clarify the setup. Some guidance has been offered regarding the construction of a probability table and methods for calculating the expectation value, but there is no explicit consensus on the correct approach yet.

Contextual Notes

There is a noted ambiguity regarding the notation used for the joint probability function and the values assigned to it. Participants are encouraged to ensure all information is accurately provided, as the completeness of the data is crucial for solving the problem.

ParisSpart
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we have variables X,Y with f(m,n)=P(X=m,Y=n) with f(0.1)=0.1 f(1.0)=0.1 f(1.1)=0.344
find the expectation value E(X+Y)


i need help because i don't how to start to solve this , if i begin with the definition of the expected value i can't do anything any ideas?
 
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If your function f requires two arguments, as you indicate by f(m,n), how can you calculate f(0.1) = 0.1 and so on? do you mean f(0,1) = 0.1, f(1,0) = 0.1, f(1,1) = 0.344?
What values other than 0 and 1 are possible for X and Y, or are these the only possibilities? Make sure you've provided ALL the information accurately.
 
f(m.n) the m and n takes 0 and 1 values ...
 
I will assume my comment about you meaning f(0,1) instead of f(0.1) was correct.
Think about it this way: if X, Y can each be only 0 or 1 there are only 4 possibilities: (0,0), (1,0), (0,1), (1,1). You have the probabilities assigned to three of the four, so you can make a table with two columns (labeled 0 and 1) and two rows (also labeled 0 and 1). In each cell put the appropriate probability:
In cell (0,1) put .1, in cell (1,0) put .1, in cell (1,1) put 0.344. (Since the four probabilities have to sum to 1 you can find the probability for cell (0,0) yourself.)

Now you have two ways to go.
Method A: To calculate the expectation work out the sum

[tex] E(X+Y) = \sum f(m,n) (m+n)[/tex]

- it will have as many terms as there are cells in the table.

Method B:
The rows of your table have the values for X, the columns the values for Y, so the edges of the table give the marginal distributions of X and Y. Then

[tex] E(X+Y) = E(X) + E(Y)[/tex]

can be calculated using the appropriate marginal distributions.

Methods A and B give the same answer (when applied correctly) so you can use whichever you prefer.

Set things up and try the work before posting more questions.
 
Last edited:
i don't understand the commands of the table... the commands...
 
can you write again the table?
 
ParisSpart said:
can you write again the table?

More important: can YOU write the table? If so, do it. If you cannot, tell us why.
 
if i wanted to find the P(X=0) i will find f(0,0) and if yes , why?
 

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