Independence of Random Variables

In summary, the question asks if two discrete random variables, X and Y, are independent based on their probability mass functions. The solution involves calculating the marginal mass functions for both X and Y and then finding the joint mass function pX,Y. If the joint distribution does not factorize, then X and Y are not independent.
  • #1
kingwinner
1,270
0

Homework Statement


Suppose X is a discrete random variable with probability mass function
pX(x)=1/5, if x=-2,-1,0,1,2
pX(x)=0, otherwise
Let Y=X2. Are X and Y independent? Prove using definitions and theorems.

Homework Equations


The Attempt at a Solution


The random variables X and Y are independent <=> pX,Y(x,y)=pX(x)pY(y) for ALL x,y E R

But the trouble here is that we don't have pX,Y(x,y) and pY(y). What can we do?

Thanks for any help!
 
Physics news on Phys.org
  • #2
You will first have to ccalculate pY and pX,Y.
For example
[tex]
p_Y(1)=P(Y=1)=P(X=1\vee X=-1)=P(X=1)+P(X=-1)=1/5+1/5=2/5
[/tex]
IN the same way calculate pY(y) for the remaining values in the range of Y, and similarly for pX,Y. Of course you would suspect X and Y not to be independent, so it would suffice to find one case where the joint distribution does not factorize.
 
  • #3
How can I find the joint mass function pX,Y from their marginal mass functions?
 
  • #4
[tex]
p_{X,Y}(1,2)=P(X=1\vee Y=2)=P(X=1\vee X^2=2)...?
[/tex]
Which X values contribute to this probabbility. They should be such that X=1 and X^2=2:smile:
 
  • #5
Pere Callahan said:
[tex]
p_{X,Y}(1,2)=P(X=1\vee Y=2)=P(X=1\vee X^2=2)...?
[/tex]
Which X values contribute to this probabbility. They should be such that X=1 and X^2=2:smile:

How is this possible? Is the probability 0?
What other cases do I have to do?
 
  • #6
kingwinner said:
How is this possible? Is the probability 0?
What other cases do I have to do?
Yes, the prob. is zero. Either do all remaining cases or just compare
[tex]
p_{X,Y}(1,2)=0
[/tex]
to
[tex]
p_{X}(1)p_{Y}(2)=?
[/tex]
 

1. What is the concept of independence of random variables?

The independence of random variables refers to the idea that the values of two or more random variables do not affect each other. In other words, the occurrence of one variable does not impact the likelihood of the other variable occurring.

2. How is independence of random variables different from correlation?

Independence of random variables means that there is no relationship between the variables, while correlation measures the strength and direction of the linear relationship between variables. Two variables can be independent but not correlated, or they can be correlated but not independent.

3. How can I determine if two random variables are independent?

To determine if two random variables are independent, you can calculate their joint probability distribution and compare it to the product of their individual probability distributions. If they are equal, the variables are independent. Additionally, you can use statistical tests such as the chi-square test or Pearson's correlation coefficient to assess independence.

4. What are some examples of independent random variables?

Some examples of independent random variables include the outcomes of two separate coin flips, the height and weight of individuals in a population, and the number of cars that pass through two different intersections in a city. In each of these examples, the values of one variable do not impact the values of the other variable.

5. Why is the concept of independence of random variables important?

The concept of independence of random variables is important in statistics and probability because it allows us to make assumptions and simplify calculations. When two variables are independent, we can use simpler models and techniques to analyze them. It also helps us to understand the relationship between variables and make more accurate predictions in various fields such as finance, engineering, and social sciences.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
414
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
975
  • Calculus and Beyond Homework Help
Replies
19
Views
760
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
466
  • Calculus and Beyond Homework Help
Replies
6
Views
603
  • Calculus and Beyond Homework Help
Replies
5
Views
925
Back
Top