Showing Chi squared is independent with another variable

In summary, the conversation discusses the distribution of Y and Z, where Y is found to be chi squared with degrees of freedom 2 and Z has a distribution that can be determined using a shortcut. It is also mentioned that Y and Z are not independent due to their dependence on each other. The idea of using polar coordinates is brought up as a possible approach to further analyze the relationship between Y and Z. It is emphasized that simply finding an equation where Y and Z appear together is not enough to prove dependence.
  • #1
torquerotates
207
0
So I have X1 and X2 are iid standard normal.

Then I have Y=X1^2+ X2^2

and

Z=X1/(X1^2+x2^2)

I'm supposed to find the distribution of Y and Z and then determine if they are independent.

Clearly Y is chi squared with degrees of freedom 2.

But I have no idea how to find the distribution of Z. I know there is a shortcut without using the jacobian, like I did with the Chi Squares, but I'm not sure how to do it.
I know Y and Z are not independent because with some algebra,

Y=Y*Z^(2)+X2
So Y depends on Z and vice versa, therefore they cannot be independent. Is that true?
 
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  • #2
(I moved this thread to the homework section)

I could be interesting to go to polar coordinates (in the X1-X2-plane).

Just finding an equation where Y and Z appear is not sufficient to show a dependence.
Imagine Y=X1, Z=X2, then Y+Z=X1+X2 but they are clearly independent.
 

1. What is the significance of showing that Chi squared is independent with another variable?

Understanding the relationship between two variables is important in statistical analysis. By showing that Chi squared is independent with another variable, we can determine if there is a significant relationship between the two variables. This can help us make predictions and draw conclusions about the data.

2. How do you determine if Chi squared is independent with another variable?

To determine if Chi squared is independent with another variable, we use the Chi squared test. This test compares the expected frequencies of a variable to the actual frequencies observed in the data. If the test shows a significant difference between the expected and observed frequencies, we can conclude that Chi squared is not independent with the other variable.

3. What is the null hypothesis in a Chi squared independence test?

The null hypothesis in a Chi squared independence test is that there is no relationship between the two variables. This means that any observed differences between the expected and observed frequencies are due to chance.

4. How do you interpret the results of a Chi squared independence test?

If the Chi squared test shows a p-value less than 0.05, we can reject the null hypothesis and conclude that there is a significant relationship between the two variables. On the other hand, if the p-value is greater than 0.05, we fail to reject the null hypothesis and cannot conclude that there is a significant relationship between the variables.

5. What are some limitations of using Chi squared to show independence with another variable?

One limitation of using Chi squared to show independence with another variable is that it only measures linear relationships. This means that a significant Chi squared result does not necessarily indicate a strong relationship between the variables. Additionally, Chi squared does not take into account the strength or direction of the relationship, only whether or not there is a relationship at all.

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