Cov(X,Y) Distribution Function

In summary: If there were a change of variables, I would transform the original density function to a new function f(z) = e-z and then integrate f(z). If there weren't a change of variables, I would calculate a double integral fX,Y(x,y) = e-(u+v) with du and dv from 0 to x and 0 to y respectively. If there were a change of variables, I would transform the original density function to a new function f(z) = e-z and then integrate f(z).
  • #1
Shackleford
1,656
2
For (4), when I calculated the Distribution Function, I had to break it up into two cases whether or not the point was above or below the line.

For (6.c), I got the right answer for the Cov(X,Y), but I didn't break up the double integral into two cases. Is this because since the density function has the condition that y is less than or equal to x, it is inherent to the double integral? That's why I was able to set the limits of integration of y from 0 to x?

http://i111.photobucket.com/albums/n149/camarolt4z28/2-2.png?t=1302966321

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-3.png?t=1302966341

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110416_100322.jpg?t=1302966359
 
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  • #2


Shackleford said:
For (4), when I calculated the Distribution Function, I had to break it up into two cases whether or not the point was above or below the line.
I couldn't see the picture for 4.
For (6.c), I got the right answer for the Cov(X,Y), but I didn't break up the double integral into two cases. Is this because since the density function has the condition that y is less than or equal to x, it is inherent to the double integral? That's why I was able to set the limits of integration of y from 0 to x?
Yes, that's correct since the restriction on x is that it must be positive but the restriction on y is that it is bounded by x so we could write that 0 < x < ∞ and 0 < y < x and it turns out from there we get our limits of integration with respect to x and y. I'm assuming you found E(X) = 2 and E(Y) = 1 since you found E(XY) = 3(which was calculated perfectly).
 
  • #3


Leptos said:
I couldn't see the picture for 4.

Yes, that's correct since the restriction on x is that it must be positive but the restriction on y is that it is bounded by x so we could write that 0 < x < ∞ and 0 < y < x and it turns out from there we get our limits of integration with respect to x and y. I'm assuming you found E(X) = 2 and E(Y) = 1 since you found E(XY) = 3(which was calculated perfectly).

Yes. I found the means all right.

I'm not sure how to start (a). The problem looks easy enough, though.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-4.png?t=1302983928

I assume in the uv-plane I draw the line v = z - u. I then setup a double integral ∫∫e-z dudv.

I'm looking for P(Z ≤ z). This would be everything integrated above the line in the shaded region.
 
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  • #4


Shackleford said:
Yes. I found the means all right.

I'm not sure how to start (a). The problem looks easy enough, though.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-4.png?t=1302983928

I assume in the uv-plane I draw the line v = z - u. I then setup a double integral ∫∫e-z dudv.

I'm looking for P(Z ≤ z). This would be everything integrated above the line in the shaded region.
Actually the change of variables would transform the original density f(x,y) = e-(x+y) to a function of z alone, i.e, f(z) = e-z whose only restriction is that 0 < z and now all you have to do is apply the definition of the cumulative density function: F(Z) = P(X ≤ x).
 
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  • #5


Leptos said:
Actually the change of variables would transform the original density f(x,y) = e-(x+y) to a function of z alone, i.e, f(z) = e-z whose only restriction is that 0 < z and now all you have to do is apply the definition of the cumulative density function: F(Z) = P(X ≤ x).

I assume I integrate f(z) = e-z. What the heck are my variables and limits of integration? 0 to infinity? dx and dy? Double or single integral?

If there weren't a change of variables, I would calculate a double integral fX,Y(x,y) = e-(u+v) with du and dv from 0 to x and 0 to y respectively.
 
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Related to Cov(X,Y) Distribution Function

1. What is the Covariance Distribution Function?

The Covariance Distribution Function is a statistical measure that describes the relationship between two random variables, X and Y. It measures how these variables change together, or how they are related to each other.

2. How is the Covariance Distribution Function calculated?

The Covariance Distribution Function is calculated by taking the product of the deviations of each variable from their respective means and then summing them up. This value is then divided by the total number of observations.

3. What does a positive/negative Covariance Distribution Function indicate?

A positive Covariance Distribution Function indicates a positive relationship between the two variables, meaning that as one variable increases, the other also tends to increase. A negative Covariance Distribution Function indicates a negative relationship, meaning that as one variable increases, the other tends to decrease.

4. What does a Covariance Distribution Function of zero mean?

A Covariance Distribution Function of zero indicates that there is no relationship between the two variables. This does not necessarily mean that the variables are independent, but rather that there is no linear relationship between them.

5. How is the Covariance Distribution Function used in data analysis?

The Covariance Distribution Function is used in data analysis to determine the strength and direction of the relationship between two variables. It can also be used to identify patterns and trends in data, as well as to assess the effectiveness of a statistical model.

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