Conditional Variances

  • Thread starter Scootertaj
  • Start date
  • #1
Scootertaj
97
0
1. Given f(x,y) = 2, 0<x<y<1, show V(Y) = E(V(Y|X)) + V(E(Y|x))



Homework Equations



I've found [tex]V(Y|X) = \frac{(1-x)^2}{12}[/tex] and [tex]E(Y|X) = \frac{x+1}{2}[/tex]




The Attempt at a Solution


So, [tex]E(V(Y|X))=E(\frac{(1-x)^2}{12}) = \int_0^y \frac{(1-x)^2}{12}f(x)dx[/tex], correct?
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
1. Given f(x,y) = 2, 0<x<y<1, show V(Y) = E(V(Y|X)) + V(E(Y|x))



Homework Equations



I've found [tex]V(Y|X) = \frac{(1-x)^2}{12}[/tex] and [tex]E(Y|X) = \frac{x+1}{2}[/tex]




The Attempt at a Solution


So, [tex]E(V(Y|X))=E(\frac{(1-x)^2}{12}) = \int_0^y \frac{(1-x)^2}{12}f(x)dx[/tex], correct?

As written, no it is not correct: f is a function of two variables. Perhaps you do not really mean f(x) in what you wrote, in which case you should re-write it by saying what you do mean. (I can guess, but you should not ask me to do that, nor should you ask that of the person who will mark the work.)

RGV
 
  • #3
Scootertaj
97
0
Well, my thinking was that the solution for V(Y|X) is not dependent on the value of y, thus we would only need to use the marginal dist [tex]f(x) = \int_{-\infty}^{\infty} f(x,y)dy[/tex]

Even though V(Y|X) contains no y, should we still use the joint pdf?

Moreover, I started thinking that we should be using dy instead of dx for the expectation.
So, my thinking is we would get [tex]E(V(Y|X)) = \int_x^1 \frac{(1-x)^2}{12}f(x)dy[/tex]
Should we instead get [tex]E(V(Y|X)) = \int_x^1 \frac{(1-x)^2}{12}f(x,y)dy[/tex] ?
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Well, my thinking was that the solution for V(Y|X) is not dependent on the value of y, thus we would only need to use the marginal dist [tex]f(x) = \int_{-\infty}^{\infty} f(x,y)dy[/tex]

Even though V(Y|X) contains no y, should we still use the joint pdf?

Moreover, I started thinking that we should be using dy instead of dx for the expectation.
So, my thinking is we would get [tex]E(V(Y|X)) = \int_x^1 \frac{(1-x)^2}{12}f(x)dy[/tex]
Should we instead get [tex]E(V(Y|X)) = \int_x^1 \frac{(1-x)^2}{12}f(x,y)dy[/tex] ?

No, no, no. Just use a different name for f(x), such as g(x) or fX(x). It is bad form to use the same letter to stand for two different functions in the same problem. That is something you should learn once and for all, because not observing it is a good way to lose marks on an assignment and on a test.

RGV
 
Last edited:
  • #5
Scootertaj
97
0
No, no, no. Just use a different name for f(x), such as g(x) or fX(x). It is bad form to use the same letter to stand for two different functions in the same problem. That is something you should learn once and for all, because not observing it is a good way to lose marks on an assignment and on a test.

RGV

RGV

That is the way we are instructed to "name" it in class. f(x) is the joint density function of f(x,y).
fx(x) is equivalent, but 99.9% of the time the Professor uses f(x).

I'm still stuck on whether we should be integrating with respect to y or x (use dy or dx).

Intuitively, dy makes more sense to me since we are taking the expectation of the variance of Y given x.
 

Suggested for: Conditional Variances

Replies
4
Views
220
Replies
4
Views
465
Replies
3
Views
90
Replies
4
Views
724
  • Last Post
Replies
2
Views
407
Replies
58
Views
910
Replies
8
Views
645
Top