How Does Law of Total Variance Apply in Calculating Conditional Variances?

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Homework Help Overview

The discussion revolves around applying the Law of Total Variance in the context of conditional variances, specifically focusing on the joint probability density function f(x,y) defined for the region 0

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of E(V(Y|X)) and question the correct use of the joint versus marginal distributions in their integrals. There is uncertainty about whether to integrate with respect to x or y, and some participants express confusion over the notation used for the probability density function.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the assumptions made regarding the integration variables and the notation of the probability density function. Some guidance has been offered regarding naming conventions, but no consensus has been reached on the integration approach.

Contextual Notes

Participants are working under the constraints of specific homework instructions that dictate the use of certain notations for the joint density function. There is also a noted ambiguity regarding the dependence of V(Y|X) on the variable y, which is being actively explored.

Scootertaj
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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?
 
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Scootertaj said:
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
 
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] ?
 
Scootertaj said:
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:
Ray Vickson said:
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.
 

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