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

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The discussion revolves around calculating conditional variances using the Law of Total Variance, specifically for the joint density function f(x,y) = 2, where 0<x<y<1. Participants have derived V(Y|X) and E(Y|X) but are debating the correct approach to compute E(V(Y|X)). There is confusion regarding whether to use the marginal distribution or the joint probability density function for integration, as well as the appropriate variable of integration (dy or dx). The importance of clearly distinguishing between functions in notation is emphasized to avoid losing marks on assignments. The conversation highlights the complexities of conditional variance calculations in probability theory.
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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 V(Y|X) = \frac{(1-x)^2}{12} and E(Y|X) = \frac{x+1}{2}

The Attempt at a Solution


So, E(V(Y|X))=E(\frac{(1-x)^2}{12}) = \int_0^y \frac{(1-x)^2}{12}f(x)dx, 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 V(Y|X) = \frac{(1-x)^2}{12} and E(Y|X) = \frac{x+1}{2}




The Attempt at a Solution


So, E(V(Y|X))=E(\frac{(1-x)^2}{12}) = \int_0^y \frac{(1-x)^2}{12}f(x)dx, 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 f(x) = \int_{-\infty}^{\infty} f(x,y)dy

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 E(V(Y|X)) = \int_x^1 \frac{(1-x)^2}{12}f(x)dy
Should we instead get E(V(Y|X)) = \int_x^1 \frac{(1-x)^2}{12}f(x,y)dy ?
 
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 f(x) = \int_{-\infty}^{\infty} f(x,y)dy

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 E(V(Y|X)) = \int_x^1 \frac{(1-x)^2}{12}f(x)dy
Should we instead get E(V(Y|X)) = \int_x^1 \frac{(1-x)^2}{12}f(x,y)dy ?

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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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