Conditional Variances

• Scootertaj
In summary: However, I am not sure if this is the right way to do it.In summary, Homework Equations state that V(Y|X) = \frac{(1-x)^2}{12} and E(Y|X) = \frac{x+1}{2}

Scootertaj

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?

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.

1. What is a conditional variance?

A conditional variance is a statistical measure that represents the variability or spread of a data set around its conditional mean. It measures the deviation of the data points from the conditional mean and can be used to assess the level of risk or uncertainty in a given set of data.

2. How is conditional variance calculated?

Conditional variance is calculated by taking the sum of the squared differences between each data point and its conditional mean, and then dividing that sum by the number of data points. This is also known as the mean squared difference or the mean squared error.

3. What is the difference between conditional variance and unconditional variance?

Conditional variance is calculated based on a specific condition or subset of data, such as a particular time period or group. Unconditional variance, on the other hand, is calculated for the entire data set without any conditions. It represents the overall variability of the data.

4. How is conditional variance used in statistical modeling?

Conditional variance is an important measure in statistical modeling, particularly in time series analysis and financial modeling. It is used to assess the accuracy of predictions and to identify patterns and trends in the data. It is also used to determine the appropriate level of risk in investment decisions.

5. Can conditional variance be negative?

No, conditional variance cannot be negative. It represents the squared differences between data points and the mean, so it will always be a positive value. If the calculated value is negative, it is usually due to an error in the calculation or the data being used.