Understanding the Expectation of Random Variables: EEXYZ ≠ EEXZY for Dependent Variables
In the realm of probability and statistics, it is often assumed that E(E(XYZ)) E(E(XZY)). However, this is not always the case, especially when the random variables involved are dependent. This article aims to explore this concept through a concrete example, providing a comprehensive understanding of the properties of expectations in such scenarios.
Background and Common Misconceptions
When dealing with the expectation of the product of random variables, a common misconception is that the order in which these variables are grouped does not affect the expectation. This is true for independent variables, but for dependent variables, the grouping can significantly alter the outcome. The primary goal of this article is to clarify this important distinction through a detailed example.
The Example: A Dependent Random Variable Set
Consider the set S {-1, 0, 1} from which we randomly draw a number. We define three indicator random variables as follows:
Rationale for Defining Indicator Variables
Total numbers: S {-1, 0, 1}
Variable X: Takes the value 1 if the number drawn is 1, otherwise 0. Variable Z: Takes the value 1 if the number drawn is -1, otherwise 0. Variable Y: Takes the value 1 if the number drawn is 1, otherwise 0. Note that Y X, meaning Y and X are the same.Calculating the Expectations
We aim to compute both E(E(XYZ)) and E(E(XZY)) and demonstrate their differences.
First Case: E(E(XYZ))
Since Y X, the expression simplifies as follows:
E(E(XYZ)) E(E(XXZ))
To understand this, let's break down the expected value:
E(XXZ) E(XYZ)
We now proceed to calculate E(XYZ). Since Y X, the probability distribution for the product XYZ can be simplified. When XYZ 0 or XYZ -1 or XYZ 1, we can calculate the expected value.
Second Case: E(E(XZY))
Again, using the relationship Y X, we can rewrite:
E(E(XZY)) E(E(XYZY))
Let's denote W YZ. The value of W will determine the expected value. Since Y X, we have:
E(Wy) E(XYZ)
Given that W YZ, we need to calculate E(1 - W / 2). Let's break down the possible values of W:
When W 1, the probability of this occurring is 1/3. When W 0, the probability of this occurring is 2/3.
The expected value E(1 - W / 2) can be calculated as follows:
E(1 - W / 2) (1 - (1 / 2)) * (1 / 3) (1 - (0 / 2)) * (2 / 3) 1/2
Thus, the final values are:
E(XYZ) 1/2 E(XZY) 1/2Conclusion
From the above demonstration, it is clear that the expectations E(XYZ) and E(XZY) are indeed the same in this specific example. However, this does not generalize the property E(E(XYZ)) E(E(XZY)) for all dependent variables. The key takeaway is that for dependent variables, the grouping of the variables in the expectation can lead to different results. This highlights the importance of understanding the implications of the underlying dependencies in probability theory.