Exercise: Expected value rule and total expectation theorenm 8 points possible (graded) Let X, Y, and Z be jointly continuous random variables. Assume that all conditional PDFs and expectations are well defined. E.g., when conditioning on X, assume that is such that fx(x) > 0. For each one of the following formulas, state whether it is true for all choices of the function g or false (i.e., not true for all choices of g). True False True 4 True
The Correct Answer and Explanation is:
Let’s analyze each of the four formulas one by one using the Expected Value Rule and the Law of Total Expectation, also known as the Tower Property. While the exact formulas are not listed in your message, from the context and typical exercises, we can infer common forms involving expectations and conditional expectations of functions like g(X,Y,Z)g(X, Y, Z). Let’s assume the four statements to evaluate are as follows:
(1) E[g(X,Y,Z)]=E[E[g(X,Y,Z)∣X]]\mathbb{E}[g(X,Y,Z)] = \mathbb{E}[\mathbb{E}[g(X,Y,Z)\mid X]]
✅ True
This is the law of iterated expectations (Total Expectation Theorem). It states that the expectation of a random variable (or function of random variables) equals the expectation of its conditional expectation, as long as the expectation is defined. This holds for any measurable function gg, provided E[∣g(X,Y,Z)∣]<∞\mathbb{E}[|g(X,Y,Z)|] < \infty. This is a fundamental identity in probability theory.
(2) E[g(X,Y,Z)]=E[E[g(X,Y,Z)∣Y]]\mathbb{E}[g(X,Y,Z)] = \mathbb{E}[\mathbb{E}[g(X,Y,Z)\mid Y]]
✅ True
Same as above, but conditioning on YY. The law of total expectation also holds regardless of which variable you condition on, as long as the joint distribution is well-defined and expectations are finite. So this is also true for all functions gg under the stated conditions.
(3) E[g(X,Y,Z)∣X]=E[g(X,Y,Z)]\mathbb{E}[g(X,Y,Z)\mid X] = \mathbb{E}[g(X,Y,Z)]
❌ False
This is not true in general. The conditional expectation E[g(X,Y,Z)∣X]\mathbb{E}[g(X,Y,Z)\mid X] is a random variable that depends on XX, while the right-hand side is a constant (the unconditional expectation). These two are only equal if g(X,Y,Z)g(X,Y,Z) is independent of XX, which is not guaranteed for all gg. So this is false in general.
(4) E[E[g(X,Y,Z)∣Y,Z]]=E[E[g(X,Y,Z)∣X,Y]]\mathbb{E}[\mathbb{E}[g(X,Y,Z)\mid Y,Z]] = \mathbb{E}[\mathbb{E}[g(X,Y,Z)\mid X,Y]]
❌ False
The left-hand side conditions on (Y,Z)(Y,Z), while the right-hand side conditions on (X,Y)(X,Y). Unless g(X,Y,Z)g(X,Y,Z) has some symmetry or specific dependence structure, these two inner expectations will differ, and thus their expectations will not necessarily be equal. So this is also false in general.
✅ Final Answers:
- True
- True
- False
- False
💡 Explanation
The expected value rule and the law of total expectation (also called the tower property) are fundamental in probability. They help break down complex expectations into more manageable conditional expectations. Specifically, for random variables X,Y,ZX, Y, Z and a measurable function gg, the law says: E[g(X,Y,Z)]=E[E[g(X,Y,Z)∣X]]\mathbb{E}[g(X,Y,Z)] = \mathbb{E}[\mathbb{E}[g(X,Y,Z) \mid X]]
This identity is always true under basic regularity conditions, such as E[∣g(X,Y,Z)∣]<∞\mathbb{E}[|g(X,Y,Z)|] < \infty. The conditioning variable could be any subset of the random variables involved—XX, YY, or even (X,Y)(X,Y)—and the law will still hold. Therefore, both formulas (1) and (2) are universally true.
However, formula (3) falsely equates a conditional expectation with an unconditional one. Conditional expectations depend on the conditioning variable and can vary with its values, while the unconditional expectation is a constant. These are only equal if the conditioned variable provides no additional information—i.e., when the function is independent of the variable being conditioned on—which is not guaranteed for all functions gg.
Formula (4) incorrectly assumes that conditioning on different variables (e.g., (Y,Z)(Y,Z) vs. (X,Y)(X,Y)) yields the same result. This isn’t generally true unless the function gg and the joint distribution of the variables have special properties (like symmetry or conditional independence). Hence, it cannot hold for all functions gg, making it false.
In conclusion, only statements (1) and (2) universally hold.
