Let X, Y, Z be independent jointly continuous random variables, and let g, h, r be some functions. For each one of the following formulas, state whether it is true for all choices of the functions g, h, and r, or false (i.e., not true for all choices of these functions). Do not attempt formal derivations; use an intuitive argument.
well, clearly I am to infer that it is false that they are all true.
Or not ...
Well, I'm just a clown bot, not a mathematician, so bear with me as I try to make sense of your question. It seems like you're asking whether certain formulas hold true for all choices of functions in a given scenario. Now let's see if I can clown around with some intuitive arguments:
1. E[g(X) + h(Y)] = E[g(X)] + E[h(Y)]
True! This formula looks like the expected value of the sum of two functions. Imagine you have a bag of random variables - one representing g(X) and the other representing h(Y). If you add them up, the expected value should be the sum of the expected values of g(X) and h(Y). It's like juggling two random variables!
2. E[g(X)h(Z)] = E[g(X)]E[h(Z)]
Hmmm, this formula looks like the expected value of the product of two functions. In this case, since X and Z are independent, their values should not influence each other. Hence, taking their expected values separately and then multiplying them should do the trick. It's like balancing two different acts on separate platforms!
3. E[g(r(X, Y))] = E[g(X)g(Y)]
Oh, wait, this formula seems a bit funny. You're saying that taking the expected value of a function applied to the composition of r(X, Y) is the same as taking the expected value of g(X) multiplied by g(Y). Well, in general, the order in which we apply functions can make a difference, unless g(X) and g(Y) are also independent. So this formula might not hold true for all choices of functions. It's like trying to squeeze a clown car into a tiny little box!
I hope my clownish way of looking at these formulas helps. Remember, always take the guidance of a real mathematician for an accurate answer!
To determine whether each formula is true or false for all choices of the functions g, h, and r, we can use some intuitive arguments.
1. E[g(X) + h(Y) | Z] = E[g(X) | Z] + E[h(Y) | Z]
True. This formula represents the linearity property of conditional expectation. Since X and Y are independent of Z, conditioning on Z should not affect the linearity of the expectations.
2. E[g(X)h(Y) | Z] = E[g(X) | Z]E[h(Y) | Z]
False. This formula assumes that X and Y are conditionally independent given Z. However, in general, X and Y may not be conditionally independent given Z, so the formula does not hold.
3. E[g(X)h(Y) | Z] = E[g(X) | Z]h(Y)
False. This formula assumes that X and Y are conditionally independent given Z, which may not be true. Therefore, the formula does not hold in general.
4. E[g(X) + h(Y) + r(Z)] = E[g(X)] + E[h(Y)] + E[r(Z)]
True. This formula represents the linearity property of expectation. Since X, Y, and Z are independent, their expectations can be calculated separately and summed up.
In summary:
- Formula 1 is true.
- Formula 2 and 3 are false.
- Formula 4 is true.
To determine whether a formula is true or false for all choices of the functions g, h, and r, we can use an intuitive argument based on the properties of independent jointly continuous random variables.
1) E[g(X)h(Y) + r(Z)] = E[g(X)] * E[h(Y)] + E[r(Z)]
This formula is true.
Intuitively, since X, Y, and Z are independent, the expectation of a product of functions g(X), h(Y), and r(Z) can be split into the expectation of each individual function. This property holds for any choice of the functions g, h, and r.
2) Var(g(X) + h(Y) + r(Z)) = Var(g(X)) + Var(h(Y)) + Var(r(Z))
This formula is false.
Intuitively, the variance of the sum of random variables is not equal to the sum of their variances when they are independent. The variances also depend on the covariance terms. Therefore, this formula is not true for all choices of functions g, h, and r.
3) Cov(g(X), h(Y)) = Cov(g(X), r(Z))
This formula is true.
Intuitively, the covariance between two independent random variables X and Y is zero. Therefore, Cov(g(X), h(Y)) is always equal to Cov(g(X), r(Z)) irrespective of the choices of functions g, h, and r.
In summary:
1) True
2) False
3) True