Final answer:
To show that f(X) and g(Y) are independent, we need to show that their joint probability distribution can be factorized into the product of their individual probability distributions. By using the definitions of expected value and independence of X and Y, we can show these conditions are satisfied and thus prove the independence of f(X) and g(Y).
Step-by-step explanation:
Let X and Y be independent random variables and f,g be Borel measurable functions. We want to show that f(X) and g(Y) are independent as well. To prove this, we need to show that the joint probability distribution of f(X) and g(Y) can be factorized into the product of their individual probability distributions.
Since X and Y are independent, their joint probability distribution is given by P(X=x, Y=y) = P(X=x) * P(Y=y) for all x and y.
Now, let's consider the joint probability distribution of f(X) and g(Y). Let c be any real number. We have:
P(f(X) ≤ c, g(Y) ≤ c) = P(X ∈ x ) * P(Y ∈ y ) (1)
Since f and g are Borel measurable functions, the sets x and y are Borel sets. Therefore, we can rewrite equation (1) as:
P(f(X) ≤ c, g(Y) ≤ c) = ∫ f(x) ≤ c P(X=x) dx * ∫y P(Y=y) dy
Using the definition of the expected value of a function of a random variable, we can rewrite this as:
E[1 f(x) ≤ c(X)] * E[1y (Y)]
However, since X and Y are independent, we know that E[1x (X)] = P(X ∈ f(x) ≤ c) and E[1y (Y)] = P(Y ∈ g(y) ≤ c). Therefore, we can further simplify the expression:
P(f(X) ≤ c, g(Y) ≤ c) = P(X ∈ f(x) ≤ c) * P(Y ∈ y )
This shows that the joint probability distribution of f(X) and g(Y) can be factorized into the product of their individual probability distributions, which means that f(X) and g(Y) are independent. Therefore, we have shown that if X and Y are independent random variables and f,g be Borel measurable functions, then f(X) and g(Y) are independent as well.