Define the random variable g() specifically here to be Q(, i), where i is a random variable that is of uniform distribution in the integer set [1 : k]. So, g() is to be Q(, i). Further, we define the noise z = Q(, i) f(), hence g() = Q(, i) = f() + z. That's why we call g is the noisy version of f. Given this definition of f and g, prove that equations E[g()] = f(), E[g()] = f(), and E[z] = 0 are satisfied.