Let X and Y be continuous random variables with joint density function f(x,y) and marginal density function fx and fy respectively, that are nonzero only on the interval (0,1). Which one of the following statements is always true**?
(A)
$$E[X^{2}Y^{3}]=( \int_{0}^{1}x^{2} \, dx ) ( \int_{0}^{1}y^{3} \, dy ) $$
(B)
$$E[X^{2}]= \int_{0}^{1}x^{2}f(x,y) \, dx $$
(C)
$$E[X^{2}Y^{3}]=( \int_{0}^{1}x^{2}f(x,y) \, dx ) ( \int_{0}^{1}y^{3}f(x,y) \, dy ) $$
(D)
$$E[X^{2}]= \int_{0}^{1}x^{2}f(x) \, dx $$
(E)
$$E[Y^{3}]= \int_{0}^{1}y^{3}f(x) \, dx $$
Q1. The question said fx and fy are nonzero only on the interval (0,1), so 0<=x<=1,0<=y<=1, but the question didn't say f(x,y)=1, So, we can't say it's a joint uniform distribution right?
Q2. With the information provided, how do we know whether x,y are independent?
Q3. If f(x,y)=1, then (A) is correct?
Q4. If x,y are independent, is (C) correct? Can we use f(x,y) instead of f(x)?
Best Answer
I'm not sure why you concern with whether $f(x,y)=1$ or anything about independence and uniform distribution. The question is about what is always true. So for each of the 5 choices, just check whether the LHS always equals RHS, not just in special cases.
Hint: In general, $$ E[g(X,Y)] = \int_0^1\int_0^1g(x,y)f(x,y)dydx=\int_0^1\int_0^1g(x,y)f(x,y)dxdy. \tag{$*$} $$ For example, (E) is incorrect because: $$ \text{LHS}=E[Y^3]=\int_0^1\int_0^1y^3f(x,y)dxdy=\int_0^1y^3\left[\int_0^1f(x,y)dx\right] dy=\int_0^1y^3f_Y(y)dy $$ which is not the RHS of (E). You can also use ($*$) to verify / refute the smaller questions that you have.
Hint 2: the answer is