Question-66

Suppose the joint probability density function (PDF) of \(X\) and \(Y\) is given as:

\(f_{XY}(x,y) = \begin{cases} 6xy^2, &~ 0 \leq x \leq 1, 0 \leq y \leq 1\\ 0, &~\text{otherwise}. \end{cases}\)

Choose the correct options. (Select all that apply)

• \(f_X(x) = 3x, ~ 0\leq x \leq 1\)

• \(f_{Y|X}(y | x) = 3xy^2, ~ 0 \leq y \leq 1\)

• \(f_{Y|X}(y | x) = 3y^2, ~ 0 \leq y \leq 1\)

• \(E[Y | X = x] = \dfrac{3}{4}\)

Hint

Answer

• \(f_X(x) = 3x, ~ 0\leq x \leq 1\)

• \(f_{Y|X}(y | x) = 3xy^2, ~ 0 \leq y \leq 1\)

• \(f_{Y|X}(y | x) = 3y^2, ~ 0 \leq y \leq 1\)

• \(E[Y | X = x] = \dfrac{3}{4}\)

Solution

\(f_{XY}(x,y) = \begin{cases} 6xy^2, &~ 0 \leq x \leq 1, 0 \leq y \leq 1\\ 0, &~\text{otherwise}. \end{cases}\)

\(f_X(x) = \int_{y=0}^{1}f_{XY}(x,y) dy = \int_{y=0}^{1} 6xy^2 dy = 6x \int_{y=0}^{1}y^2dy = 6x\left[\dfrac{y^3}{3}\right]_{0}^{1} = 2x\) for \(0 \leq x \leq 1\).

Thus, first option is incorrect.

Now, \(f_{Y|X}(y | x) = \dfrac{f_{XY}(x,y)}{f_X(x)} = \dfrac{6xy^2 }{2x} = 3y^2\). Thus, the third option is correct and second option is incorrect.

\(E[Y | X = x] = \int_{0}^{1}y f_{Y|X}(y | x) dy = \int_{0}^{1} y. 3y^2 dy = 3 \int_{0}^{1} y^3 dy = 3 \left[\dfrac{y^4}{4}\right]_{0}^{1} = \dfrac{3}{4}\). Hence, fourth option is correct.