Let (X, Y) have joint density f(x, y) = c x² y on the region x² ≤ y ≤ 1, and 0 otherwise. Find c, both marginal densities, and decide whether X and Y are independent.

Step-by-step solution

1. 1

   Describe the support before integrating. The condition $x^{2}\le y\le 1$ requires $x^{2}\le 1$, so $-1\le x\le 1$: the region is the area between the parabola $y=x^{2}$ and the horizontal line $y=1$. Missing the implicit range of $x$ is the usual first error.

2. 2

   Normalise, integrating y first.

   $\int_{-1}^{1}\!\!\int_{x^{2}}^{1}cx^{2}y\,dy\,dx=\int_{-1}^{1}cx^{2}\cdot\frac{1-x^{4}}{2}\,dx=\frac{c}{2}\int_{-1}^{1}\left(x^{2}-x^{6}\right)dx$

   Both integrands are even, so

   $\frac{c}{2}\left(\frac23-\frac27\right)=\frac{c}{2}\cdot\frac{8}{21}=\frac{4c}{21}=1\quad\Longrightarrow\quad c=\frac{21}{4}$

3. 3

   Find the marginal density of X. For each $x\in[-1,1]$ integrate $y$ from $x^{2}$ to $1$:

   $f_X(x)=\int_{x^{2}}^{1}\frac{21}{4}x^{2}y\,dy=\frac{21}{4}x^{2}\cdot\frac{1-x^{4}}{2}=\frac{21}{8}x^{2}\left(1-x^{4}\right),\qquad -1\le x\le 1$

4. 4

   Find the marginal density of Y. For each $y\in[0,1]$ the condition $x^{2}\le y$ means $-\sqrt y\le x\le\sqrt y$:

   $f_Y(y)=\int_{-\sqrt y}^{\sqrt y}\frac{21}{4}x^{2}y\,dx=\frac{21}{4}y\cdot\frac{2y^{3/2}}{3}=\frac{7}{2}y^{5/2},\qquad 0\le y\le 1$

5. 5

   Decide independence. The support is bounded by a parabola, not a rectangle — the possible values of $X$ shrink as $Y$ decreases — so the variables cannot be independent. Confirming with the formulas:

   $f_X(x)f_Y(y)=\frac{21}{8}x^{2}\left(1-x^{4}\right)\cdot\frac72y^{5/2}\neq\frac{21}{4}x^{2}y=f(x,y)$

   $\boxed{c=\tfrac{21}{4};\ f_X(x)=\tfrac{21}{8}x^{2}\left(1-x^{4}\right),\ f_Y(y)=\tfrac72y^{5/2};\ \text{not independent}}$

6. 6

   Check both marginals integrate to 1. $\int_{-1}^{1}\tfrac{21}{8}x^{2}\left(1-x^{4}\right)dx=\tfrac{21}{8}\cdot 2\left(\tfrac13-\tfrac17\right)=\tfrac{21}{4}\cdot\tfrac{4}{21}=1$ ✓ and $\int_{0}^{1}\tfrac72y^{5/2}dy=\tfrac72\cdot\tfrac{2}{7}=1$ ✓.