1. **Problem statement:** Given the joint probability density function (pdf) $$f(x,y) = \begin{cases} \frac{3}{2} y^2 & 0 < x < 2, 0 < y < 1 \\ 0 & \text{otherwise} \end{cases}$$ Find: (a) The marginal pdf of $X$. (b) The marginal pdf of $Y$. 2. **Recall the formulas:** - The marginal pdf of $X$ is found by integrating the joint pdf over all values of $y$: $$f_X(x) = \int_{-\infty}^{\infty} f(x,y) \, dy$$ - The marginal pdf of $Y$ is found by integrating the joint pdf over all values of $x$: $$f_Y(y) = \int_{-\infty}^{\infty} f(x,y) \, dx$$ 3. **Find $f_X(x)$:** Since $f(x,y) = \frac{3}{2} y^2$ for $0 < x < 2$ and $0 < y < 1$, and zero otherwise, $$f_X(x) = \int_0^1 \frac{3}{2} y^2 \, dy = \frac{3}{2} \int_0^1 y^2 \, dy$$ Calculate the integral: $$\int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3}$$ So, $$f_X(x) = \frac{3}{2} \times \frac{1}{3} = \frac{1}{2}$$ This holds for $0 < x < 2$, and $f_X(x) = 0$ otherwise. 4. **Find $f_Y(y)$:** Similarly, $$f_Y(y) = \int_0^2 \frac{3}{2} y^2 \, dx = \frac{3}{2} y^2 \int_0^2 dx = \frac{3}{2} y^2 \times 2 = 3 y^2$$ This holds for $0 < y < 1$, and $f_Y(y) = 0$ otherwise. 5. **Summary:** - Marginal pdf of $X$: $$f_X(x) = \begin{cases} \frac{1}{2} & 0 < x < 2 \\ 0 & \text{otherwise} \end{cases}$$ - Marginal pdf of $Y$: $$f_Y(y) = \begin{cases} 3 y^2 & 0 < y < 1 \\ 0 & \text{otherwise} \end{cases}$$