Marginal Distributions 394A4C

1. **State the problem:** We have joint density function $$f(x,y) = \frac{1}{3}x + \frac{1}{3}y$$ for $$0 \leq x \leq 2$$ and $$0 \leq y \leq 1$$, zero elsewhere. We want to find: (a) Marginal distributions of $X$ and $Y$. (b) Probability $P(X > 1.5, Y > 0.2)$. --- 2. **Marginal distribution formulas:** - Marginal distribution of $X$: $$g_X(x) = \int_{y} f(x,y) \, dy$$ - Marginal distribution of $Y$: $$g_Y(y) = \int_{x} f(x,y) \, dx$$ Since $f(x,y)$ is defined on $0 \leq x \leq 2$ and $0 \leq y \leq 1$, integrals are over these intervals. --- 3. **Find marginal distribution of $X$:** $$g_X(x) = \int_0^1 \left( \frac{1}{3}x + \frac{1}{3}y \right) dy = \int_0^1 \frac{1}{3}x \, dy + \int_0^1 \frac{1}{3}y \, dy$$ Calculate each integral: $$\int_0^1 \frac{1}{3}x \, dy = \frac{1}{3}x \times (1 - 0) = \frac{1}{3}x$$ $$\int_0^1 \frac{1}{3}y \, dy = \frac{1}{3} \times \frac{y^2}{2} \Big|_0^1 = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$$ Add them: $$g_X(x) = \frac{1}{3}x + \frac{1}{6}$$ Domain: $$0 \leq x \leq 2$$. --- 4. **Find marginal distribution of $Y$:** $$g_Y(y) = \int_0^2 \left( \frac{1}{3}x + \frac{1}{3}y \right) dx = \int_0^2 \frac{1}{3}x \, dx + \int_0^2 \frac{1}{3}y \, dx$$ Calculate each integral: $$\int_0^2 \frac{1}{3}x \, dx = \frac{1}{3} \times \frac{x^2}{2} \Big|_0^2 = \frac{1}{3} \times \frac{4}{2} = \frac{1}{3} \times 2 = \frac{2}{3}$$ $$\int_0^2 \frac{1}{3}y \, dx = \frac{1}{3}y \times (2 - 0) = \frac{2}{3}y$$ Add them: $$g_Y(y) = \frac{2}{3} + \frac{2}{3}y$$ Domain: $$0 \leq y \leq 1$$. --- 5. **Find $P(X > 1.5, Y > 0.2)$:** $$P = \int_{x=1.5}^2 \int_{y=0.2}^1 \left( \frac{1}{3}x + \frac{1}{3}y \right) dy \, dx$$ First integrate w.r.t. $y$: $$\int_{0.2}^1 \left( \frac{1}{3}x + \frac{1}{3}y \right) dy = \int_{0.2}^1 \frac{1}{3}x \, dy + \int_{0.2}^1 \frac{1}{3}y \, dy$$ Calculate each: $$\int_{0.2}^1 \frac{1}{3}x \, dy = \frac{1}{3}x (1 - 0.2) = \frac{1}{3}x \times 0.8 = \frac{4}{15}x$$ $$\int_{0.2}^1 \frac{1}{3}y \, dy = \frac{1}{3} \times \frac{y^2}{2} \Big|_{0.2}^1 = \frac{1}{6} (1^2 - 0.2^2) = \frac{1}{6} (1 - 0.04) = \frac{1}{6} \times 0.96 = 0.16$$ Sum: $$\int_{0.2}^1 \left( \frac{1}{3}x + \frac{1}{3}y \right) dy = \frac{4}{15}x + 0.16$$ Now integrate w.r.t. $x$ from 1.5 to 2: $$P = \int_{1.5}^2 \left( \frac{4}{15}x + 0.16 \right) dx = \int_{1.5}^2 \frac{4}{15}x \, dx + \int_{1.5}^2 0.16 \, dx$$ Calculate each: $$\int_{1.5}^2 \frac{4}{15}x \, dx = \frac{4}{15} \times \frac{x^2}{2} \Big|_{1.5}^2 = \frac{2}{15} (4 - 2.25) = \frac{2}{15} \times 1.75 = \frac{3.5}{15} = \frac{7}{30} \approx 0.2333$$ $$\int_{1.5}^2 0.16 \, dx = 0.16 (2 - 1.5) = 0.16 \times 0.5 = 0.08$$ Sum: $$P = 0.2333 + 0.08 = 0.3133$$ --- **Final answers:** - Marginal distribution of $X$: $$g_X(x) = \frac{1}{3}x + \frac{1}{6}, \quad 0 \leq x \leq 2$$ - Marginal distribution of $Y$: $$g_Y(y) = \frac{2}{3} + \frac{2}{3}y, \quad 0 \leq y \leq 1$$ - Probability: $$P(X > 1.5, Y > 0.2) = 0.3133$$