1. **Problem statement:** Given the joint PDF of random variables $X$ and $Y$ as $$f(x,y) = \frac{2}{75}(2x + y^2), \quad 0 \leq x \leq 3, 1 \leq y \leq 2$$ Find: (1) $P[1 \leq X \leq 2, \frac{1}{3} \leq Y \leq \frac{5}{3}]$ (2) Joint distribution function $F(x,y)$ (3) Marginal distribution function $F_X(x)$ (4) Marginal PDF $f_X(x)$ for $x \in [0,3]$ --- 2. **Step 1: Calculate $P[1 \leq X \leq 2, \frac{1}{3} \leq Y \leq \frac{5}{3}]$** First, note the support of $Y$ is $[1,2]$, but the interval $[\frac{1}{3}, \frac{5}{3}]$ intersects with $[1, 2]$ only over $[1, \frac{5}{3}]$ because $\frac{1}{3} < 1$. Thus, effective interval for $Y$ is $[1, \frac{5}{3}]$ Calculate the probability: $$ P = \int_{x=1}^2 \int_{y=1}^{5/3} \frac{2}{75}(2x + y^2) \, dy \, dx $$ Integrate w.r.t. $y$: $$ \int_1^{5/3} (2x + y^2) dy = 2x(y) \Big|_1^{5/3} + \frac{y^3}{3} \Big|_1^{5/3} = 2x \left(\frac{5}{3} - 1\right) + \frac{1}{3} \left(\left(\frac{5}{3}\right)^3 - 1\right) $$ Simplify: $$ = 2x \left(\frac{2}{3}\right) + \frac{1}{3} \left(\frac{125}{27} - 1\right) = \frac{4x}{3} + \frac{1}{3} \times \frac{98}{27} = \frac{4x}{3} + \frac{98}{81} $$ Now integrate w.r.t $x$ from 1 to 2: $$ \int_1^2 \left(\frac{4x}{3} + \frac{98}{81}\right) dx = \int_1^2 \frac{4x}{3} dx + \int_1^2 \frac{98}{81} dx = \frac{4}{3} \frac{x^2}{2} \Big|_1^2 + \frac{98}{81} (2-1) = \frac{2}{3}(4 - 1) + \frac{98}{81} $$ Simplify: $$ = \frac{2}{3} \times 3 + \frac{98}{81} = 2 + \frac{98}{81} = \frac{162}{81} + \frac{98}{81} = \frac{260}{81} $$ Multiply by $\frac{2}{75}$: $$ P = \frac{2}{75} \times \frac{260}{81} = \frac{520}{6075} \approx 0.0856 $$ --- 3. **Step 2: Find the joint distribution function $F(x,y) = P(X \leq x, Y \leq y)$** By definition: $$ F(x,y) = \int_0^x \int_1^y f(u,v) dv du $$ with $0 \leq x \leq 3$ and $1 \leq y \leq 2$ (zero outside these ranges). Calculate inner integral: $$ \int_1^y (2u + v^2) dv = 2u (y - 1) + \frac{y^3 - 1}{3} $$ Hence, $$ F(x,y) = \int_0^x \frac{2}{75} \left[ 2u (y - 1) + \frac{y^3 - 1}{3} \right] du = \frac{2}{75} \left[ (y - 1) \int_0^x 2u du + \frac{y^3 -1}{3} \int_0^x du \right] $$ Evaluate integrals over $u$: $$ \int_0^x 2u du = x^2, \quad \int_0^x du = x $$ Thus: $$ F(x,y) = \frac{2}{75} \left[ (y-1) x^2 + \frac{y^3 - 1}{3} x \right] = \frac{2}{75} x^2 (y-1) + \frac{2}{225} x (y^3 - 1) $$ --- 4. **Step 3: Find marginal distribution function $F_X(x)$** By definition: $$ F_X(x) = P(X \leq x) = \int_1^2 F(x, dy) = \int_1^2 \int_0^x f(u,y) du dy $$ Switch integrals: $$ F_X(x) = \int_0^x \int_1^2 f(u,y) dy du $$ Integrate $f(u,y)$ w.r.t $y$: $$ \int_1^2 \frac{2}{75} (2u + y^2) dy = \frac{2}{75} \left( 2u(y) + \frac{y^3}{3} \right) \Big|_1^2 = \frac{2}{75} \left( 2u(1) + \frac{8}{3} - (2u + \frac{1}{3}) \right) = \frac{2}{75} \left( 0 + \frac{7}{3} \right) = \frac{14}{225} $$ Note the integral of $2u$ terms cancels out because limits are substituted, check carefully: Actually: $$ 2u(y) \Big|_1^2 = 2u(2) - 2u(1) = 4u - 2u = 2u $$ For the $y^2$ term: $$ \frac{y^3}{3} \Big|_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} $$ So total integral w.r.t. $y$: $$ \frac{2}{75} (2u + \frac{7}{3}) $$ Integrate w.r.t. $u$: $$ F_X(x) = \int_0^x \frac{2}{75} (2u + \frac{7}{3}) du = \frac{2}{75} \left( \int_0^x 2u du + \int_0^x \frac{7}{3} du \right) = \frac{2}{75} \left( x^2 + \frac{7x}{3} \right) $$ --- 5. **Step 4: Find marginal PDF $f_X(x)$ for $0 \leq x \leq 3$** Take derivative w.r.t $x$ of $F_X(x)$: $$ f_X(x) = \frac{d}{dx}F_X(x) = \frac{d}{dx} \left[ \frac{2}{75} \left( x^2 + \frac{7x}{3} \right) \right] = \frac{2}{75} \left( 2x + \frac{7}{3} \right) = \frac{4x}{75} + \frac{14}{225} $$ --- **Final answers:** 1. $P[1 \leq X \leq 2, \frac{1}{3} \leq Y \leq \frac{5}{3}] = \frac{520}{6075} \approx 0.0856$ 2. Joint CDF: $$F(x,y) = \frac{2}{75} x^2 (y-1) + \frac{2}{225} x (y^3 - 1), \quad 0 \leq x \leq 3, 1 \leq y \leq 2$$ 3. Marginal CDF of $X$: $$F_X(x) = \frac{2}{75} \left(x^2 + \frac{7x}{3} \right), \quad 0 \leq x \leq 3$$ 4. Marginal PDF of $X$: $$f_X(x) = \frac{4x}{75} + \frac{14}{225}, \quad 0 \leq x \leq 3$$