Avaliação Resolvida-2022 2

Questão 01 (Pontuação: 2,0) Assuma que X e Y são variáveis aleatórias independentes, onde a função de distribuição de probabilidade de X e Y é dada respectivamente por: f_X(x) = 2xI_X(0,1) e f_Y(y) = 2(1 - y)I_Y(0,1) Ache a distribuição de X + Y. Questão 02 (Pontuação: 2,0) Seja f(x, y) = (x + y) no intervalo [0 < X < 1/2, 0 < Y < 1/2] uma função de distribuição conjunta contínua. Qual o valor da probabilidade abaixo: P(0 < X < 1/2; 0 < Y < 1/2) = \int_{0}^{1/2} \int_{0}^{1/2} (x + y)dxdy. Com relação à distribuição conjunta calculada pela probabilidade acima, pode-se dizer que as variáveis X e Y são independentes? Sugestão: Verifique usando: f_{X,Y}(x,y) = f_X(x)f_Y(y) Questão 03 (Pontuação: 2,0) A densidade conjunta de X e Y é dada por f_{X,Y}(x,y) = (x + y)I_X(0,1)I_Y(0,1). Determine: a) E(X/Y) e E(Y/X) b) VAR(X/Y) e VAR(Y/X) Questão 04 (Pontuação: 2,0) A variável aleatória bidimensional (X, Y) tem a seguinte tabela de distribuição de probabilidades: | X | Y | |---|----| | | 1 | 2 | 3| | 2 |0,10|0,30|0,20| | 3 |0,06|0,18|0,16| Calcular: a) E(2X + Y) b) VAR(Y/X = 2) Questão 0 f(x, y) = \frac{1}{8} (6 - x - y) I_X(x)(0,2) I_Y(y)(0,2) a) E[Y/x=x] f_X(x) = \int_0^2 f(x,y)dy = \int_0^2 \frac{1}{8} (6 - x - y)dy = \frac{1}{8}(6y -xy - \frac{y^2}{2})|_0^2 = \frac{1}{8} (12 - 2x - 2) = \frac{1}{8} (10 - 2x) = \frac{1}{4} (5 - x) I_X(x)(0,2) f_{Y/x} (y/x = x) = \frac{f_{X,Y}(x,y)}{f_X(x)} = \frac{\frac{1}{8} (6 - x - y)}{\frac{1}{4} (5 - x)} = \frac{6 - x - y}{2(5 - x)} E[Y/x=x] = \int_0^2 y f_{Y/x} (y/x) dy = \int_0^2 y \frac{6 - x - y}{2(5-x)} dy = \int_0^2 \frac{6y - xy - y^2}{2(5-x)} dy = \frac{1}{2(5-x)} \left( \frac{3y^2}{2} - \frac{xy^2}{2} - \frac{y^3}{3} \right)|_0^2 = \frac{1}{2(5-x)} (12 - 2x - \frac{8}{3}) I_X(x)(0,2) b) E(X/y=y) f_Y(y) = \int_0^2 \frac{1}{8} (6 - x - y)dx = \frac{1}{8} (6x - \frac{x^2}{2} - xy)|_0^2 = \frac{1}{8} (12 - 2 - 2y) = \frac{1}{4} (5 - y) I_Y(y)(0,2) f_{X/y} (x/y) = \frac{1}{8} (6 - x - y) = \frac{6 - x - y}{1/4 (5-y)} = \frac{6 - x - y}{2(5-y)} E[X/y=y] = \int_0^2 x \frac{6 - x - y}{2(5-y)} dx = \frac{1}{2(5-y)} \int_0^2 (6x - x^2 - xy) dx = \frac{1}{2(5-y)} \left( \frac{3x^2}{2} - \frac{x^3}{3} - \frac{xy^2}{2} \right) |0^2 = \frac{1}{2(5-y)} (12 - \frac{8}{3} - 2y) I_Y(y)(0,2) c) Var (Y/x=x) E[Y^2/x] = \int_0^2 y^2 f_{Y/x} (y/x) dy = \int_0^2 y^2 \frac{6 - x - y}{2(5-x)} dy = \frac{1}{2(5-x)} \int_0^2 (6y^2 - xy^2 - y^3) dy = \frac{1}{2(5-x)} \left( 2y^3 - xy^3 \frac{y^4}{4} \right) \Bigg|_0^{12} \\ = \frac{1}{2(5-x)} \left( 12 \cdot \frac{8x}{3} - 4 \right) = \frac{1}{2(5-x)} \left( 12 - \frac{8x}{3} \right) \\ Var(y/x=x) = E[y^2/x=x] - (E[y/x=x])^2 \\ = \frac{1}{2(5-x)} \left( 12 - \frac{8x}{3} \right) - \left( \frac{1}{2(5-x)} \left( 12 - 2x - \frac{8}{3} \right) \right)^2 \\ = \frac{36 - 8x}{6(5-x)} - \frac{36 - 6x - 8}{6(5-x)} \right)^2 \\ = \frac{18 - 4x}{3(5-x)} - \frac{14 - 3x}{3(5-x)} \right)^2 \\ = \frac{18-4x}{3(5-x)} - \frac{196 - 84x + 9x^2}{(3(5-x))^2} \\ = \frac{(18-4x)(225 - 90x + 9x^2) - 196 + 84x - 9x^2}{9(5-x)^2} \\ = \frac{4050 - 1620x + 162x^2 - 900x + 36x^2 - 36x^3 - 196 + 84x - 9x^2}{9(5-x)^2} \\ = \frac{3854 - 2436x + 513x^2 - 36x^3}{9(5-x)^2} d) Var(X/Y=y) \\ E[x^2/Y=y] = \frac{1}{2(5-y)} \left( 12 - \frac{8y}{3} \right) \\ Var(X/Y=y) = \frac{1}{2(5-y)} \left( 12 - \frac{8y}{3} \right) - \left[ \frac{1}{2(5-y)} \left( 12 - 2y - \frac{8}{3} \right) \right]^2 \\ = \frac{3854 - 2436y + 513y^2 - 36y^3}{9(5-y)^2} \\ \\ Questão 1: \\ f(x_1, x_2, x_3) = 8x_1^3 x_2^3 x_3^2 I(x_1)(0,1) I(x_2)(0,1) I(x_3)(0,1) \\ f_{x_1}(x_3) = \int_0^1 \int_0^1 \int_0^1 8x_1^3 x_2^3 x_3^2 dx_2 dx_3 \frac{8x_1^3 x_2^3 x_3^3}{3} \Bigg|_0^1 \right) dx_3 \\ = \int_0^1 \frac{8x_1^3 x_3^3}{3} dx_3 = \frac{8x_1^3 x_3^3}{9} \Bigg|_0^1 = \frac{8x_1^2}{9} \\ E[x_1] = \int_0^1 x_1 \frac{8x_1^2}{9} dx_1 = \int_0^1 \frac{8x_1^3}{9} dx_1 = \frac{2x_1^4}{9} \Bigg|_0^1 = \frac{2}{9} \\ E[x_2] = \frac{2}{9} \\ E[x_3] = \frac{2}{9} a) E[3x_3 + 2x_2 + 6x_3] = 3E[x_1] + 2E[x_2] + 6E[x_3] \\ = 3 \cdot \frac{2}{9} + 2 \cdot \frac{2}{9} + 6 \cdot \frac{2}{9} \\ = \frac{6 + 4 + 12}{9} = \frac{22}{9} \\ b) E[x_1, x_2, x_3] = \int_0^1 \int_0^1 \int_0^1 x_1, x_2, x_3 8x_1^2 x_2^3 x_3^3 dx_1 dx_2 dx_3 \\ = \int_0^1 \int_0^1 \left[ \int_0^1 8x_1^3 x_2^3 x_3^3 \bigg| dx_1 \right] dx_2 dx_3 \\ = \int_0^1 \int_0^1 \left( 2x_1^4 x_2^3 \bigg| dx_2 dx_3 \right) \\ = \int_0^1 \left[ \frac{x_2^3 x_3^3}{2} \right] dx_2 \\ = \int_0^1 \left( \frac{x_2^4 x_3^3}{2} \right) dx_3 \\ = \int_0^1 \left( \frac{x_3^3}{2} \right) dx_3 = \frac{x_3^4}{8} \bigg|_0^1 = \frac{1}{8} \\ c) E[x_1, x_2] f_{x_1,x_2}(x_1, x_2) = \int_0^1 8x_1^2 x_2 x_3^2 dx_3 = \frac{8x_1^2 x_2 x_3^3}{3} \bigg\vert _{0}^{1} = \frac{8x_1^2 x_2}{3} E[X, x_2] = \int_0^1 \left[\int_0^1 x_1 f_{x_1, x_2} - \frac{8x_1^2 x_2^2}{3} dx_1 \right] dx_2 = \int_0^1 \left[\int_0^1 \frac{8x_1^2 x_2^3}{3} dx_1 \right] dx_2 = \int_0^1 \left( \frac{2x_1^4 x_2^3}{3} \bigg\vert_0^1 \right) dx_2 = \int_0^1 \frac{2x_2^3}{3} dx_2 = \frac{x_2^4}{6} \bigg\vert_0^1 = \frac{1}{6} Questão 2 f(x,y) = \frac{1}{2x}; \ I(x)(0,2) \ I(y)(0,x) E(x) = 1 \quad E(y) = \frac{1}{2} a) E[x|y] = \int_0^2 \int_0^x \frac{y}{2x} dy dx = \int_0^2 \left[ \int_0^x \frac{y}{2} dy \right] dx = \int_0^2 \left( \frac{y^2}{4} \bigg\vert_0^x \right) dx = \int_0^2 \frac{x^2}{4} dx = \frac{x^3}{12} \bigg\vert_0^2 = \frac{8}{12} = \frac{2}{3} b) \text{Cov}(x,y) = E(xy) - E(x) \cdot E(y) = \frac{2}{3} - 1 \cdot \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} Questão 1 f_x(x) = 2x \ I(x)(0,1) f_y(y) = 2(1-y) \ J(y)(0,1) z = x + y F_z(z) = P(z \leq z) = P(x+y \leq z) = P(X \leq z-y) = \int_0^{z-y} 2x dx = x^2 \bigg\vert_0^{z-y} = (z-y)^2 f_z(z) = \frac{dF_z(z)}{dz} = 2(z-y) F_z(z) = P(z \leq z) = P(x+y \leq z) = P(y \leq z-x) = \int_0^{z-x} 2(1-y) dy = 2 \left( y - \frac{y^2}{2} \right) \bigg\vert_0^{z-x} = 2y - y^2 \bigg\vert_0^{z-x} = 2(z-x) - (z-x)^2 f_z(z) = \frac{dF_z(z)}{dz} = 2 - 2(z-x) f_2(z) = f_2(z) 2(z-y) = 2 - 2(z-x) 2z - 2y = 2 - 2z + 2x 4z = 2x + 2y z = \frac{x + y}{2} f_{x,y}(x,y) = \frac{x + y}{2}; \ I(x)(0,1) \ J(y)(0,1) Questão 2. P(0 \leq x \leq \frac{1}{2}, 0 \leq y \leq \frac{1}{2}) = \int_0^{\frac{1}{2}} \left[\int_0^{\frac{1}{2}} (x+y)dx \right] dy = \int_0^{\frac{1}{2}} \left( \frac{x^2}{2} + xy \right)_{0}^{\frac{1}{2}} dy = \int_0^{\frac{1}{2}} \left( \frac{1}{8} + \frac{y}{2} \right) dy = \frac{1}{8} y + \frac{y^2}{4} \bigg\vert _{0}^{\frac{1}{2}} = \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8} Resp: Pela probabilidade calculada acima não é possível afirmar que X e Y são independentes. Questão 3: f_{X,Y}(x,y) = (x+y) \ I(x)(0,1) \ I(y)(0,1) \ f_Y(y) = \int_0^1 (x+y) dx = \frac{x^2}{2} + xy \Bigg|_0^1 = \frac{1}{2} + y \ f_{X,Y}(x/y) = \frac{x+y}{\frac{1}{2}y} = \frac{x+y}{\frac{1+2y}{2}} = \frac{2(x+y)}{1+2y} \ E[X/Y] = \int_0^1 x \left(\frac{2(x+y)}{1+2y}\right) dx = \frac{2}{1+2y} \int_0^1 (x^2 + xy) dx \ = \frac{2}{1+2y} \left(\frac{x^3}{3} + \frac{x^2y}{2}\right) \Bigg|_0^1 = \frac{2}{1+2y} \left(\frac{1}{3} + \frac{y}{2}\right) \ = \frac{2+3y}{3(1+2y)} f_X(x) = \int_0^1 (x+y) dy = xy + \frac{y^2}{2} \Bigg|_0^1 = x + \frac{1}{2} \ f_{Y/X}(y/x) = \frac{x+y}{x+\frac{1}{2}} = \frac{2(x+y)}{1+2x} E[\frac{Y}{X}] = \int_0^1 y \left(\frac{2(x+y)}{1+2x}\right) dy = \frac{2+3x}{3(1+2x)} \ E[X^2/Y] = \int_0^1 x^2 \left(\frac{2(x+y)}{1+2y}\right) dx = \frac{2}{1+2y} \int_0^1 (x^3 + x^2y) dx \ = \frac{2}{1+2y} \left(\frac{x^4}{4} + \frac{x^3y}{3}\right) \Bigg|_0^1 = \frac{2}{1+2y} \left(\frac{1}{4} + \frac{y}{3}\right) \ = \frac{3+4y}{6(1+2y)} \ Var(X/Y) = \frac{3+4y}{6(1+2y)} - \left(\frac{2+3y}{3(1+2y)}\right)^2 = \frac{3+4y}{6(1+2y)} - \frac{4+12y+9y^2}{9(1+2y)^2} \ = \frac{(3+4y)(3+8y+12y^2) - (4+12y+9y^2)}{18(1+2y)^2} \ = \frac{9 + 36y + 36y^2 + 12y + 48y^2 + 48y^3 - 8 - 24y - 18y^2}{18(1+2y)^2} \ = \frac{1 + 24y + 66y^2 + 48y^3}{18(1+2y)^2} E[Y^2/X] = \frac{3+4x}{6(1+2x)} \ Var(Y/X) = \frac{3+4x}{6(1+2x)} - \left(\frac{2+3x}{3(1+2x)}\right)^2 = \frac{1+24x+66x^2+48x^3}{18(1+2x)^2} \ Questão 4: x \vdots 2 \quad 3 \ P(x=x) \vdots 0,60 \quad 0,40 \ y \vdots 1 \quad 2 \quad 3 \ P(y=y) \vdots 0,16 \quad 0,48 \quad 0,36 \ a) \ E[2x+y] \ E[X] = 2 \cdot 0,60 + 3 \cdot 0,40 = 2,4 \ E[Y] = 1 \cdot 0,16 + 2 \cdot 0,48 + 3 \cdot 0,36 = 2,2 \ E[2x+y] = 2 \cdot E[X] + E[Y] = 2 \cdot 2,4 + 2,2 = 7 \ b) \ Var(Y/X=2) P(Y/X=2) = ? \ P(Y=1/X=2) = \frac{0,10}{0,60} = 0,1667 \ P(Y=2/X=2) = \frac{0,30}{0,60} = 0,5 \ P(Y=3/X=2) = \frac{0,20}{0,60} = 0,3333 P(y/x=2) 3 2 3 y 0,1667 0,5 0,3333 E[Y/x=2] = 3·0,1667 + 2·0,5 + 3·0,3333 = 2,1667 E[Y^2/x=2] = 1^2·0,1667 + 2^2·0,5 + 3^2·0,3333 = 6,1667 Var(Y/x=2) = 6,1667 - (2,1667)^2 = 1,4722