Example 3.12. Let X, Y and Z be jointly continuous random variables with joint PDF is given by: fx.y.z(x, y, z) = (12r°yz)I0,1)(z)I(0,1)(y)I0,1)(2) Recall that fx(x) =(3r²)I(0,1)(z) fy(y) =(2y)I(0.1)(») fz(2) =(2=)I(0,1)(2) Are X,Y and Z independent? To show that X, Y and Z are independent, you need to show that fx,y(r, y) =fx(r) × fy(y) fx,z(x, z) =fx(r) × fz(z) fy,z(y, 2) =fy(y) × fz(2) fx.y.z(x, y, z) =fx(1) × fy(y) × fz(z)
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- Example 2.17 Let X be a continuous type random variable with PDF given as 1 = (x)f 27 Let Y be another random variable such that Y = X². Find the PDF of Y.Example 2.7. Let X and Y be jointly continuous random variables with joint PDF is given by: fx,Y (x, y) = (6/5)(æ² + y))I(o,1)(x)I(0,1)(4) 1. Show the marginal PDF of X. 2. Find P(X > Y). 3. Find P( < X < }).Example 3.6. Let X and Y be jointly continuous random variables with joint PDF is given by: 3 fx.x(x, y) =(1+ a²y)!(0,1)(x)/(0.2)() Note that fx(x) = (1+ x²) I0,1) (2) fy(y) : (1+ y) I(0,2)(y) Solve for fx|y(x | y). Remark. It follows immediately from the definition of conditional probability density function that fx,y(r,y) = fxjy(x | y) · fy (y), fx.x (x, y) = fy|x(y | x) · fx(x), -00Suppose that Y₁ and Y₂ are uniformly distributed over the triangle shaded in the accompanying figure. 3₂ (0, 1) (-1,0) (a) Find Cov(Y₁ Y₂). Cov(Y₁, Y₂) = (b) Are Y₁ and Y₂ independent? Yes O No (1, 0) (c) Find the coefficient of correlation for Y₁ and Y₂. P= y/₁ (d) Does your answer to part (b) lead you to doubt your answer to part (a)? Why or why not? O Even though Cov(Y₁Y₂) # 0, Y₁ and Y₂ are not necessarily dependent. Since Cov(Y₁ Y₂) # 0, we should expect Y₁ and Y₂ to be dependent. O Since Cov(Y₁, Y₂) = 0, we should expect Y₁ and Y₂ to be independent. O Even though Cov(Y₁Y₂) = 0, Y₁ and Y₂ are not necessarily independent.Let X1, X2,... , Xn be independent Exp(A) random variables. Let Y = X(1)min{X1, X2, ... , Xn}. Show that Y follows Exp(nA) dis- tribution. Hint: Find the pdf of Y). If X is a continuous random variable whose pdf is f(x): (a) 0 (b) 1 (c) 1/4 (d) 4 (e) None of the above C (1+x)5 ¹ x > 0, what is c?Example 3.6. Let X and Y be jointly continuous random variables with joint PDF is given by: 3 fx,x(r, y) = (1 + a°y)[(0,1)(=)[(0,2) (1) Note that fx(x) =÷ (1 + a²) I0,1)(x) fr(y) =- (1+y) I(0,2) (y) Solve for fxjy(æ | y). Remark. It follows immediately from the definition of conditional probability density function that fx,x(x, y) = fxjy(x | y) · fy(y), -o∞ < x < ∞ fx,y (x, y) = fy|x(y | æ) · fx(x), -The premise pdf of the random variable X is X ∼ U (0, 2). The pdf of the Y random variable under the condition X Geometric (x). That is, since fY | X (y | x) ∼ Geometric (x) and Y = 3 observations fX | Y (x | 3) =?by the way f(y|x)=x(1-x)^y-1 right? if not tell me whY?Assume that X and Y are independent random variables where X has a pdf given by fx(x) = 2aI(0,1)(x) and Y has a pdf given by fy(y) = 2(1– y)I(0,1)(y). Find the distribution of X + Y.3. Let X be a continuous random variable on the interval [0, 1] having cdf Fx(x) = 3x² − 2x³. Let W = 4 + 3X. (a) What is the range of W? (b) Find a formula for Fw(w). (c) Find a formula for fw(w).3. Suppose that X is a continuous random variable with pdf 3x², 0Suppose X and Y are random variables with joint PDF fx,y(x, y) = 4xyI(0,1)(x)I(0,1) (y) Let U = ln X and V = √Y. Find the joint PDF of U and V using the method of transformation.SEE MORE QUESTIONSRecommended textbooks for youAlgebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:CengageTrigonometry (MindTap Course List)TrigonometryISBN:9781337278461Author:Ron LarsonPublisher:Cengage LearningAlgebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:CengageTrigonometry (MindTap Course List)TrigonometryISBN:9781337278461Author:Ron LarsonPublisher:Cengage Learning