ke *+y)0 x<y<o 6. Suppose f(x, y) ' 0 otherwise What is the value of k?...
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Sk(x+y) 0<x<1, 0<y</ 14. Determine k, so that fx.y(x, y)= otherwise is a joint pdf. 10 15. Determine k, so that fxy(x,y)= kry 0<x<1, 0<y<1. 6 otherwise is a joint pdf. k(xy?) 0<x<1, 0<y<1. is a joint pdf. Determine k, so that fx.x(x,y)= 1 otherwise 17. Determine k, so that fx.y(x,y)= kr 0<x<y<1 O otherwise is a joint pdf. k(x + y) 0<x< y<1 18. Determine k, so that fx. (x,y)= 1 0 otherwise is...
4. Suppose X and Y have the joint pdf f(x,y) = 6x, 0 < x < y < 1, and zero otherwise. (a) Find fx(x). (b) Find fy(y). (c) Find Corr(X,Y). (d) Find fy x(y|x). (e) Find E(Y|X). (f) Find Var(Y). (g) Find Var(E(Y|X)). (h) Find E (Var(Y|X)]. (i) Find the pdf of Y - X.
Let (X,Y) have joint pdf given by f(rw)-y <x, 0 < x < 1, | 0, 0.W., (a) Find the constant c. (b) Find fx (x) and fy(y) (c) For 0 < x < 1, find fy|x=r(y) and My X=r and oỉ x=x (d) Find Cov(X,Y). (e) Are X and Y independent? Explain why.
Let f(x,y)= K(x^2+y^2 ) in 0≤x≤1, 0≤y≤1. Determine the value of the constant K that makes f(x,y) a joint density function. (a) Find fx(x) (b) Find fy(y) (please answer (a) and (b))
4. The joint distribution of X and Y is given by 0 otherwise (a) Are X and Y independent? Explairn. (b) Find the marginal probability function (pdf) of Y, fy (). (c) Provide the integral for finding P(X < Y), but DO NOT evaluate.
Suppose X andY have joint density f(x,y)=6*x*y^2 for 0<x<1, 0<y<1. (a) What is P(X+Y ≤1)? (b) Compute the marginal densities fX , fY of X, Y .
X and Y are random variables with the joint PDF fx.^(t,y)-65536 0 otherwise. (a) What is the marginal PDFfx(x)? ㄑㄨ 8 5xA4/65536 fx(x) 0 otherwise (b) What is the marginal PDF fy(v)? (5 * 843)/(3*655 0 〈y〈 64 fy(y) = 0 otherwise
Let (X, Y) have joint pdf given by f(r, y)= < a, 0 < < 0, О.w., (a) Find the constant c (b) Find fx(x) and fy(y) (c) For 0 x< 1, find fyx=r (y) and py|x=x and oyx= (d) Find Cov(X, Y) (e) Are X and Y independent? Explain why
2. Suppose X and Y are independent continuous random variables. Show that P(Y < X) = | Fy(x) · fx (x) dx -oo where Fy is the CDF of Y and fx is the PDF of X [hint: P[Y E A] = S.P(Y E A|X = x) · fx(x) dx]. Rewrite the above equation as an expectation of a function of X, i.e. P(Y < X) = Ex[•]. Use the above relation to compute P[Y < X] if X~Exp (2)...
Consider the following pdf: ; 0<x<1 f(x)-2k ; l<x<2 0 otherwise (i)Determine the value of k. (ii) Find P(X 0.3) (iii) Find (0.1 〈 X 1.5).