Problem 6: Random variables X and Y have the joint CDF0.5 (y +1) (1 – e-2r ), r 2 0 and |y| < 1,1-e-2r,Fx,y(r, y)x 20 and y > 1,0,otherwise.(a) What is P[X < 1,Y < 0]?(b) What is P[X < 0, Y < 1]?(c) What is P(0 < X < 1,0 < Y < 1]?(d) What is the marginal CDF Fy(y)? (d) What is the marginal CDF Fy(y)?(e) What is the marginal CDF Fx(x)?(f) What is the joint PDF fxy(x, y)?(g) Are X and Y independent? Justify your answer mathematically.

Since you have posted a question with multiple subparts, we will solve first 3 parts for you. If you want any specific subpart to be solved, then please resubmit the question and mention the subparts to be solved. If probability of X≤x,Y≤y is required to compute then its value equals to the value of CDF at (x,y). So P[X≤1,Y≤0]=FX,Y(1,0).The CDF for x≥0, |y|≤1 is defined by 0.5(y+1)(1-e-2x), for x≥0, y>1 is defined by 1-e-2x and other then these the function has value equal to 0. For condition x=1,y=0, first condition is satisfied, thus the value of CDF will be computed by substituting x=1,y=0 in function 0.5(y+1)(1-e-2x). FX,Y(1,0)=0.5(0+1)(1-e-2×1)=0.5×1(1-e-2)=0.5(1-0.1353)≈0.4323 So P[X≤1,Y≤0]=0.4323Similarly, P[X≤0,Y≤1]=FX,Y(0,1), the value of x=0,y=1 satisfies the first condition. Thus the value of CDF will be computed by substituting x=0,y=1 in function 0.5(y+1)(1-e-2x). FX,Y(0,1)=0.5(1+1)(1-e-2×0)=0.5×2(1-e0)=1(1-1)=0 So P[X≤0,Y≤1]=0To compute the value of P[x1<X≤x2,y1<Y≤y2], use the following formula. Px1<X≤x2,y1<Y≤y2=Fxyx2,y2-Fxyx1,y2-Fxyx2,y1+Fxyx1,y1 So in P0<X≤1,0<Y≤1 the values are x1=0,x2=1,y1=0,y2=1. Substitute these in the formula. P0<X≤1,0<Y≤1=Fxy1,1-Fxy0,1-Fxy1,0+Fxy0,0 Use the function 0.5(y+1)(1-e-2x) to find the CDF at each value of x and y. P=Fxy1,1-Fxy0,1-Fxy1,0+Fxy0,0=0.5(1+1)1-e-2(1)-0.5(1+1)1-e-2(0)-0.5(0+1)1-e-2(1)+0.5(0+1)1-e-2(0)=0.5×2×0.865-0.5×2×0-0.5×1×0.865+0.5×1×0=0