Question

Write clearly and neatly your final answers in the table below: # Points Question Answer Plot Vx, x and x2 Show your plot here 1- 5 Bonus 2- 10 Determinec Plot fxy(x,y) Show your plot here 3- 5 Bonus 4- 10P 6- 10 7- 10 8-15 5 0<x<0.25, 0 <Y < 0.5) P(Y<X fx(x) EX fry) Are X and Y independent? E(Y) E(XY) VOX) V(Y) cov(X,Y) 10 10 Show your detailed solutions below

Answer 1

y=/x y=x^2 0.757 0.257 y=1/12*x -0.5 -0.25 0.25 0.75 1.25 1.5 -0.25

C1,1) - Given try (ay) a cay over the range you & se 2) txy (ay) il a joint probability density function Ara under the joint PDF curve 21 yan? & l I can dy dr a un -100 - For the limits of a&y assume a vertical strip in the range of txy (14) on this strip y malues range from se to x2 & if we sweep this strip from left to right & values will range from oto 1 O V & -f'sen den sydy) as a close da ye na ang s trani daw Stuen CE-24, fry (14) = -24x4 --('scadas com as 1) 3 - 15 45 da) - - Content AD) = wrę (- » C: +12 : Dow (my) - +12MY

1404 plot txyly over y: x2 & y = re. 10,1) y=va (1,0) e 2 & PCOLX20-25, OL 420.5) d if we need placxcb, ccycd) - e sll try (2,4) da) dy, & PLOCAL 0.25 , OLYCOG) - 124 dy lada o 0-5 0.5 PloLX20-25, ok Y605) = 0.046875 u 0.047 140249)

6) f(a) – required. – marginal pdf of t. = I flagyldy, yo ranges from x2 to see de las = 12xy dy & 1200 un 1 o de Cass GM [x - 307 1,602(6[32_48] , ocak 0,437 minimum value of x=0 o g otherwise maximum value of x= point of intersection of yo re d y = 1 il a= 0.437 12x

) F1*5= ? Empe.code. Wat ons und DE: faceus) dx any do 0.437 0-437 0437 EM:6 74_97 **lo [0.0091 - 0.00042] ū lo E[K] = 0.052 - proti 8 fyl) - marginal Pot of y filz - af fcrin de range from a to sy. - fyty) - wypony dx . by x2 Ver by Cay-yo? XY 24 d uly) - Slyt-ys) , o Lyco.437, 4.0.437 is the internection. l o otherwise point of (12xy = 15e (yor) o the maximum value of y... a) To check whether two continuous random variable area independent cor not , we should check the condition: blo y Caly) - ) fyly) y ca) fyly) & (x²-x5) & 6cyt-yr) dearly 1 & byly) 1244 fy are dependent variables. Moreover, If the range of two sandom variables are dependent, then they cannot be independent 320d mot be independe

alla col hiphit tol chilift and now so ne to this shra ->E[4 [-091- 0.000427 = 0.052 11) E3 I lay fayldy doe of Elxy] u definition 0437 ny 0.437 0.437 Mandal pinay). caso, ya oma fade e arsyogurt [courts'] o 7 m » E[x4], 0.1092 , . vex) - Hariance of X = E [CX-4323 where 4: E[x] cor vex)=E(X). (ECxTJ? od vitamina e(pie triunde { ecums par zoda 0.437 0.427 (4) 80, Oca 20.437) 20, otherwise 1 x2(x2-x1) don 8,0.419 32-0.000166) 5 8 0 6 .022 - 0.000161] 6[0-0030147 3 0.0181 >> Vex) 310-0181-(0.0523 = 0.0154 13) VIY)- variance of4 € 4)-(E [y]) je 47 » E[UJS 0.062 €14). Tu tuluy dychowy3dy sy 0-437 - 00

es (0.0032- 0.000161] = 0.013) Now & [42] E [y] E(X)= ECHI - >> vey) E Cup ?) - (€ 14392 , VX) 0.0154. e con exit). n covariance of xay cox covex,y) = E(44) - EXECU) ECX49 = 0.1092, E CXJ - tty)=0-052 & COVcX,Y) I 6. 1092- 20.0 527² Covexel 0.1065 0.2886 x 0.2886 5) PCycx ,we have to find the shaded area The point of intersection of 0.2887, 0.2887) between You & y a 1 divides the ana into a parts In the firt part, if we consider a vertical strip, as values sange from x2 to x & we sweep the strip across the part I's area as values range from a to 0.2886. · Area of part 0 = 12 spay day de 12 Poxdm yalam - [ 8x dx (22-24)] = al por es, de] • . [60-2463" - 60.180034] = 0.0098 Area of part & & & ranges from an to es ranges from 0.2886 to 0.487 0-437 x 2 Area of part 12 fede ydy – 12 xdx y2, Tex 4 0.437 0-286 2

auß7 0.437 ob 0427 oll ons sda] - [ to enla in 6.2886 6.2886 0-28867 il (120.437) - Un (9:280) ).- [60.4373'6094603"). (-0.8283 + 1-9425] - [ 0.0064) - 0-2071 -0.0064 » Prycx) = 0.201 b )

x2 €0.2887,0.2887) Yare co.u37) al 4 a Part 3 part lag-strip - x2 to a a o to 0-2887 part © x2 to y-strip- a-strip - 0-2888 to 0.437