Question

6. Suppose X and Y have the joint pdf fr,y) = 2 exp(-:- 0 ) 0< <y otherwise o a. Find Px.x, the correlation coefficient between X and Y. b. Let U = 2X-1 and V=Y +2. What is pu.v, the correlation coefficient between U and V? c. Repeat (b) if U = -TX and V = Y + In 2. d. Let W = Y - X. Compute Var (W). e. Refer to (d). Find an interval that will contain W with probability at least 15/16. N

Answer 1

solution::

© Solution - - Given that, Suppose X and Y have the goint pdf f(x,y) = $ a exp(-x-4) 0<x+y 1 0 otherwise. @ fx (x) = aeopl-zy) dy a = d expl-x)ſ exply) dy = 20"[-e" = dé" Le tex] e de* (0+2) = aceae, «20 4 fyly) = 1 a exp(-x-4) dx = dezply) s*e* dix = raee =-delet:) = 2e7(1-2y) , yzo fy () = dety(181)

E [x] = à os xn exp (mean = ) E[N] = va * (-e-M) ay = a yetu dy-a I yeny dy =ar (+1-(aybe *y dy = af (a)- 41 56da = 3-4() = a- E[V = test 24-2,ady = ds. E[MV] =-4ae tern er dy [[(a) = (2-1)!=)! =1] =a[4"[-*e*)' + l'e+du] dy = 2 [ye[640 40.) -e *[*]dy = aſye [-ye ()]dy = aſ yet B-ft ) e 14 ] dy

gdy = dy] =Qq yeddy - afy (94Det dy = 81cm -a ſ t'e*1 by - 364e" day = arc)-19) "204 -- [4*899 dy [Toksing oy=2, = a[ca) -> p (a) – ağ xé*4 dy = arca) barca) – Fly - 25 = 2-5-= 2- = [XY) = is coulx,y) = E[XY]-E[x]. E[Y] = - x - 2 / 4 = 2 4 cov (X,Y) FAT = Vall(x) =

ob = Var(y) = E[72] – [y] E[v+] = f ve ae Y -ety dy => [t* r *dy - af yte84 dy = ar 6) - 2. = axa - = 4-Y = Liu Vam y) =- 1 Cov(x,y) tercer ezy - - 15 O U = 2x-1, V = y +2 eu, v = (097 (0,) = loro (@x-1, y +2) Cov(@x-1,4+2) = Covlax, y) + covlax, 2) + Cov (-1,7) + cov (-1,2)

Covlax-1, Y+ a) = 2 Cov (x,y)+ 0+0+0 ☺ [ cov blw Riv. and constants are o] = 2 x 5 = = cov (u,v) scov (ax-1,8+2) = Coulu.v) = 1|| Vaul (0) = Var (ax-1) = H voor (x) + vor (1-1) + a cor (ax, 1) = 4 Var (x) +0+0 = H Var(x) Covariance blo Random = 4x+ variable & Constants Var(0) = 1 are o] Varl V) = Vor (y+2) = varcy) lar[v)= 5 - Cor(UV) . Hence, Cu,v = Ivar (V) var (V) Teur = 1 / 3 © UR-VT * , V = Ty + Ina cov (0,0) = cov (-VTX, TY+Ina) = cov (-VTX, TTY) + cov [VIT X. Ina)

= -TIT Cor (x,y) +o cov (v.v) = -TUT Van (0) = van (–1T x) = 4veyvan(x) = IT Vor (v) = van (Ty + loa) = VonCY) = sq? Hence, vor = Cov (W) = -TT/2 svorco) va (V) 1 17 x ² IV a.w=y-x Var(W) = Vot(Y-X) = var (Y)-& cor (y,x) + Varcx) == -axta + 4 = -1 = -1 = Vair (W) = 2 1

© E[W]= E[y-x] = E[y]- E[x] = Ž - de E(w) = 2 Assuming Was a normal random Variable, we have, the CI with probability atleast 15 is given by, - Min = 8 - 15 ) Var (w) = 1+1"10ta Mw=L - 9" (012), it 0467] - o-

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