hw12solns

# hw12solns - Solutions to Homework 11 6.49 For a ﬁxedx...

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Unformatted text preview: Solutions to Homework 11. 6.49 For a ﬁxedx 6 [0,1] , we have0 S y S l —x , then the marginal pdf ofX is f. (x) = I f(x,y)dy = l:dey 0 =2(l—x), 03x31 Then, the conditional probability density function forY given X = 0.25 is f“ (x,0.25) _ 2 4 —, 0950.75 frW l x =0”): fX(0.25) _ 2(1—0.25) : 3 0.25 P(Y < 0.25 I X = 0.25) = j fm (y | x = 0.25)dy 0 0,25 4 of 3 3 b. the mean proportion of chemicalY in the insecticide if 25% of the insecticide is chemicalX is E[Y | X = 0.25] = j yan (y l x = 0.25)dy 0,75 4 = ——d J3J’y 3 8 c. the variance is V[Y|X=0.25]=E|:Y2|X=0.25:|—{E[Y|X=0.25]}2 3 = jyzfmbll x = 0.25)dy_ (E) Then the standard deviation is 14 = 0.2165 6.56 Let P denote the probability of observing a defective item, we know that P ~ uniform(0,%), X l P ~ binomial(n, p) a. the joint pdfofX and P is fX.p(xaP) : fxyp(xl P)‘fp(p) =4[n]px (l—p)"_x, x =O,l,...,n 0< p <— x b. the marginal distribution of X is fx(x) = Ifx,/J(xap)dp =‘i4lilp‘<‘—p>"‘*dp an m = i:l'ﬁi:<-I>kl";xlp“*dp 6.68 a. For the next ﬁve cars entering this intersections, let X 1 denote the number of cars that turn left, X 2 denote the number of cars that turn right, X3 denote the number of cars that continue straight ahead, then X ],X 2,X3 has the multinomial distribution with p] = 0.4, p2 = 0.25, p3 = 0.35 , then the probability of interest is 5! P(Xl =1,X2 =1,X3 = 3) =ﬁ-l—EO.4-0.25-0.353 = 0.08575 b. X 2 has the binomial distribution withn = 5,122 = 0.25 ,then P(X2 21)=1—P(X2 = 0) 5 = l—[0]O.25°(1—0.25)S = 0.7627 c. For the 100 cars entering this intersections, letY denote the number of cars that turn left, Y has the binomial distribution with p] : 0.4 , then E(Y)=np] =40 V(Y)=npl(1_pl)=24 The assumption is that the trials are independent. 6.70 a. Let Xl ~N(,u,,oq), X2 ~N(,u2,o-2) , then the moment generating function of Y=aX, +sz is M, (t) = E(e”) : E(exax, era/v2 ) : E(etuX1)E(etaX2) = MXl (00sz (at) 2 2 2 2 1+6 I /2 (+0 I /2 = eﬂi 1 6142 2 (#1 +yz )l+(o‘,1 +022 )12 /2 _ 8 This is the moment-generating function of a normal distribution, thusY ~ N(;11+,uz,‘l0'l2 +022) b. 1 dt = #1 + #2 E(Y) = My (t)|1=0 =[(#1+ #2)+(o'12 + 0'22 )t] e0"+”2)”(U'Zmzz)'2/2 I=O V(Y) :d_t:My(t)|’=0 : [(0.12 +022)+((#l +#2)+(o_]2 +022)t)2:le(#1+ﬂz)l+(0.2+0'21)12/2 (=0 = 0'12 + 0'22 6.86 a. for a ﬁxedX=xe[-—1,l], we have —\/1—x2 Sysxll—x2 , the marginal density function of X is fx(x)= jf(x,y)dy b. deﬁne the set A={(x,y):x2+y2 SLxSy}, thenAis the shadow area in the ﬁgure below. The probability of interest is P(X s Y) = Hf(x,y)dxdy = ism) va—nlwa .1 2 7.3 a. X have the negative binomial distribution with r = 4, p = 0.4 , then the distribution oﬂXB x+r—1 P(X=x)=[ Jp'ﬂwv)‘ r—l +3 =[x )0.4“.0.6X, x=0,1,2,... 3 hY=X+4 c. the distribution of Y is P(Y=y)=P(X+4=y) = P(X = y —4) _[y—4+3 J04“ ~06"4 3 —1 =[y3 )044 -0.6y'4, y=4,5,6,... 7.9 Fory 6 (Le) , Fy (y) = P(Y < y) = P(eX < y) = P(X < 1n y) lny =j1dx O =lny WhenySl,Fy(y)=0,whenyZe,Fy(y)=1 Thus the probability density function of Y is 1 d —,1<y<e fy(y) =-Fy(y) = y dx . 0, otherw1se 7.11 ForyZO, Fy(y)=P(Y <y) =P(cX<y) 1 =P(X <—y) c 'l‘)’ 0 _1X 2 Ile 9 dx 0 0 l =1—e_5y Then the probability density function of Y is 1 —1y d —e 09 , 20 my>=—F,(y)= 06 y dx 0, otherwise Thus Y ~ exponential(c6) 7.13 For y > 0 , Fy (y) = P(Y < y) = P(eX < y) = P(X < ln y) : FX (1n y) Differentiation this equation, we have my)=ifx(lny) y 1 1 -—"“""’Z 2 _ e 2::2 ’ y > 0 y V250 Otherwise, f, (y) = 0 7.15 a. Deﬁne the setA={(xl,x2):OSxl \$2,03xZ \$1,2x2 le,)c1—x2 <y}, When 0 < y <1 , A is the shadow area in the ﬁgure below x2 xl’xz =y A can be equivalently written as A ={(xl,x2):()Sx2 S y,2x2 le <x2 +y} Then FY (y) = IIf(x1’x2)dx1dx2 x1_x2 =y A can be equivalently written as A={(x1,x2):OSxl \$2,03x2 \$1,2x2 le}\{(xl,x2):OSx2 SZ—y,x2+nyl 32} Then FY (y) = IIf(x1’x2)dx1dx2 A 2—y 2 =1— j [161x]de 0 x2+y Z—y =l— [(2_y_xz)dx2 0 l 2 =1-— 2— 2( y) Therefore, the probability density function forY is d y, 0 < y <1 fy(y)=;x-Fy(y)= 2—y, ISySZ 0, otherwise E(Y) = nyy(y)dy 2 = lIyzdy + Iy(2 -y)dy I 0 a 6.56 gag/La) vcx) : vLELx1p)+ ECVLXJW a n! Nev-Iv *2/94): hfgﬁl‘fa), anX/yzb, Ma. Q How/m UGO: V E (“Purl”) = hLVLp + hE.(},) - nap) 'e a E % i 3 | 7 i 1 3 1 € AA Ap‘mé.awh-. wMAmmmu‘p» a.“ “K. m ‘ gm“; w- 3 n‘ 1-” + n (01%) w (Vtmmn‘ﬂ a?— --—"“‘ '4 é __..__A..< .49... ...
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hw12solns - Solutions to Homework 11 6.49 For a ﬁxedx...

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