Hw6 - l I 8.2.8 Let then 1000 tobe the portion of the $1000...

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l. 3.2.15. (AR) J bd E[sr(x) . sz(Y)l = [ [ nrll. [email protected]). f(r,y) d,y d,x orto tl I I s'@) . gz(v) . [email protected]) . f2(y) dy dr JJ Note that /"(C) = t [email protected],y)da : [email protected]) t [email protected])dv and similarly fvfu) = [email protected] Il fi(c) dr. Thus' First, 8.2.8. Let tobe the portion of the $1000 ,t ", i, t.""rt"ffi inu ,irt y *,"t;' then 1000 - to is put into the bond. The yield, in dollars, is then Y = (1000 - tr)(.04) + u(.01)R i because .R is in units of cents. Using the graph, you can find that the density of .R is [email protected])={i$a'1, f ;E[l:li The expectation of R can be found in a straightforward way: Sg f1 f'l ElRl = l r' ib-3) dr+ I "'(--(" -s)) dr =t7/3' -L--, J 6.- -, -- 'J ' 12' 35 .E[R2] can be found similarlY: S 9, r - 1 f ^ 'l EIR'I : l r'. *t' - 3) d,r + l r'z' ?i('- e)) da :toll3' JOJ 35 Therefore a2 =Var(R) : l}l/3 - (17 F)2 = 1419' The mean of Y minus half of the variance of Y is Etyt - |var(r) : fl,[ff.j;rt]li? J,:.tiilfr_ itXli$"{, : aO + $ur - Oo5oo.'. setting the derivative of the last function with respect to tu equal to 0 produJe, the critical point tu : $107-14 for the amount to invest in the stock. and therefore $892.86 in the bond' Y; chb *fu) = E(D- {tb,+{D = (toaa - u)/'u 5 i --,Y rt /rgo)=-!*i-f f I f,(r)fs(y) : fl /r(c)' fz(i' I f'@)dr' I f2(y)dy .JJ ob d c ff [email protected])'
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