Since we know that Xt +ι = Xt + eD+1 — eB+1, we can write:
- γYβκ(ɪ - q) + [ɪ - κ(ɪ - q)] χ + χt + ⅜1 - ⅜1
YyβI - κ(1 - q)] t ɪ - к(1 - q)
And, sinceEt [et-1] = 0:
-Yyβκ(1 - q) - I - κ(1 - q)] + γγβ χ
Yyβ[ɪ - к(ɪ - q)] 11
It can be simplified as:
Yy β - ɪ
Yy β
Xt.
The other intercept term is - -γ~~1E [ɪZt-ɪ]. We can finally write the value of Dt as:
E[Dt +ι] - [ɪ + γγ ] Dt + ɪDt-1 = -ɪXt - ~^~E[ ɪ Zt+11.
(83)
(84)
(85)
Lγγ β J β γγ β γγ β La J
Deposits
We can obtain the value of Dt from:
E[Dt +1] - [ɪ + γγ ] Dt + ɪDt-1 = γγβ-1 Xt - E [ ɪ Zt+11.
Lγγ β J β γγ β γγ β La J
The solution is given by:
∞ ∞ -Ixi
Dt+j+1 = γγβDt+j + βYY ∑ (YY) Et+i [ O1 ] - βγγa(ɪ - Yy)C +
- Y-1 ∑ ( γY ) lEt-+i [ Xt+j+i+1],
Under the assumption that interest rates follow a random walk process, we can for simplicity
assume that deterministic component of the rates remain constant and treat the values of the rates
at time t and t + ɪ that enter in the solution as constants. In this case the result is the following:
Dt+j +1 = γγΓβDt+j - βγγa(ɪ - γγ) [(ɪ - K)(γγβ - H) + (ɪ - q)κH][rB+j+1 + COV (rB, ɪ)] +
- βγγa(ɪ - γγ) K(rR+ jq - rD+ j) + γγβ(1 - γγ) (g3rD+j - g4rB+j) - βγγa(ɪ - γγ) C +
- jγγβ - ɪ)K COV (Ld, ɪ ); (86)
βγY (ɪ - γY ) a
it can finally be simplified as:
ɪ κ[(ɪ- d)(γγβ- h) + (1 - q)kH]+ (γγβ- ɪ)ag4h b
Dt + ’ +' = γγβ Dt + j + a T) r‘ + ’ ' +
κ R κ2 + a(γγβ - ɪ)g3 d i4 (βγγ - ɪ) [Z - a] - κ^}
~rrt q+ +
βγγa(γγ - ɪ) + j βγγa(γγ - ɪ) + j βγγa(γγ - ɪ)
+ d----K----U [(ɪ - k)(Yyβ - ɪ) + (ɪ - q)K] COV(rB, ɪ) - (γγβ ɪ)K COV(Ld, ɪ ). (87)
βγγa(γγ - ɪ) L ∖ b/ a β a/ βγγ(γγ - ɪ) ∖ a/
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