Question

Question 1. Suppose t ≤ T1 ≤ T2 ≤ T3, where t is the current time, and ∆ > 0. Recall that Z(T1, T2) is the price at time T1 of a ZCB with maturity T2 and F(T1, T2, T3) is the forward price at time T1 for a forward contract with maturity T2 on a ZCB with maturity T3. a) For each of the pairs of A and B in the table, choose the most appropriate relationship out of ≥, ≤, = , ?, where ? means the relationship is indeterminate. Give brief reasoning. A ≥, ≤, = , ? B (i) Z(t, T1) 1 (ii) Z(T1, T1) 1 (iii) Z(t, T2) Z(t, T3) (iv) Z(T1, T2) Z(T1, T3) (v) Z(T1, T3) Z(T2, T3) (vi) Z(T1, T1 + ∆) Z(T2, T2 + ∆) (vii) F(t, T1, T2) F(t, T1, T3) (viii) F(t, T1, T3) F(t, T2, T3) (ix) limT→∞ Z(t, T) 0 Hint: Remember that at current time t, F(t, ·, ·) is known but Z(T, ·) is a random variable. b) What can you say about interest rates between T1 and T2 if i) Z(t, T1) = Z(t, T2)? ii) Z(t, T1) > 0 and Z(t, T2) = 0?

Step 1

Part (a)

Let's assume a continuous compoundde discount rate of r for the purpose of illustration.

Please note that

Z (t, T_{1}) = e^{-r(T1 – t)} < 1

Please see the white board.

- Exponent of a negative number is always less than 1. This explains the answer to (i)
If we put t = T

_{1}in the equation above, we get Z (t = T_{1}, T_{1}) = e^{-r(T1 – T1)}= 1. This explains answer to (ii)T

_{2}≤ T_{3}; Hence, T_{2}– t ≤ T_{3}– t; Hence, - (T_{2}– t) ≥ - (T_{3}– t); Hence, Z (t, T_{2}) = e^{-r(T2 – t)}≥ = e^{-r(T3 – t)}= Z (t, T_{3}). This explains the answer to (iii)- Explnation to (iv) is same as that to (iii) above. Put t = T
_{1}in the expression above and we end up getting what we are looking for in (iv) - For (v) there will be no sure shot, certain relationship.

Step 2

(vi) the relationship is indeterminate. It can be either ways.

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