为什么计算T=0时的value是要乘0.966184？而不是除1.025？

问题如下：

Question

An analyst values a 2-year European-style put option where the underlying is the periodically compounded 1-year spot interest rate. The notional amount of the option is $1 million, and the put exercise rate is 3.5% of par. The risk-neutral probability is 5.0% and these options cash settle at time step 2 based on the observed rates as shown in the exhibit below.

At time step 0, the value of the interest rate put option is closest to:

选项：

A.A.$4,623. B.B.$4,713. C.C.$4,878.

解释：

B Correct because using the expectations approach, per $1 of notional value, the values of the put options at time step 2 are:

p⁺⁺ = Max(0,X-S⁺⁺) = Max(0,0.0.035-0.055) = 0
p^+– = Max(0,X-S^+– ) = Max(0,0.035 – 0.035) = 0
p^{– –} = Max(0,X-S^{– –}) = Max(0,0.035 – 0.0015) = 0.0200

At time step 1, the put values are:

p⁺ = PV[πp⁺⁺ + (1 – π)p^+–] .
p⁺ = 0.956938[0.50(0) + (1 – 0.50)(0)] =0
p^– = PV[πp^+– + (1 – π)p^{– –}].
p^– = 0.97561[0.50(0) + (1 – 0.50)(0.02)] = 0.009756

At time step 0, the put option value is:

p = PV[πp⁺ + (1 – π)p^–].
p = 0.966184[0.50(0) + (1 – 0.50)(0.009756)] = 0.004713

The value of the put option is:

0.004713 x 1,000,000 = 4,713