# Put-Call Parity on Currency and Binomial Trees

I tried solving the below problem without knowing the shortcut of thinking about this in terms of a put versus a call. I can't seem to arrive at the correct answer using my method and I'm wondering where I'm going wrong.

$u = e^{(0.03 - 0.05)/12 + 0.05\sqrt{1/12}} = 1.012848937$

$d = e^{(0.03 - 0.05)/12 - 0.05\sqrt{1/12}} = 0.984028496$

$p^* = \frac{1}{1 + e^{0.05\sqrt{1/12}}} = 0.496391623$

The call pays off at nodes: $u^6, u^5d, u^4d^2, u^3d^3, u^2d^4, u^1d^5$.

The desired price is then:

$\big[(0.8u^6 - 1.35^{-1}){p^*}^6(1 - p^*)^0 + (0.8u^5d^1 - 1.35^{-1}){p^*}^5(1 - p^*)^1 + (0.8u^4d^2 - 1.35^{-1}){p^*}^4(1 - p^*)^2 + (0.8u^3d^3 - 1.35^{-1}){p^*}^3(1 - p^*)^3 + (0.8u^2d^4 - 1.35^{-1}){p^*}^2(1 - p^*)^4 + (0.8u^1d^5 - 1.35^{-1}){p^*}^1(1 - p^*)^5\big]$

$= 0.005760287$, multiplied by the discount factor $e^{-0.03 \cdot 0.5}$ and $\$1,000,000$to arrive at$5,760.28$euros. ## 1 Answer You have forgotten the combinatorial factors for binomial probabilities on your terms. You need $${n\choose k} p^n(1-p)^{n-k},$$ not just $$p^n(1-p)^{n-k}.$$ The second term should have a factor of$6$and the third should have a factor of$15,$etc. • Is$d^6\$ missing, too? Or is that just a typo? – lebelinoz Sep 23 '17 at 10:15
• @levelinoz no they're correct to leave that term out (that's where the call finishes out of the money). – spaceisdarkgreen Sep 23 '17 at 13:34
• Of course. Silly me. – lebelinoz Sep 23 '17 at 14:56