Ron Murray - Binomial Valuation of a Derivative

# Binomial Valuation of a Derivative

Ron Murray
Fin 622
Extra Credit

I found Chapter 7 (Pricing Derivatives) to be the most challenging chapter to date. I will now go over one of the problems assigned to us from this chapter.

**Using risk-neutral valuation (the problem can be checked using the risk-neutral valuation, but I will only use the binomial valuation), derive a formula for a derivative that pays cash flows over the next two periods. Assume the risk-free rate is 4% per period.

The underlying asset, which pays no cash flows unless sold, has a market value that is modeled as follows: \$1 current, up \$1.10 yr 1, up \$1.20 yr 2 or dn \$1.05 yr 2, and \$1 current, dn \$.95 yr 1, up \$1.05 yr 2 or dn \$.90 yr 2.

The cash flows of the derivative that correspond to the underlying asset are as follows: \$0 current, up \$.10 yr 1, up \$\$2.20 yr 2 or dn \$3.05 and \$0 current, dn \$.90 yr 1, up \$3.05 yr 2 or dn \$0 yr2.

Find the present value of the derivative.**

To solve this you must solve for each "node" (up node and down node).

The equation for the up node is ΔSu + Β(1+rf) = Vu, where:

Δ is the number of units of the underlying asset
Su is the value of the underlying asset in the “up” node
Β is the number of dollars in the risk free security
rf is the risk free rate
Vu is the value of the derivative in the “up” node.

So, for the up node the first equation would be Δ1.2+B(1.04)=2.2. The second equation would be Δ1.05+B(1.04)=3.05. Subtract the second equation from the first (which eliminates B) and you get Δ.15 = -.85 or Δ = -5.67. Plug Δ into either of the first 2 equations to solve for B. You will find that B = 8.65. Now that you have solved for both Δ and B, you can plug them into the equation V=ΔS+B. Here you will find that Vu=2.42.

Now, for the down node the equation is ΔSd + Β(1+rf) = Vd. You follow the same steps as above and eventually work your way down to Vd=1.72.

Now that you have solved for both the up node (Vu) and the down node (Vd) you can solve for the present value of the derivative as follows:
Δ1.1+B(1.04)=.1+2.42
Δ.95+B(1.04)=.9+1.72
Δ.15 = -.1 (again, subtract eq 2 from eq 1)
Δ = -.6667

Now use either of the first two equations and plug in your new Δ value and solve for B. B = 3.28.

Plug these into V=ΔS+B, and you get Vo = 2.46. This is the present value of the current derivative.

page revision: 2, last edited: 02 Dec 2008 07:19