Hi,
I’m going through FI stuff and saw this youtube link:
Is it possible to solve this question below with pen and paper? Or do you need excel/solver?
“Assume that one-year forward rate follows a lognormal random walk with volatility σ=10%, annual. Given that current on-the-run, one-year yield Ro = 8% and our target two year option-free corporate bond is paying 7% coupon rate annually, currently trading at par $100”
Thanks in advance
If by solve you mean construct a Binomial tree, then yes:
that problem only has 2 steps so it’s possible to do it with pencil and paper.
The interest rate at the start is R_{0}=8\%. At the end of year 1, there will be two possible interest rates which you need to find, R_{1}^{(u)} and R_{1}^{(d)}=R_{1}^{(u)}e^{-2\sigma\Delta t}, with \Delta t=1 because it’s a year.
You value the bond by discounting the cashflows. There’s a 7% coupon C at the end of year 1 and year 2 and return of principal P at the end of year 2, so the value is
\frac{C}{1+R_{0}}
+\frac{1}{2}\left(P+C\right)\left[\frac{1}{\left(1+R_{0}\right)\left(1+R_{1}^{(u)}\right)}
+\frac{1}{\left(1+R_{0}\right)\left(1+R_{1}^{(d)}\right)}\right]
The factor of \frac{1}{2} is there because the probability of each branch of the tree is 50%.
The value has to be equal to 100, because the bond is trading at par
\frac{C}{1+R_{0}}
+\frac{\left(P+C\right)}{2\left(1+R_{0}\right)}
\left[\frac{1}{\left(1+R_{1}^{(u)}\right)}
+\frac{1}{\left(1+R_{1}^{(u)}e^{-2\sigma}\right)}\right]=P
You can rearrange that as a quadratic in R_{1}^{(u)}
-2e^{-2\sigma}(P+PR_{0}-C)R_{1}^{(u)2}-(e^{-2\sigma}+1)(P+2PR_{0}-3C)R_{1}^{(u)}
-2PR_{0}+4C=0
The solution to the quadratic ax^{2}+bx+c=0 is x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}
Put values for P, C, \sigma and R_{0} in the quadratic equation above and solve it and you find R_{1}^{(u)}=6.536324\%, along with another root ($-11.008236%) which you discard because it’s negative.
Then you can write down R_{1}^{(d)}=5.3514895\%
I’m not sure you even need a Binomial tree. The only piece of information you aren’t given is the interest rate R_{1} for the second year, and you can find that because you know the price of the bond
\frac{C}{1+R_{0}}
+\frac{\left(P+C\right)}{\left(1+R_{0}\right)\left(1+R_{1}\right)}=P
The only thing you don’t know in this equation is R_{1} (so solve the equation for R_{1})
The solution is R_{1}=\frac{2C-PR_{0}}{P+PR_{0}-C}=5.940594059\%
That makes sense in some way but ax^2+bx+c=0, which values do you plug in here? Do you just decide that c= 7% for example and b=10%?
Can you show how you got r1=5,351% please? This is math from like pre university haha, long time ago.
Thank you so much!
I had an equation
-2e^{-2\sigma}(P+PR_{0}-C)R_{1}^{(u)2}-(e^{-2\sigma}+1)(P+2PR_{0}-3C)R_{1}^{(u)}
-2PR_{0}+4C=0
This is a quadratic aR_{1}^{(u)2}+bR_{1}^{(u)}+c=0 with a=-2e^{-2\sigma}(P+PR_{0}-C), b=-(e^{-2\sigma}+1)(P+2PR_{0}-3C), and c=-2PR_{0}+4C.
You have volatility \sigma=0.1 and \Delta t=1, so
R_{1}^{(d)}=R_{1}^{(u)}e^{-2\sigma\Delta t}=R_{1}^{(u)}e^{-0.2}=e^{-0.2}\times 6.536324\%
=5.3514895\%
Thanks, I get we lower node to be calculated with e^-0,2 x 6,536324=5,35 but how did you get 6,53%? Like how do I plug in the values etc do derive that number?
The solution to the quadratic aR_{1}^{(u)2}+bR_{1}^{(u)}+c=0 is R_{1}^{(u)}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.
First you need to calculate a, b, and c by plugging the constants into the expressions for a, b, and c above.
Then you plug the values of a, b, and c into the quadratic formula to get R_{1}^{(u)}
Interesting! But how do I know which values should be plugged in where? Like is -b the volatility and ac the coupon rate and a the coupon yield or?
I had an equation
-2e^{-2\sigma}(P+PR_{0}-C)R_{1}^{(u)2}-(e^{-2\sigma}+1)(P+2PR_{0}-3C)R_{1}^{(u)}
-2PR_{0}+4C=0
This is a quadratic aR_{1}^{(u)2}+bR_{1}^{(u)}+c=0 with a=-2e^{-2\sigma}(P+PR_{0}-C), b=-(e^{-2\sigma}+1)(P+2PR_{0}-3C), and c=-2PR_{0}+4C.
Here \sigma=0.1 is the volatility, C=7 is the coupon, P=100 is the principal, and R_{0}=8\%
The solution is R_{1}^{(u)}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}
with
a=-2e^{-2\sigma}(P+PR_{0}-C)=-2e^{-0.2}(100+100\times 8\%-7)=-202e^{-0.2}=-165.3836121,
b=-(e^{-2\sigma}+1)(P+2PR_{0}-3C)=-(e^{-0.2}+1)(100+200\times 8\%-21)=-95(e^{-0.2}+1)-172.7794215
and c=-2PR_{0}+4C=-200\times 8\%+28=12.
Cool, how did you learn this?
quadratic equations ax^{2}+bx+c=0 I learned in high school
Me too but I’ve forgotten it until now. But I mean, how did you learn how to put this equation to work out the discount rate for binominal trees? I wouldn’t have thought about doing that…