4
$\begingroup$

I'm trying to simulate a 3-factor HJM model. I got the algorithms from Glasserman book. In my case, I have $3$ maturity:$ 0.25y, 0.5y, 0.75y$. So my time grid is: $t_0=0,t_1=0.25,t_2=0.5,t_3=0.75$.

I'm trying to price the zero-coupon bonds with:

$B(0,t_j)=\exp(-\sum_{i=0}^{j-1}r(t_i)h_i),\quad \text{where}\quad h_i=t_i-t_{i-1}$

formula. $h_i=\frac{3}{12} \forall i$

$D(t_j)=B(0,t_j)$ and the starting value of $D$ is $1$, as its written in the algorithm.

In the first "for" cycle the index is going only untill $M-1$, which is $2$, so it wont be multipled with $r(t_2)$.

In other words:

$i=1$ $\rightarrow$ $D=D*\exp(-r(t_0)\frac{3}{12} )$,

$i=2$ $\rightarrow$ $D=D*\exp(-r(t_1)\frac{3}{12} ). $

After that the algorithm stops, so I either have to change the max indext to $M$ or just multiple with $\exp(-r(t_2))$ in the end, but i dont think these are the good solutions. Or if in the $i=1$ case, my first update on bondprice would be $r(t_1)$, that would be good too, but then I dont know where $\exp(-r(0))$ is in the algorithm.

$\endgroup$
2
  • $\begingroup$ Could you please use latex for the formulas. $\endgroup$
    – Richi W
    Apr 11, 2014 at 15:53
  • $\begingroup$ Sorry, I corrected it. $\endgroup$
    – user7778
    Apr 11, 2014 at 18:36

1 Answer 1

2
$\begingroup$

I am not sure about this specific algorithmic implementation, but I am a bit confused by your indexes and suspect you might be as well (e.g. $M$ not defined, you're showing cases of $i$ looping when it seems you mean $j$). I think it would be useful to revisit the basics:

Let $D_t \in (0,1]$ be the present value factor for a cash flow at time $t$. By construction, $D_0 = 1$.

Often, $B(0,t_j) = \text{exp}\{\int_0^{t_j} r(t) \ dt\}$ denotes a money market account as a numeraire. The discrete equivalent, as in your implementation, is $B(0,t_j) = \text{exp}\{\sum_{i=0}^{j} r(t_i) \Delta t_i\}$.

If you are building your rates from data from zeroes, you would take the reciprocal $B(0,t_j)^{-1}$, and equate this with the price for the zero with maturity $t_j$ (adjusting for quoting by multiplying by 100). It looks like this is what your $B(0,t_j)$ denotes, from the $-1$ in the exponential. This makes sense (since Bond begins with a "b"), but it could be initially confusing if you are reading papers with the other notation. Anyway, I will use your notation hereafter.


Anyway, in your implementation: Since your $\Delta t$ is constant, $$ B(0,t_j) = \text{exp}\left(-\Delta t \sum_{i=0}^{j} r(t_i)\right) \\ $$ where $r(0) = 0$ and $r$ is the annualized rate.

For clarity, the first iteration, $j=1$, yields: $$ B(0,t_1) = \text{exp}\{-0.25(0+r(t_1))\} \\ $$ and the second iteration, $j=2$, yields: $$ B(0,t_1) = \text{exp}\{-0.25(0+r(t_1)+r(t_2))\} \\ $$ Iterating this for all of your $t$ values should give you what you are looking for.

$\endgroup$
1
  • $\begingroup$ Can I use the inst. forward rates, that I simulate in the algorithm with $f(t_i,t_j)=f(t_{i-1},t_j)+\mu(t_{i-1},t_j)+\sum_{k=1}^{d}\sigma_k(t_{i-1},t_j)*Z_i$ formula to price a bond? The price of bind is $B(t_i,t_j)=\exp\left(-\sum_{l=i}^{j-1}f(t_i,t_l)(t_{l+1}-t_l)\right)$. Thanks for help in advance. $\endgroup$
    – user7778
    Apr 26, 2014 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.